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\begin{center}
{\mymagthreebf MATHEMATICS 300\ ---\  SPRING 2004}

\medbreak

{\mymagtwoit Introduction to Mathematical Reasoning}\\

\medbreak

{\mymagtwoit  H. J. Sussmann}

\medbreak

\vspace{0.12cm}

{\Large{\bf  INSTRUCTOR'S NOTES, PART 2}}\\
{\large{\em (February 10, 2004)}}

\end{center}

%\vspace{2cm}


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\section{Reading,  homework and other announcements}

% for the first 7 days (January 20, 22, 27, 29, and February 3, 5 )} 

\subsection{Reading for  February 10, 12, 17 and 19}

\begin{itemize}
\item{} this handout,
\item{} the book's  Chapter 2 ({\bf all of it!}).
\end{itemize}

\subsection{GOOD \ NEWS \ ABOUT \ THE \ SUPPLEMENTARY PROBLEMS!!!!}

You can hand in the supplementary problems any time during the week of 
February 16 to 20. (For example, you can hand them in  class on 
Thursday February
19, or bring them over to the Hill Center and put them in my mailbox
on February 20.)

\subsection{Homework assignment due on February 19} 

{\em NOTE: this is a very short homework assignment, because you have 
the supplementary problems due the same week.}

\begin{itemize}
\item[A.] Problems 1 and 2 on page \pageref{thepagez11} of this handout.

\item[B.] Prove that the equation $x^2+1=0$ has a solution
in $\zz_{29}$ (i.e., that the statement ``$\exists x,x^2+1=0$'' is  
true in $\zz_{29}$).
\item[C.] Prove that the equation $x^2+1=0$ has a solution
in $\zz_{31}$ (i.e., that the statement ``$\exists x,x^2+1=0$'' is  
true in $\zz_{31}$).
\end{itemize}	


\section{$\zz_{11}$ and ISBN numbers}

As explained earlier, $\zz_{11}$ is a set consisting of $11$ members, 
namely, the integers $0,1,2,3,4,5,6,7,8,9$ and $10$. 

The members of $\zz_{10}$ are called {\bf\em digits}. The members
of $\zz_{2}$ are called {\bf\em bits}. By analogy with these two 
widely used names, we are going to call the members of
$\zz_{11}$ {\bf\em elvits}. So an {\bf\em elvit}
is one of the numbers $0,1,2,3,4,5,6,7,8,9,10$. Notice that 
{\bf\em there are eleven elvits}. 

\subsection{ISBN numbers}

{\bf ISBN} numbers are used to catalog and identify books. Each book 
has an ISBN number. 

{\bf An ISBN number is a sequence 
$$
(m_1,m_2,m_3,m_4,m_5,m_6,m_7,m_8,m_9,m_{10})
$$
of ten elvits.}  When we work with ISBN numbers, the name 
of the elvit ``$10$'' is ``$X$'', so
$$
\zz_{11}=\Big\{\,0,1,2,3,4,5,6,7,8,9,X\,\Big\}\,.
$$
The \ first \ nine \ of \ these \ ten \ elvits \ are \ required \ to 
\ belong \ to \ the \ set 
$\Big\{\,0,1,2,3,4,5,6,7,8,9\Big\}$, so  $m_{10}$ is the only
elvit in an ISBN number that is permitted to have the value $X$. 

Furthermore, $m_{10}$ is a {\bf redundant symbol}, included for
error-correcting purposes. It is always given by the formula
\begin{equation}\label{formten}
m_{10}=\sum_{j=1}^9jm_j\,,
\end{equation}
that is,
$$
m_{10}=m_1+2m_2+3m_3+4m_4+5m_5+6m_6+7m_7+8m_8+9m_9\,.
$$

Why do we use $\zz_{11}$ rather than $\zz_{10}$? And why do 
we use Equation (\ref{formten})? I will explain this eventually. 
But before we do 
that, let us verify the formula for some books.

\medbreak\noindent EXAMPLE 1. Our textbook has the ISBN number
0-201-43724-4.  Let us verify (\ref{formten}), making sure we remember 
that we are working in $\zz_{11}$, so all the operations are performed
modulo $11$.
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
m_1+2m_2+3m_3+4m_4+5m_5+6m_6+7m_7+8m_8+9m_9\\
&=&0+2\times 2+3\times 0+4\times 1+5\times 4+6\times 3+7\times 7
+8\times 2+9\times 4\\
&=&0+4+0+4+20+18+49+16+36\\
&=&4+4+9+7+5+5+3\\
&=&37\\
&=&4\,.
\end{eqnarray*}
So {\Large it works!}

\medbreak\noindent EXAMPLE 2. The book ``An Introduction to Analysis,''
by William R. Wade, has the ISBN number
0-13-093089-X.  Let us verify (\ref{formten}).
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
m_1+2m_2+3m_3+4m_4+5m_5+6m_6+7m_7+8m_8+9m_9\\
&=&0+2\times 1+3\times 3+4\times 0+5\times 9+6\times 3+7\times 0
+8\times 8+9\times 9\\
&=&0+2+9+0+45+18+0+64+81\\
&=&11+1+7+9+4\\
&=&1+7+9+4\\
&=&21\\
&=&10\\
&=&X\,.
\end{eqnarray*}
Once again, {\Large it works!}

\bigbreak I suggest you look at a few books that you can find at home
or in the library, look at heir ISBN numbers,
and check that in all cases Formula (\ref{formten}) is true. 

\medbreak
Now that you are convinced that ISBN numbers do indeed
obey Formula (\ref{formten}), the three questions that you
are probably asking
yourself are: 
\begin{itemize}
\item[1.] {\em Why do we use $\zz_{11}$?,} 
\item[2.] {\em Why do we include $m_{10}$, which contains absolutely no 
new information,
since it is completely determined by the first $9$ elvits?,} 
\item[3.] {\em Why do we use Formula (\ref{formten})?}
\end{itemize}

{\em You will get the answers to these questions a little bit further in
this handout.}  But it would be a good idea if you started thinking about
them right now.  Here is a hint:

\medbreak {\Large The crucial distinction is that
$\zz_{11}$ is a field, but $\zz_{10}$ is not.}
(The definition of ``field'' is given in the textbook, pages 180, 181.) 

\subsection{Why is $\zz_{10}$ not a field?}

The number system  $\zz_{10}$ is not a field
because in
it is not true that ``every number which is not equal to zero
has a multiplicative inverse.'' Or, in symbols, if you prefer,
the statement
$$
\forall x,(x\neq 0\Rightarrow \exists y,x\cdot y=1)
$$
is not true in $\zz_{10}$.

How do we know that this is not true? Let us find a counterexample.
I will show that $x=2$ is a counterexample. I have to show that
$$
2\neq 0\Rightarrow \exists y,2\cdot y=1
$$
isn't true. Notice that 
``$2\neq 0$'' {\em is} true, fortunately for us. (Why 
``fortunately for us''? Because if 
``$2\neq 0$'' wasn't true, then the implication 
that we are trying to prove, which says that
``$2\neq 0\Rightarrow \exists y,2\cdot y=1\,$,'' would be true!)

Since ``$2\neq 0$'' is true, what we need is to prove that
``$\exists y,2\cdot y=1$'' isn't true. Let us do it {\large
\em by contradiction.}  Suppose that there exists such a $y$.
Pick one and call it $\bar y$. Then $2\cdot\bar y=1$ in $\zz_{10}$. 
This means that in $\zz$ the number $2\cdot\bar y$ is equal to 
a multiple of $10$ plus $1$.  But a multiple of $10$ plus $1$ must be odd,
and $\bar y$ is even. So we have reached a contradiction, $y$ doesn't
exist, and we are done.\ \ {\bf END OF PROOF}

\subsection{Why is $\zz_{11}$ a field?}

Let us show that, at least, it is true that
``every nonzero member of $\zz_{11}$ has a multiplicative inverse.''
Here is the proof:

\begin{eqnarray*}
1\times 1&=&1\\
2\times 6&=&1\\
3\times 4&=&1\\
4\times 3&=&1\\
5\times 9&=&1\\
6\times 2&=&1\\
7\times 8&=&1\\
8\times 7&=&1\\
9\times 5&=&1\\
10\times 10&=&1\,.
\end{eqnarray*}

Now, to prove that $\zz_{11}$ is a field we also have to prove all the other
field axioms, such as, for example, the commutative law of addition
(i.e., $\forall x,\forall y,x+y=y+x$), and many other things, listed 
in the book. I leave that up to you, if you feel like doing it. 
But the main part is the existence of multiplicative inverses, 
because this is what makes $\zz_{11}$ from, say $\zz_{10}$ or $\zz_{12}$. 

\subsection{Why do we use elvits for ISBN numbers?}

In principle, we would like an ISBN number to consist of nine digits.
(This would allow us to have a total of $10^9$---i.e. one billion---ISBN
numbers, which presumably is more than the number of books ever
published or to be published up to, say, the year 2100. For example,
as this is being written, the U.S. Library of Congress has $14$
million books, and a total of $88$ million items including manuscripts,
maps, music, art prints, photographs, motion pictures, videotapes,
newspapers, pamphlets, recordings and other materials.) 

Now, as you well know, ISBN numbers are transmitted, copied, rewritten, etc.
Each of these operations can introduce {\bf\em transmission errors}. 
So it is customary to introduce some {\bf\em redundancy} into 
a signal to use it to {\bf\em check for errors}. (For example, 
when you change your password for a computer system you use, 
you are asked to type in the new password and then to type it in 
again. In principle, if you type it in correctly the first 
time, there is no need to type it in again, because this
would convey no new information. However, if you made a mistake
the first time you typed it in, you will almost certainly type 
in a different string of symbols the second time, because it is extremely 
unlikely that you will make exactly the same mistake the second time.
And then the computer will know that you made a mistake somewhere,
and ask you start all over again.)

For ISBN numbers, we introduce redundancy by adding an extra symbol
that carries no new information but will enable us to detect the most
frequent types of errors.
A simple example of how this could be done, if we were using $\zz_{10}$, 
would be to take $m_{10}$ to be the sum of the other
$9$ digits, i.e., to take
\begin{equation}\label{chch}
m_{10}=m_1+m_2+m_3+m_4+m_5+m_6+m_7+m_8+m_9\,.
\end{equation}
{\bf\em This would detect errors consisting of getting one 
digit wrong.} (Indeed, if you make a mistake 
in one of your first nine digits, that will change the sum, so 
the tramsitted $m_{10}$ will no longer satisfy (\ref{chch}).
For example, if the correct $m_5$ is $2$, and you send
a $7$ instead, then the sum $m_{10}$ will be increased by $5$
in $\zz_{10}$.) 

Naturally, {\bf\em this method will  not detect more 
complicated errors such as, for example, if you send two 
wrong symbols.} (For example, if the correct string 
is $0636274165$, and you send is $0686774165$ instead,
getting  two digits wrong, then (\ref{chch}) will still hold,
because the errors in $m_3$ and $m_5$ are both $5$, so their sum
in $\zz_{10}$ is $0$.) But two mistakes are a lot less likely 
than one mistake, so this imperfect  method will enable us
to detect most errors, even though some errors will still go 
undetected. 

Another limitation of the method is that 
it will detect errors but {\bf\em it will not tell you how to correct 
them,} because you will not know which digit is wrong. 
(For example, if you receive the string $0636774165$, then you know
for sure that it has to be wrong,
because $0+6+3+6+7+7+4+1+6=0$---in $\zz_{10}$---whereas $m_{10}=5$, 
so (\ref{chch}) is not true. But you do not know which 
digit was wrong.) Again, this is
a limitation of the method that we are willing to accept, because 
once we know that there is a mistake we can ask the sender to 
send us the number again.

The real problem is that there is another very common 
type of error that this method will not detect, namely, 
{\bf\em transposition errors}, in which two 
symbols get exchanged. (For example, if the correct 
$9$-digit string is $0636274165$, and you send instead 
$0616274365$, so that $m_3$ and $m_7$ are interchanged,
the sum of (\ref{chch}) will be the same.)

If we want to detect transposition errors as well as 
the simpler ones consisting of getting one symbol wrong,
then we need a formula 
\begin{equation}\label{chch-a}
m_{10}=a_1m_1+a_2m_2+a_3m_3+a_4m_4+a_5m_5+a_6m_6+a_7m_7+a_8m_8+a_9m_9\,,
\end{equation}

\medbreak\noindent 
where the coefficients $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9$ 
are all different.  But then, if we insist on working in $\zz_{10}$, 
and we use (\ref{chch-a}) taking the coefficients $a_j$ to be
all different, we run into another difficulty:
at least one of the $a_j$ will have to be even (why?), and then
$5a_j=0$ in $\zz_{10}$, which means that if we make 
an error of $5$ in the $j$-th digit, then we will not 
detect this error. In other words: {\bf\em if we work in 
$\zz_{10}$ there is no way to add an extra ``error-detecting''
digit $m_{10}$ given by a formula (\ref{chch-a}) so that 
we will detect both (a)~errors in the transmission of a single digit,
and (b)~transposition errors.}

On the other hand, let me prove to you that 
{\bf\em in $\zz_{11}$ {\Large it is possible} to add an extra 
``error-detecting''
digit $m_{10}$ given by a formula (\ref{chch-a}) so that 
we will detect both (a)~errors in the transmission of a single digit,
and (b)~transposition errors.}  Indeed, we can use the formula
that is actually used in real life, namely,
\begin{equation}\label{chch-b}
m_{10}=m_1+2m_2+3m_3+4m_4+5m_5+6m_6+7m_7+8m_8+9m_9\,.
\end{equation}
Notice that this corresponds to taking
$a_1=1$, $a_2=2$, $a_3=3$, $a_4=4$, $a_5=5$, $a_6=6$, $a_7=7$,
$a_8=8$, $a_9=9$.
These are $9$ different elvits, and this is exactly as it 
should be, since we have already seen that in order to detect transposition
errors you do need that $a_j$ to be all different. 

I will now prove to you rigorously (with your help) that {\bf\em if 
the $m_j$ are elvits chosen so that (\ref{chch-b}) holds, then 
every error consisting of one wrong elvit or two transposed elvits
will be detected.}  

Here is the first half of the proof. Consider the first type of error,
namely, one wrong elvit. Let $j$ be the place where this error is
made, so $j$ is one of the numbers $1,2,3,4,5,6,7,8,9,10$ and the
correct elvit $m_j$ is replaced by the wrong elvit $\tilde m_j$, but
all the other elvits $m_i$ (for $i\neq j$) are correct. Let us deal
separately with the cases when $j<10$ and $j=10$. If $j<10$, then 
the right-hand side $m_{10}^{computed}$ of (\ref{chch-b}) computed 
from the transmitted elvits will be $m_{10}+j(\tilde m_j-m_j)$,
but the received $10$-th elvit will be the correct $m_{10}$. 


\begin{center}
\fbox{\fbox{\begin{minipage}{11.3cm}
{\Huge Since $j\neq 0$ and $\tilde m_j-m_j\neq 0$, the product 
$j(\tilde m_j-m_j)$ is also different from $0$}
\end{minipage}}}\end{center}

\noindent so $m_{10}^{computed}$ will not be equal to $m_{10}\,$. 
This means that the received ten-elvit string will not satisfy 
(\ref{chch-b}), so the error will be detected. 

Now let us look at the remaining case, when the wrong elvit 
is the tenth one. In that case, the transmitted $m_j$ (for 
$j=1,2,3,4,5,6,7,8,9$) will be correct, but the transmitted
value $\tilde m_{10}$ will be different from the true value
$m_{10}$. The computed tenth elvit will be $m_{10}$, but the 
one received will be $\tilde m_{10}$, which is different. 
So once again the error will be detected.

This completes the first half of the proof, having to do with
detecting errors in the transmission of a single elvit. Now we have to
do the second part, namely, prove that the formula used in real life
also detects transposition errors. And here I have a real treat for you! 
{\Large\sl I am leaving this part of the proof for you to do 
as a homework problem!}  (No need to thank me.)

\medbreak\noindent PROBLEM 1 (ON $\zz_{11}$): Prove that
Formula (\ref{chch-b}) for ISBN numbers detects
transposition errors. That is, prove the following:
if a sequence ${\bf m}=(m_1,m_2,\cdots,m_{10})$ of ten elvits 
satisfies (\ref{chch-b}) then any sequence of 
$\tilde{\bf m}$ obtained from ${\bf m}$ by exchanging 
two different elvits (that is, any $\tilde m=
(\tilde m_1,\tilde m_2,\cdots,\tilde m_{10})$ such 
that, for some $i,j$ for which $i\neq j$ and 
$m_i\neq m_j$, we have $\tilde m_k=m_k$ if
$j\neq k\neq i$, $\tilde m_i=m_j$ and $\tilde m_j=m_i$) 
will fail to satisfy (\ref{chch-b}).  (WARNING: you will have to be careful
and consider separately two types of interchanges, namely, those 
that do not involve $m_{10}$ and those that do. Alternatively, there is a
cute trick that will enable you to deal with both cases in one swoop.
See if you can figure it out.)

\label{thepagez11}

\medbreak\noindent PROBLEM 2 (ON $\zz_{11})$: Suppose we wanted to increase 
the number of ISBN numbers by using $11$ elvits instead of $10$.
That is, we would define an ISBN number to be a sequence
$(m_1,m_2,\cdots,m_{10},m_{11})$ of $eleven$ elvits, and we would use 
the error-detecting formula
\begin{equation}\label{chch-c}
m_{11}=m_1+2m_2+3m_3+4m_4+5m_5+6m_6+7m_7+8m_8+9m_9+Xm_{10}\,.
\end{equation}
(Recall that $X$ is the name of $10$ in $\zz_{11}$.) Would this still
work to detect both types of errors?


\subsection{A remark on fields}

The crucial step in the proof
of the first half of our theorem or the error-detecting properties
of $\zz_{11}$-based ISBN numbers was the assertion that 

\begin{center}
\fbox{\fbox{\begin{minipage}{11.3cm}
{\huge In $\zz_{11}$, 
if $a\neq 0$ and $b\neq 0$ then $ab\neq 0$.}
\end{minipage}}}\end{center}

\medbreak\noindent How do we know that this is true? Well, we have to 
{\bf\em prove it}. One possible proof would go as follows:
just compute all possible products $ab$ in $\zz_{11}$ 
of two nonzero members of $\zz_{11}$ and in each case check that
the result is not zdero. This is not too hard to do: there are only
$100$ such products (why?), so you can do it in a few minutes.
And you can reduce the work if you allow yourself to use the 
commutative law of multiplication ($\forall a,\forall b(ab=ba)$),
because this would reduce the number of products to $55$
(why $55$?). 

If you do not want to work hard, there is an alternative,
called {\Large thinking}. The idea is to use a fact that
we have already proved, namely, that 
{\bf\em $\zz_{11}$ is a field}. Recall that this means that,
in $\zz_{11}$, 
\begin{equation}\label{fiel}
\forall a,(a\neq 0\Rightarrow \exists b,ab=1)\,,
\end{equation}
(or, in words, ``if you take a member of $\zz_{11}$ which is not equal
to zero then there exists a member of $\zz_{11}$ that multiplied by the first 
member yields one'' or, if you prefer, ``every nonzero member of $\zz_{11}$ 
has a multiplicative inverse''). 

Now let me prove: 

\medbreak\noindent THEOREM: {\em In a field, the statement
``$\forall a,\forall b,({\rm if}\,\,a\neq 0\,\,{\rm  and}
\,\,b\neq 0\,\,{\rm then}\,\, ab\neq 0)$'' \ is true.}

\medbreak\noindent PROOF: Let $a,b$ be arbitrary nonzero members
of the field. We want to prove that $ab\neq 0$. Suppose that
$ab=0$. (So we are doing a proof by contradiction!) 
Since $b\neq 0$, the field axiom (\ref{fiel}) tells us that
there exists a $c$ such that $bc=1$. Pick one such $c$ and call it
$\gamma$. Then $b\gamma=1$. So
$$
a=a\cdot 1=a(b\gamma)=(ab)\gamma=0\cdot\gamma=0\,.
$$
So $a=0$, contradicting the fact that $a\neq 0$. {\bf THE END}

Obviously, you must have lost of questions at this point.
 For example, {\em how do we know that $0\cdot\gamma=0$?}
It turns out that this can be proved, as  a consequence of the field axioms.
But I am not in a position to prove this for you now, because I have not yet
given you the complete list of the field axioms. (Be patient, this 
will be done soon!)

\medbreak\noindent REMARK: If you are curious, and want to see right now
the proof of ``$0.\gamma=0$'', here it goes. I will show that 
$\forall x,0.x=0$. {\em Proof:} Let $x$ be arbitrary. We have
$0+0=0$, using the axiom that says that $\forall y,y+0=y$, with $y=0$.
So $(0+0).x=0.x$. But $(0+0).x=0.x+0.x$, by the distributive law.
Hence $0.x+0.x=0.x$. Now, the additive inverse axiom says that 
$\forall y,\exists z,y+z=0$. Apply this with $0.x$ in the role of $y$. 
Then $\exists z,0.x+z=0$. Pick one such $z$ and call it $\zeta$.
Then $0.x+\zeta=0$. So
\begin{eqnarray*}
0.x&=&0.x+0\\
&=&0.x+(0.x+\zeta)\\
&=&(0.x+0.x)+\zeta\\
&=&0.x+\zeta\\
&=&0\,.
\end{eqnarray*}
So $0.x=0$, and we are done.\proofend

\section{Propositional Calculus}

In a propositional calculus, the {\bf symbols} are:
\bi
\item[1.] the {\it propositional variables} (usually letters such as $P$,
$Q$, $R$, $S$, etc., or maybe $A$, $B$, $C$, $D$, etc.),
\item[2.] the {\it logical connectives} $\vee$, $\wedge$, $\Rightarrow$, 
$\Leftrightarrow$, $\sim$,
\item[3.] the {\it parentheses} $\,\,\,(\,\,\,$ and $\,\,\,)\,\,\,$.
\ei
REMARK: In the above list, we have cheated
a little bit, by being somewhat 
vague about the list of propositional variables. (The word ``etc.'' 
is a strong symptom of vagueness!!!) Exactly what are the 
propositional variables? Just $A$, $B$, $C$, $D$? Or just
$P$, $Q$, $R$, $S$? That is, do we have only four propositional
variables? Suppose that for a particular proof or definition
we need formulas with 6 variables. Can we make new variables as we go
along?


It would be much better if we had a completely prespecified list of
variables, for example, ``$P$, $Q$, $R$ and nothing else.'' The
trouble with this is that you may want in some cases to use $4$, $5$,
$6$, $\ldots$, $457$, or any number of variables, so we really need an
{\em infinite} collection of propositional variables. We can achieve
that with finitely many symbols, as follows: we pick just one letter
(for example, the $P$) and a ``marker'' symbol (for example, the
vertical bar $|\,\,$), and then let the propositional variables be
$P$, $P|$, $P||$, $P|||$, $P||||$, $P||||$, and so on. Now we have
a propositional calculus with {\em exactly $9$ symbols}, namely,

\medbreak\noindent 
\fbox{\fbox{$
P\quad\qquad |\quad \qquad\vee\quad\qquad\wedge\quad\qquad\Rightarrow
\quad\qquad\Leftrightarrow\quad\qquad\sim\quad\qquad
(\quad\qquad)\,$}}$\,\,\,$.


\medbreak\noindent In this calculus, instead of writing something like 
``$(A\wedge B)\Rightarrow(C\vee D)$'', we would write
``$(P\wedge P|)\Rightarrow(P||\vee P|||)$''. Thus, 
the formulas we end up writing get a little bit 
longer, but we have achieved something that in this course we are always 
striving for, namely, {\bf complete precision}. 

\bigbreak\noindent Using the basic symbols of our propositional calculus,
we want to write {\bf propositional forms} (a.k.a.
{\bf sentence forms}, or {\bf statement forms}). that is,
expressions such as ``$(A\wedge B)\Rightarrow A$'', where we can plug in 
a statement for each of the variables and get 
a ``compound statement.'' (For example, if we plug in
``I am rich'' for $A$ and ``I am smart'' for $B$, we get 
the statement ``if I am rich and I am smart then I am rich,''
which is incredibly stupid but undoubtedly true.)

Not every string of symbols will be ``acceptable'' to us. For example,
``$AB$'' should not be acceptable, since we would not know how 
to plug in statements for $A$ and $B$ and get something meaningful. 
(If you think you can plug in ``I am rich'' and
``I am smart'' for $A$ and $B$, think again! What you get if you do that
is ``I am rich I am smart,'' which is certainly not a correct
English sentence.)  

We are going to call those strings that we would like to accept as
being really ``forms of compound statements'' {\bf well formed}.

\medbreak\noindent REMARK: Every time you use some basic symbols
to create a language, there is going to be a distinction between
strings that are acceptable and strings that are not. 
For example, in English the string ``the cow is white''
(consisting of $16$ symbols; we {\em are} counting the blank space
as a separate symbol) is acceptable, but the string 
``eo wt hics htiwe'' is not. (And what do you think about
``the white is cow'', or ``the is cow white''? I leave these
ones up to you, 
but you should draw from these examples the conclusion 
that in English the notion of an ``acceptable string''
is not 100{\%} precise.) 

In the so-called ``formal languages'' there is a precise, totally
unambiguous notion of acceptability of a string. The acceptable strings 
are called ``well formed.'' 

For example, suppose we use the symbols 

\medbreak\noindent
\fbox{\fbox{$
0\quad\,\quad\, 1\quad \,\quad\, 2\quad\,\quad\, 3\quad\,\quad\, 4
\quad\,\quad\, 5\quad\,\quad\, 6\quad\,\quad\,
7\quad\,\quad\, 8\quad\,\quad\, 9\,$}}

\medbreak\noindent (called ``digits'') to name nonnegative integers. Then 
the strings ``$0$'', ``$1$'', ``$3$'', ``$70$'', 
``$63$'', ``$20954$'', ``$137828$'', are acceptable, but ``$01$''
is not. The precise rule is as follows: a string of digits
is {\bf well formed} if it is either the ``$0$'' string 
or a string whose leftmost symbol is not $0$.


\medbreak\noindent The well-formed formulas are also called 
{\bf propositional forms.} The {\bf rules} determining which strings are
well-formed formulas (i.~e. propositional forms) are:

\medbreak\noindent RULE 1: One-letter strings consisting of a propositional
variable are well-formed formulas. 

\medbreak\noindent RULE 2: If ${\cal A}$ is a well-formed formula
consisting of a single propositional variable then 
$\sim{\cal A}$ is a well-formed formula.

\medbreak\noindent RULE 3: If ${\cal A}$ is a well-formed formula
consisting of more than a single symbol then 
$\sim({\cal A})$ is a well-formed formula.

\medbreak\noindent RULE 4: If ${\cal A}$ and 
${\cal B}$ are well-formed formulas each one of which  consists
of a single propositional variable then 
${\cal A}\wedge{\cal B}$, ${\cal A}\vee{\cal B}$, 
${\cal A}\Rightarrow{\cal B}$ and ${\cal A}\Leftrightarrow{\cal B}$ and 
are  well-formed formulas.

\medbreak\noindent RULE 5: If ${\cal A}$ and 
${\cal B}$ are well-formed formulas such that ${\cal A}$
consists of a single propositional variable but ${\cal B}$ does not,
then  ${\cal A}\wedge({\cal B})$, ${\cal A}\vee({\cal B})$, 
${\cal A}\Rightarrow({\cal B})$ and ${\cal A}\Leftrightarrow({\cal B})$ and 
are  well-formed formulas.

\medbreak\noindent RULE 6: If ${\cal A}$ and 
${\cal B}$ are well-formed formulas such that ${\cal B}$
consists of a single propositional variable but ${\cal A}$ does not,
then  $({\cal A})\wedge{\cal B}$, $({\cal A})\vee{\cal B}$, 
$({\cal A})\Rightarrow{\cal B}$ and $({\cal A})\Leftrightarrow{\cal B}$ 
and are  well-formed formulas.

\medbreak\noindent RULE 7: If ${\cal A}$ and 
${\cal B}$ are well-formed formulas neither of which 
consists of a single propositional variable, 
then  $({\cal A})\wedge({\cal B})$, $({\cal A})\vee({\cal B})$, 
$({\cal A})\Rightarrow({\cal B})$ and $({\cal A})\Leftrightarrow({\cal B})$ 
and are  well-formed formulas.

\medbreak\noindent RULE 8: Only those strings obtained by applying 
Rules 1, 2, 3, 4, 5, 6, 7 and 8 once or repeatedly
are well-formed formulas. 

\medbreak\noindent EXAMPLE: the string 
$$
((P\vee Q)\wedge(\sim Q))\Rightarrow(P\vee Q)
$$ 
is well formed, because one can arrive at it by applying the rules as follows:

\medbreak\noindent STEP 1: $P$ and $Q$ are well formed, by Rule 1.

\medbreak\noindent STEP 2: Applying Rule 2 with $Q$ in the role of ${\cal A}$,
we see that $\sim Q$ is well formed. 

\medbreak\noindent STEP 3: Applying Rule 4 with $P$ in the role of 
${\cal A}$ and $Q$ in the role of ${\cal B}$, we see that 
$P\vee Q$ is well formed. 

\medbreak\noindent STEP 4: Applying Rule 7 with $P\vee Q$ in the role of
${\cal A}$ and $\sim Q$ in the role of ${\cal B}$, we see that 
$(P\vee Q)\wedge(\sim Q)$ is well formed. 

\medbreak\noindent STEP 5: Applying Rule 7 with $(P\vee Q)\wedge(\sim Q)$
in the role of ${\cal A}$ and $P\vee Q$ in the role of ${\cal B}$, we 
we see that $((P\vee Q)\wedge(\sim Q))\Rightarrow(P\vee Q)$ is well formed.

\medbreak\noindent EXAMPLE: I claim that the string
$((P\wedge Q)\Rightarrow Q$ is {\bf\em not} well formed. Let us
try to understand {\it why}.
First of all, you can try to produce $((P\wedge Q)\Rightarrow Q$ using the 
rules, and no matter how hard you try you will not be able to.  But,
{\mymagonerm just because you have not been able to do something it does
not follow that it cannot be done.}  So, how do we know that $((P\wedge
Q)\Rightarrow Q$ {\bf cannot} be obtained using the rules?  We need an {\it
argument}. We need {\it reasoning}.  We need a {\bf proof.}

Here is a proof. First, we prove a small theorem.

\medbreak\noindent THEOREM 1. {\sl In every well-formed formula the number
of left parentheses is equal to the number of right parentheses.}

\medbreak\noindent PROOF.  Call a string ``balanced'' if it has the same
number of left and right parentheses. Any string obtained using Rules 1, 2
or 4 is
balanced, because it has zero left parentheses and 
zero right parentheses. Any application of Rule 3 or 5 or 6
will add an extra left
parenthesis and an extra right parenthesis, so when you apply any of these 
rules to a balanced string you get a balanced string.  
In any application of Rule 7, the number of left
parentheses in the result is that of ${\cal A}$, plus that of ${\cal B}$, 
plus two, 
and the same is true for the number of right parentheses. So when you apply
Rule  7 to two balanced strings you get a balanced string.   So the
application of Rules 1 through 7 can never create unbalanced strings. END OF
PROOF. 

\medbreak\noindent{\bf Remark.}  Theorem 1 gives an example 
of a {\bf necessary condition} for a string to be well formed.  (We say
that $A$ is a {\it necessary condition} for $B$ if $A$ happens whenever $B$
does, that is, if $B$ implies $A$.  For example: ``he took the final exam''
is a necessary condition for ``he passed the course,'' because if he
passed the course then he must have taken the final exam. But is of course
not a ``sufficient condition.'' Just taking the final exam is not enough to
pass the course.)  The theorem says that if a string is well formed then it
has to have the same number of left and right parentheses. This condition
is not sufficient:  there are many strings---such as,
for example, the string ``$P)($''---that have 
the same number of left and right parentheses but are not well formed.  
How can we tell if a string is well
formed? Even 
more precisely, can one give a {\bf computer algorithm} to check?  
The answer is ``yes,'' but this is a hard thing to do, and 
if you are interested you should look it up in a mathematical logic 
book. \proofend



\medbreak\noindent PROBLEM 3. 
Is $((P\vee Q)\wedge\sim Q)\imply(P\vee Q)$ well formed?  Justify your
answer. 

\medbreak\noindent PROBLEM 4. 
Is $((P\vee Q)\wedge(\sim Q))(\imply(P\vee Q))$ well formed?  Justify your
answer. 

\end{document}

\bigbreak\noindent When we work with a propositional calculus, we usually
do not want to have to write too many parentheses. So we add the

\medbreak\noindent PARENTHESIS DELETION OPTION:   If there is only one way
to put the parentheses back in to produce a well-formed formula, then you
are allowed to omit them. 

\medbreak\noindent For example, you can write 
$P\wedge Q$ rather than $(P\wedge Q)$, or $\sim P$ rather than $(\sim P)$. 

\medbreak\noindent{\bf Remark}. The book, in the bottom half of page 6, 
gives you extra leeway to omit parentheses. For example, the book gives you
{\bf extra rules} so that, if you write $\sim P\wedge Q$, it is understood
that this stands for $((\sim P)\wedge Q))$. (In principle, it could also
have stood for $(\sim (P\wedge Q))$.  The rules given in the book exclude
this second possibility, because ``$\sim$ applies to the smallest
proposition following it.'')  

Notice that {\bf if you want to make the formulas simpler by allowing
deletion of lots of parentheses, as the book does, then the rules become
more complicated.} In my own personal view, it is better to keep the rules
simple, even at the cost of writing formulas with more parentheses. {\bf
You are free to do it my way, or the book's way, as you prefer, long as you
do it correctly.} \proofend

\end{document}




\begin{center}
{\large{\bf  1.\ FIRST-ORDER PREDICATE CALCULI}} \label{predca}

\end{center}

\medbreak\noindent First of all, a predicate calculus involves a list of 
{\bf basic symbols} (or ``primitive symbols''). 

\begin{center}
\fbox{
\begin{minipage}{13cm}
\begin{center}
{\large{\bf THE BASIC SYMBOLS\\
OF A FIRST-ORDER PREDICATE CALCULUS}}\bf
\end{center} 
A first-order predicate calculus uses {\bf five} kinds of basic symbols: 
\bi
\item[1.] individual symbols\footnote{WARNING: the phrase ``individual
symbol'' is not meant to refer to ``a single symbol,'' as in the 
statement ``if you only look at the individual symbols then you might not 
understand the words.''  ``Individual symbol'' here means 
``symbol for an individual,'' in the same way as further down 
we use ``predicate symbol'' in the sense of ``symbol for a predicate.'' 
In other words, ``individual'' here is a {\bf noun}, not 
an {\bf adjective.}}$\,\,$;
\vspace{-2.5mm}
\item[2.] function symbols$\,\,$;
\vspace{-2.5mm}
\item[3.] predicate symbols$\,\,$;
\vspace{-2.5mm}
\item[4.] the logical connectives $\forall$, $\exists$, $\wedge$, $\vee$,
$\imply$, $\lto$, $\sim$$\,\,$;
\vspace{-2.5mm}
\item[5.] the left and right parentheses, ``$($'' and ``$)$,'' and the
comma ``,''$\,\,$.  
\ei
\end{minipage} 
}
\end{center} 

\medbreak\noindent {\bf INDIVIDUAL SYMBOLS} are usually letters such as
$x$, $y$, $z$, $u$, $r$, $a$, $b$, $c$, $n$, $p$, $q$, 
$\ldots$. In principle, you can use {\sl any symbol} to be an individual,
even if it is not a Roman letter. For example, you can use Greek
letters ($\alpha$, $\beta$, $\varepsilon$, $\pi$, etc.), Hebrew letters
(e.~g. $\aleph$), other symbols such as 
$\diamondsuit$, $\clubsuit$, or $\#$, 
and even longer and more complicated expressions that are not single
characters, such as $GNU$, $grrr$,  $oink$, $my$-$uncle$-$Jimmy$, 
$the$-$man$-$who$-$killed$-$Polonius$, etc. (If you count the blank space
as a character, then you could also use strings such as 
``{\it my uncle Jimmy,}'' or ``{\it the man who killed Polonius.}'')
Even more complicated expressions, not easily describable as 
strings of characters (for example, $\sqrt{2}$, or $\frac{3}{4}$), 
are often permitted, for historical reasons, but nowadays we prefer to
avoid them, for the very same reason that makes us avoid weird symbols 
such as $2^3$ or $\frac{2}{3}$ in programming languages, and use instead 
strings such as $2\#{}3$ or $2/3$. 

{\em The following characters are {\bf NOT} allowed to be parts of a
string used an an individual symbol: the parentheses, the comma, 
the logical connectives, predicate and function letters.} 

{\em If a string containing more than one character is introduced 
as an individual symbol, then it has to be explicitly declared 
to be so.} (For example, if I want to use the string ``{\it student}'' 
as an individual symbol, then I have to make a declaration to that effect. 
Once I have done so, I can write things such as 
``$\forall student(g(student)\to G(student))$,'' meaning---if ``$g(x)$'' 
stands for ``$x$ is good'' and ``$G(x)$'' for $x$ gets good 
grades''---that ``every good student gets good grades.'' The reason for
this stipulation ought to be clear to you: if we had not declared 
``{\it student}'' to be an individual symbol, then we might 
think of reading ``$\forall student$'' as ``for all
$s,t,u,d,e,n,t$'', where $s$, $t$, $u$, $d$, $e$, $n$, $t$ are 
six different variables. 



\medbreak\noindent {\bf CONSTANTS VS. VARIABLES}. 
An individual symbol is either a {\it constant} or a {\it variable}, and
can never be both at the same time.  Whether a symbol is a {\it constant}
or a {\it variable} depends on the context. There are various ways to {\it
declare} something to be a constant or a variable:

\bi
\item[1.] If within a proof or subproof
you say ``Let $\#$ be arbitrary'' or  ``Let $\#$ be such that
$\ldots$,''  then you are
effectively declaring  $\#$ to be a constant until the end of the proof
or subproof. (For example: 
once you say ``Let $a=3$'' then from that moment on $a$ is a constant. 
If you say ``Let $\varepsilon$ be arbitrary'' then from then on 
$\varepsilon$ is a constant. (The same applies to ``choose'' or ``pick.'') 

\item[2.] Any individual symbol that appears immediately to the right 
of a  quantifier 
is a variable. For example, if in a proof or subproof you see the formula 
$\forall x\exists y\,b(x,y)$, then you know that $x$ and $y$ are variables
throughout that proof or subproof. 
If I write $\forall x\,(x^2\ge 0)$, then 
$x$ {\sl has} to be a variable, because only variables can appear to the
right of quantifiers.

\item[3.] Some symbols are {\it reserved constants}, or {\it fixed
constants}. This means that they are declared to be
constants once and for all, and intended to have a specific value that is
fixed for ever. For example, in real-number
arithmetic the symbols $0$, $1$, $2$, $5$, $\pi$, $e$, 
are fixed constants. In set theory, the symbol $\emptyset$ is a fixed
constant. 

\ei

\medbreak\noindent At each point, it is important to know  whether a symbol
is a constant or a variable, because this helps you know what you can and
what you cannot do with it.  
For example, after you have declared
``$a$'' to be a constant by saying, for example, ``let $a$ be an arbitrary
real number,'' {\sl you cannot use quantifiers such as $\forall a$},
because $a$ is now fixed, not variable.

\medbreak Similarly, if a symbol is a variable, then there 
are things you cannot do with it.  For example, after 
you have said in a proof $\forall x\forall y(x+y=y+x)$, 
you cannot write in your proof something like $x+y=y+x$, 
because {\sl this is not a statement}.  (This means that, if $x$ and $y$
are variables, then ``$x+y=y+x$'' does not have  
a definite truth value. It doesn't make sense to say that 
$x+y=y+x$'' is true or false unless $x$ and $y$ are given specific 
values.    And, if you give $x$ and $y$ specific, definite values, then
they become constants.) 

\medbreak\noindent{\bf FUNCTION SYMBOLS} are letters such as $f$, $g$,
$F$, $G$, or other symbols such as $+$, $\cdot$, $\cap$, $\cup$,
$\in $, $-$. Each function symbol admits a {\sl fixed} number of arguments.
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} symbol. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} symbol.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} symbol. And so on.)   If you write 
a function symbol, followed by a ``$($,'' then as many arguments
as the function symbol admits, separated by commas, and then 
``$)$,'' you get a {\sl term}.  

\medbreak\noindent Most function symbols are {\bf prefix symbols}, that is,
symbols that are written {\em before} the arguments. (For example, 
when you define a function $f$ of two variables, and you write 
$f(x,y)$, you are using $f$ in prefix form.) However, some symbols 
have a different, peculiar syntax.  There are {\bf binary infix symbols}, 
that is, 2-argument function symbols that are written in the middle,
between the arguments. (Example: $+$ is a  binary infix symbol; 
we write $3+4$ rather than $+(3,4)$.)  And there are  
{\bf unary roundfix symbols}, that is, symbols that you put on both
sides of the argument. (For example, the symbol ``$|\,\,\,|$''
for ``absolute value'' is in this class. We write $|3|$, or
$|x|$, with the argument in the middle.)


The strings that can be used 
as arguments are variables, or constants, or more general ``terms.''



\bigbreak
\begin{center}
{\large\bf 2. TERMS VS. FORMULAS} \label{termfo}
\end{center}

\medbreak\noindent{\bf Terms} are supposed to be 
names of ``things.''  {\bf Formulas} are
expressions that---when we plug in values for the
open variables---become sentences or statements.

For example, the following are terms: 

\medbreak
\noindent\hbox to 3.5cm{\hfill $3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to 3.5cm{\hfill $x$}\hbox to 3.5cm{\hfill $5xyz$}

\noindent\hbox to 3.5cm{\hfill $x+3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to
3.5cm{\hfill $x^2+3y+7$}\hbox to 3.5cm{\hfill $5xyz+37-(2+x)$}

\noindent and the following are formulas:

\medbreak
\noindent\hbox to 6.3cm{\hfill $3<5$}\hbox to 6.3cm{\hfill $3+5=8$}

\noindent\hbox to 6.3cm{\hfill $x+3=7y$}\hbox to 6.3cm{\hfill$5xyz<x+z$}

\noindent\hbox to 6.3cm{\hfill $\forall x(x+3>x)$}
\hbox to 6.3cm{\hfill$\forall x(x+y>x)$}

\noindent\hbox to 6.3cm{\hfill $\exists x(x+y>xz)$}
\hbox to 6.3cm{\hfill $\forall x(x\!>\!y\imply y\!=\!z)$}

\noindent\hbox to 6.3cm{\hfill $\exists y\forall x(x\!>\!y\imply y\!=\!z)$}
\hbox to 6.3cm{\hfill $\exists x(x+y>xz)\wedge x=3$}

\noindent\hbox to 6.3cm{\hfill $\forall z(\forall x(x\!>\!y
\imply y\!=\!z)\wedge\forall w(z>w))$}
\hbox to 6.3cm{\hfill $\exists y\forall x(x\!>\!y\imply y\!=\!z)$}

\bigbreak\noindent
How can you tell whether something is a term or a formula? 

\medbreak\noindent TERMS.  A {\bf term}
stands for a {\bf definite thing} (it may be a number, or a set, or a line,
or a function, or some other entity such a person or an animal), or for a
``variable thing'' that becomes definite once you plug in values for the
variables.  (For example, ``$3+5$'' denotes a definite number, whereas
``$3+x$'' denotes a ``variable number'' that becomes definite once you plug
in values for $x$: when you plug in $x=5$, $3+x$ has the value $8$; when
you plug in $x=-3$, $3+x$ has the value $0$; when you plug in $x=-20$,
$3+x$ has the value $-17$; and so on.)  For a more complicated example 
of a term, consider the expression ``$f(x)$''.   This is a term 
that contains two letter variables, namely, $f$ and $x$, standing,
respectively, for a function and a number. So, when you plug in a specific
function for $f$, and a specific number for $x$, you get a number. 
(For example, if $f$ is the ``squaring'' function, i.e. the function such
that $f(y)=y^2$ for all $y$, and $x=3$, then $f(x)$ has the value $9$. 
If $f$ is the ``exponential'' function, i.e. the function such
that $f(y)=e^y$ for all $y$, and $x=6$, then $f(x)$ has the value $e^6$. 
If $f$ is the function such
that $f(y)=y^2+3y+7$ for all $y$, and $x=6$, then $f(x)$ has the value 
$61$.) 


\medbreak\noindent FORMULAS.  A {\bf formula} makes an {\bf assertion},
which may be 
true or false.  So a formula has a {\bf definite truth value} (``true'' or
``false,'' often abbreviated $T$ or $F$) or a ``variable truth value'' that
becomes definite once you plug in values for the variables.  (For example,
``$3+5=8$'' has a definite truth value, namely, T, and ``$3+5=6$'' has a
definite truth value, namely, F.  On the other hand, ``$3+x=y$'' has a
variable truth value, which becomes definite once you plug in values for
the variables $x$ and $y$: when you plug in $x=5$ and $y=8$, $3+x=y$ is T;
when you plug in $x=2$ and $y=8$, $3+x=y$ is F; when you plug in $x=2$ and
$y=5$, $3+x=y$ is T; and so on.)



\sepa

\begin{center}
{\large\bf 3. HOW PARENTHESES MAKE LIFE HARD} \label{parencomp}
\end{center}





\medbreak\noindent If you want to see a completely precise definition of
``term''  and ``formula'' then you have to consult a mathematical logic
textbook. In this course, we will come close, but will have to 
omit discussing in detail one of the most annoying questions, namely,
that of how to deal with {\bf parentheses}. For example: 

\bi
\item{} {\em Should ``$\,\,7-5\,$'' be a term?}
Suppose we say ``yes.'' Then, can we take 
for example, the formula $x.7=7.x$ and plug in $7-5$ for $x$?
Not really, because if we do we get 
$7-5.7=7.7-5$, i.e., $7-35=49-5$, 
which is false and is not at all what we had
in mind!   

\item{} {\em So maybe we should say that ``$\,\,7-5\,$'' is {\bf not}
a term, but $(7-5)$ is?}  Let's see.  That would take care of the
$x.7=7.x$ example, because in that case we would get
$(7-5).7=7.(7-5)$, which is O.K. But then, what do we do if we want to
``plug $7-5$ for $x$ in the formula $x=x$''?  In that case, we would
like to get $7-5=7-5$, but if we insist that the ``real term'' is
$(7-5)$ rather than $7-5$, then when we plug it in we would get
$(7-5)=(7-5)$, which is not what we wanted.

\item{} {\em A third possibility would be to say that
``in certain cases when we plug in a term
we have to put in parentheses, and in other cases we do not.''}
Unfortunately, that will not do either.  
For example, {\bf we  cannot have a rule implying that
``when you substitute a term 
for $x$ in $x.7=7.x$ you put parentheses, but when you substitute 
in $x=x$ you don't.} This will not work, because the term
``$\,\,7-5\,$'' needs parentheses when you substitute it 
for $x$ in $x.7=7.x$---so that you get 
$(7-5).7=7.(7-5)$---but the term ``$\,\,3\,$\,'' does not need 
parentheses, because you write 
$3.7=7.3$ rather than $(3).7=7.(3)$.  So maybe we have to distinguish
two kinds of terms: that require parentheses when you substitute
them into things like $x.7=7.x$, and those that do not.
\ei

If you keep analyzing the various possibilities, it gets fairly
complicated.  It's not so terribly complicated that it cannot be done,
but it is too much for us in this course.  (As explained before, this
is done carefully and precisely in mathematical logic books.)

So we will essentially ignore the problem.  In our formal definition
of ``term'' we will put lots of parentheses, but then we will omit 
many of them. The informal rule is: you can omit parentheses
as long as it's completely clear how you would put them back in. 
But when you do this, you are not ``writing down the true terms,'' 
you are just {\it referring to them by means of an abbreviated 
expression which is not itself a true term}.
For example, when you write ``$7-5$,'' this is 
just a way of naming\footnote{{\bf Naming} is a very subtle 
thing. The word ``Dumbo'' is not Dumbo. For example,
``Dumbo'' is a five-letter word, but Dumbo is a flying 
elephant. Flying elephants are clearly different kinds of objects 
from five-letter words. Flying elephants have four legs
and big ears and a trunk, but five-letter words do not. 
In mathematics, one often fails to 
distinguish carefully between objects and their names,
so for example the {\bf symbol} ``$1$'' is said to be ``the same 
thing as'' the {\bf number} $1$, even though in reality
it is not. Terms and formulas are objects---strings of 
symbols---in a formal language, but one often talks about them 
using abbreviated names. For example, accoding to the point of view
I am taking here,  the string  $((7+5)=(16-4))$ is a true term,
but in order to avoid having to write lots
of parentheses we refer to by the name $7+5=16-4$, which is not itself 
a true term, even thouh it looks almost identical. We could have taken
a different point of view, making the things we write be true terms,
but then it would be much harder to give a precise definition
of ``term.'' We are going to 
ignore all these subtleties
here, not because they are unimportant, but because we don't have time
to talk about them.}  the true term,  which is $(7-5)$. 
Similarly, when you write $x+(3-y)=7+z$, this is just a way of referring
in abbreviated form to the true formula, namely, 
$((x+(3-y))=(7+z))$.  This is not a very nice solution, but 
to do this thing really well would require too much time.

There is another related  problem that we will also ignore.
Suppose, for example, that I write
$$
3+5+7=2+4+13\,.
$$
What is the ``real'' formula behind this? Is it
$$
(((3+5)+7)=(2+(4+13)))\,,
$$
or maybe
$$
((3+(5+7))=(2+(4+13)))\,,
$$
or 
$$
((3+(5+7))=((2+4)+13))\,,
$$
or 
$$
(((3+5)+7))=((2+4)+13))\,\quad ?
$$

\medbreak\noindent Before you answer---as you are probably thinking 
of answering---that the four formulas above ``are really the same,''
so it does not matter which one we choose, think for a minute.
What do you mean by ``they are the same''?   {\bf They certainly are 
not the same string of symbols.} (For example, the second symbol 
of the first formula is a ``$($'', but the second symbol 
of the second formula is a ``$3$'', which is quite different.)

This is a serious problem, but one that can be taken care of.  It's
hard work, however, and in this course we will just not pursue this
issue.

\newpage
\begin{center}
{\large\bf 4. TERMS} \label{termdef}
\end{center}

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``TERM''}}
\end{center}

\medbreak\noindent RULE T1: constants and variables are terms;

\medbreak\noindent RULE T2: if $t_1,\ldots,t_k$ are terms 
and $F$ is a $k$-variable function symbol, then $F(t_1,\ldots,t_k)$ is a
term. (For example: if $F$ is a one-variable function symbol and $t$ is a
term, then $F(t)$ is a term; if $F$ is a two-variable function symbol and
$t$, $s$ are terms, then $F(t,s)$ is a term; 
if $F$ is a three-variable function symbol and
$t$, $s$, $u$ are terms, then $F(t,s,u)$ is a term.) If any of the $t_i$
starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $F(3+x,x+(y-z))$ rather than 
$F((3+x),(x+(y-z)))$.)

\medbreak\noindent RULE T3: if $t,s$ are terms and $F$ is a 
binary infix function symbol, then $(tFs)$ is a term.

\medbreak\noindent RULE T4: if $t$ is a term and $F$ is a 
unary roundfix function symbol, then the expression you get by putting
$t$ inside $F$ is a term. 
If $t$
starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $|3+5|$ rather than $|(3+5)|$, and 
$\sqrt{3+5}$ rather than $\sqrt{(3+5)}$.)


\medbreak\noindent RULE T5: only those expressions obtained by repeated
application of Rules T1, T2, T3  and T4 are terms. 
\vspace{2mm}
\end{minipage}
}
\end{center}

\medbreak\noindent Later, when we discuss how to {\it interpret} 
(that is, assign a meaning), function symbols will correspond to {\it
functions}, that is, expressions that have a {\bf definite value}
once you plug in values for its arguments.  For example, if we 
use a binary prefix function symbol $plus$ to stand for ``sum of two numbers,''
then $plus(3,5)$ will have the value $8$, $plus(x,3)$ will have a value 
for each value of $x$ (so $plus(x,3)$ will be $7$ if $x$ is $4$, 
$2$ if $x$ is $-1$, and so on) and $plus(x,y)$ will have a value 
for each choice of values of $x$ and $y$ 
(so $plus(x,y)$ will be $7$ if $x$ is $4$ and $y$ is $3$,  
$2$ if $x$ is $-4$ and $y$ is $6$, and so on). If we use instead 
the binary infix symbol $+$, as is commonly done in mathematics,
then of course we will write $(x+y)$ rather than $plus(x,y)$. 

\medbreak\noindent {\bf EXAMPLE 1.} Let us start setting  up a 
predicate calculus to talk about {\bf real number arithmetic}.  We
introduce the two-variable infix function symbols $+$, $.$, $-$, the
two-variable prefix function symbols $max$, $min$, and two one-variable 
function symbols, namely, $-$  (which is a prefix symbol, and should not
be confused with the binary infix symbol $-$)  and
$|\ldots|$, which is a roundfix symbol.
(``$max$'' is meant to stand for ``the largest of,'' so for example
$max(5,3)=5$ and $max(-7,2)=2$. Similarly,
``$min$'' is meant to stand for ``the smallest of,'' so for example
$min(5,3)=3$ and $min(-7,2)=-7$.  ``$\ldots|$'' stands for 
``absolute value of,'' so ``$|x|$'' means ``absolute value of $|x|$.'')

In addition, we will have {\bf reserved constants}
for the usual names of the nonnegative integers,
in decimal notation (for example: $0$, $1$, $2$, $3$, $4$, etc). 

Then the following are terms:

\def\pltw#1#2#3{\noindent#1.$\,\,\,$\hbox to 5cm{#2\hfill}$\qquad$instead
of #3} 

\def\pltwnd#1#2#3{\noindent#1$\,\,\,$\hbox to 5cm{#2\hfill}$\qquad$instead
of $\quad$ #3} 


\bi
\item[1.] $x$, 
 
\item[2.] $3$, 

\item[3.] $-3\,$,$\,\,$\footnote{The unary prefix symbol 
``$-$'' obeys the folllowing anomalous rule: if $t$ is an
individual symbol, we write $-t$ rather than $-(t)$. 
So, for example, we write $-3$ and $-x$ rather than 
$-(3)$ or $-(x)$.} 


\item[4.] $-x$, 

\item[5.] $(x+3)$, 

\item[6.] $-(x+3)\,$,$\,\,$\footnote{Do you find this puzzling?
Doesn't Rule T2 tells us that the parentheses have to be omitted?
No! Take the one-variable 
prefix symbol $-$.  If $t$ is any term, then the first part
of Rule T2 tells us that $-(t)$ is a term. Now take $t$ to be
the term $(x+3)$. Then we find that 
$-((x+3))$ is a term. But then the second part of the rule
allows us to omit the two parentheses of 
$(x+3)$, and get $-(x+3)$.}

\item[7.] $((x+y)-z)$, 

\item[8.] $|x|$, 

\item[9.] $(x+|y|)$, 

\item[10.] $(min(x,1)+|x+y|)$, 

\item[11.] $min(x+y,|y|)$, 

\item[12.] $min(x+|y|,-y)$, 

\item[13.] $(max(x,27)+min(x+y,|y|))$, 

\item[14.] $|min(x+y,|y|)+max(x+y,|y|)|$.
\ei

\medbreak\noindent In mathematics, we often use the binary infix 
symbols $\wedge$, $\vee$ instead of the prefix symbols $max$, $min$. 
With these notations, the last five terms of the above list 
would be written instead as

\bi

\item[10'.] $((x\wedge 1)+|x+y|)$, 

\item[11'.] $((x+y)\wedge|y|)$, 

\item[12'.] $((x+|y|)\wedge-y)$, 

\item[13'.] $((x\vee 27)+((x+y)\wedge|y|))$, 

\item[14'.] $|((x+y)\wedge|y|)+((x+y)\vee|y|)|$.
\ei

\medbreak\noindent {\em The binary infix symbols $\wedge$, $\vee$ of
elementary arithmetic should not be confused with the symbols
$\wedge$, $\vee$ of the propositional calculus.}



\begin{center}
\fbox{
\begin{minipage}{11cm}
{\bf The same symbol $-$ is used 
for two different things: the one-variable function $negative$-$of$, 
and the two-variable function $minus$.  For example, when we write
$5-8=-3$, the first ``$-$'' is the two-variable function symbol, and the
second ``$-$'' is the one-variable function symbol.} 
\end{minipage}
}
\end{center}

\medbreak\noindent {\bf EXAMPLE 2.} Let us start setting  up a 
predicate calculus to talk about {\bf natural number arithmetic}.  We
introduce the two-variable infix function symbols $+$, $.$, 
and the two-variable prefix function symbols 
$max$, $min$, $GCD$, $LCM$. (Here 
``$GCD$'' and ``$LCM$'' are meant to stand for
``greatest common divisor'' and ``least common multiple,'' so for example
$GCD(12,20)=4$ and $LCM(12,20)=60$.) 

In addition, we will have {\bf reserved constants} for the usual names of
the natural numbers, in decimal notation (for example: $1$, $2$, $3$, $4$,
etc). ({\sl Remember that $0$ is {\bf not} a natural number}.)

Then the following are terms:

\bi
\item[1.] $(x+y)$, 

\item[2.]$GCD(x+5,z)$,

\item[3.] $(min(x,1)+LCM(x+y,83))$.
\ei


\medbreak\noindent You could get the terms to look even nicer by also
agreeing to use $\wedge$ rather than $min$ and $\vee$ rather than
$max$, as in Example 1. Then you would write $(t\wedge s)$ instead of
$min(t,s)$ and $(t\vee s)$ instead of $max(t,s)$.  With these new
symbols, you could write, for example,

\medbreak

\pltwnd{$\,\,$}{$((x\wedge 5)+
LCM(x+y,83))$}{$(min(x,5)+LCM((x+y),83))$.}

\bigbreak\noindent {\bf EXAMPLE 3.} Let us start setting  up a 
predicate calculus to talk about {\bf kinship}.  
We introduce two 
one-variable function symbols $f$, $m$, that will be thought of as standing
for ``father of'' and ``mother of.'' Suppose we have constants such as 
$Bill$, $John$ and $Mary$, and variables $x$, $y$, $z$. 
Then, for example, the following are terms:

\bi
\item[1.] $f(x)$, 

\item[2.] $m(y)$, 

\item[3.] $f(m(Bill))$, 

\item[4.] $f(m(x))$,

\item[5.] $f(f(f(f(f(f(Mary))))))$, 

\item[6.] $m(m(m(m(m(m(Mary))))))$, 

\item[7.] $m(f(m(f(f(m(Mary))))))$.
\ei

\medbreak\noindent The term $f(Bill)$ would have a definite value, namely, 
Bill's father.  The term $f(x)$ does not have a definite value 
unless you are told what the value of $x$ is.  If $x$ is Bill then 
the value of $f(x)$ is Bill's father. If $x$ is Mary then 
the value of $f(x)$ is Mary's father. 

\medbreak\noindent QUESTION: could we introduce function symbols to stand 
for ``son,'' ``daughter,'' ``brother'' or ``sister''?  The answer is {\bf
NO!} Why? Because {\bf ``son,'' ``daughter,'' ``brother'' and ``sister''
are not functions.} They do not have a definite value for each value of the
argument. For example,  $son(x)$ might fail to have a value 
if $x$ is Bill, because it could happen that Bill does not have a son, and
if this is the case then you don't know what $son(Bill)$ means.
Also, $son(x)$ might fail to have a {\it definite} value 
if $x$ is Bill, because it could happen that Bill has more than one son,
and then once again you would not know what $son(Bill)$ means. \proofend

\sepa

\bigbreak
\begin{center}
{\large\bf 5. FORMULAS} \label{termdef}
\end{center} 



\medbreak\noindent In order ot be able to talk about about 
sons and daughters and brothers and sisters and equality and inequality
and divisibilty, we need 
{\sl PREDICATE SYMBOLS}.

\medbreak\noindent {\bf PREDICATE SYMBOLS} are letters such as $p$, $q$,
$P$, $Q$, or other symbols such as $=$, $>$, $<$, $\le$,
$\ge$, $\epsilon$, $\subseteq$, $tall$, $stupid$, $smarter$-$than$. 
Each predicate symbol admits a {\sl fixed}
number of arguments.    
(If it admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} predicate. 
If it admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} predicate.  
If it admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} predicate.  And so on.)  

Like function symbols, most predicate symbols are {\bf prefix symbols},
but some important binary predicate symbols such as $=$, $<$,
$\subseteq$, are infix sybols. 

If you write a prefix 
predicate symbol, followed by a ``$($'' and then by as many arguments
as the predicate symbol admits, separated by commas, ending, finally, with a
``$)$,'' then you get an {\sl atomic formula}.  The strings that can be used 
as arguments are {\sl terms}.  One can also get atomic formulas using infix
symbols. 

More formally, then:

\begin{center}
\fbox{
\begin{minipage}{13cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``ATOMIC FORMULA''}}
\end{center}

\medbreak\noindent RULE AF1: If $L$ is a $k$-argument prefix predicate symbol,
and $t_1,\ldots,t_k$ are terms, then $L(t_1,\ldots,t_k)$ is an atomic
formula. (For example: if $p$ is a one-variable predicate symbol and $t$ is
a term, then $p(t)$ is an atomic formula; if $p$ is a two-variable
predicate symbol and $t$, $s$ are terms, then $p(t,s)$ is an atomic
formula; if $p$ is a three-variable predicate symbol and $t$, $s$, $u$ are
terms, then $p(t,s,u)$ is an atomic formula.)
If any of the $t_i$ starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $P(3+x,x+(y-z))$ rather than 
$P((3+x),(x+(y-z)))$.)


\medbreak\noindent RULE AF2: if $t,s$ are terms and $F$ is a 
binary infix predicate symbol, then $(tFs)$ is an atomic fomula.
If $t$ or $s$ starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $(3+x=5-y)$ rather than 
$((3+x)=(5-y))$.)


\medbreak\noindent RULE AF3: only those expressions obtained by 
applying Rules AF1 and AF2 are atomic formulas. 
\vspace{2mm}
\end{minipage}
}
\end{center}


\medbreak\noindent Later, when we discuss how to {\it interpret} 
our formulas (that is, how to assign {\sl meanings} to them), predicate symbols
will correspond to {\it predicates}, that is, expressions that have a {\bf
definite truth value} once you plug in values for its arguments.  For
example, if we use a binary predicate symbol $>$ to stand for ``larger
than,'' and write $t>s$ rather than $>(t,s)$, then $3>5$ will have the
truth value $False$, $x>3$ will have a truth value for each value of $x$
(so $x>3$ will be $True$ if $x$ is $4$, $False$ if $x$ is $-1$, and so on)
and $x>y$ will have a value for each choice of values of $x$ and $y$ (so
$x>y$ will be $True$ if $x$ is $4$ and $y$ is $3$, $False$ if $x$ is $-1$
and $y$ is $0$, and so on).

\medbreak\noindent {\bf EXAMPLE 1, continued.} Let us continue our discussion
of a  language for {\bf real-number arithmetic}. Now that we know all
about predicate symbols, we introduce the following binary 
infix predicate symbols:
``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$.'' 
Then the following are examples of atomic formulas: 

\bi
\item[1.] $(x+y=7)$, 

\item[2.] $(3+7=x)$, 

\item[3.] $((x+y)+|x+3|=7-(y+z))$, 

\item[4.] $((x+y)+|(x+3)|>(7-y)+z)$,

\item[5.] $(|x|=x\wedge-x)$,

\item[6.] $(|x|+5=|(x+y)\wedge|y|\,|+((x+y)\vee|y|))$.

\ei
 
\medbreak\noindent {\bf EXAMPLE 2, continued.} Let us go on 
with our language for {\bf natural-number arithmetic}. Now that we know all
about predicate symbols, we introduce binary infix predicate 
symbols for the basic binary
predicates ``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' ``$|$'' which
are {\sl binary}. (Note: ``$|$'' stands for ``divides'', so ``$x|y$'' means
``$x$ divides $y$,'' that is $\exists z(z.x=y)$.)  

Then the following are examples of atomic formulas: 

\bi
\item[1.] $(x+y=7)$, 

\item[2.] $(3+7=x)$, 

\item[3.] $(7|3.4)$,

\item[4.] $((x+y)+GCD(x,8)>(7-y)+z)$,

\item[5.] $(x=x\wedge-x)$.

\ei
 
\medbreak\noindent {\bf EXAMPLE 3, continued.} Now let us go on with 
the predicate calculus that enables us to talk about kinship.
Now that we have predicate symbols, we could introduce one-variable
predicates $\mu$, $\varphi$---so that 
``$\mu(x)$'' would stand for ``$x$ is male''
and ``$\varphi(x)$'' for ``$x$ is female''---
and lots of two-variable predicates  such as 
$s$, $d$, $c$, $br$, $sis$, $u$, $a$,   so that 
``$s(x,y)$'' would stand for ``$x$ is $y$'s son'',
``$d(x,y)$'' would stand for ``$x$ is $y$'s daughter'',
``$c(x,y)$'' would stand for ``$x$ is $y$'s child,'' 
``$br(x,y)$'' would stand for ``$x$ is $y$'s brother,'' 
``$sis(x,y)$'' would stand for ``$x$ is $y$'s sister,'' 
``$u(x,y)$'' would stand for ``$x$ is $y$'s uncle,'' 
and ``$a(x,y)$'' would stand for ``$x$ is $y$'s aunt.'' \proofend

\sepa

\medbreak\noindent Atomic formulas can be combined with the logical
connectives to yield {\bf formulas}.  For example, let us look at the
following fairly long string of 75 characters, in a predicate calculus 
with three binary predicate symbols, namely, $=$, $s$ and $c$: 



\medbreak\noindent $\exists y\,c(y,y)\vee\exists x\forall y\exists z\,
(((s(a,x)\wedge s(a,y))\wedge((\sim x =y)\wedge
(\sim a=z)))\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$

\medbreak
Let us call this expression $F$. Then $F$ is {\sl not} an atomic formula,
but it is a formula.  This means that $F$ is obtained from atomic formulas
by combining them via the connectives $\wedge$, $\sim$ and $\exists$.  In
our case, the atomic formulas used to produce $F$ are: $c(y,y)$,
$s(a,x)$, $s(a,y)$,
$(x=y)$, $(a=z)$, $c(z,x)$, $c(z,y)$, and $c(b,z)$.

We now describe precisely the rules that allow us to combine 
atomic formulas in order to make formulas.

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``FORMULA''}}
\end{center}

\medbreak\noindent RULE F1: Atomic formulas are formulas.

\medbreak\noindent RULE F2: If $F$ is a formula then
$(\sim F)$ is a formula. If $F$ is atomic and 
starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $(\sim x=y)$ rather than $(\sim(x=y))O$.)


\medbreak\noindent RULE F3: If $F$ and $G$ are formulas then 
$(F\wedge G)$, $(F\vee G)$, $(F\to G)$, and $(F\leftrightarrow G)$ are
formulas. If $F$ or $G$ is atomic and 
starts with a ``$($'' and ends with a 
``$)$'', then these parentheses are omitted. (So, for example,
we write $(x=3\wedge y=5)$ rather than 
$((x=3)\wedge(y=5))$, but we write 
$((\sim x=3)\wedge y=5)$, because $(\sim x=3)$ is not atomic, so
the parnthesis omission rule does not apply.)

\medbreak\noindent RULE F4: If $F$ is a formula
and $v$ is a variable, then $\forall v\,F$ and  $\exists v\,F$ are
formulas.

\medbreak\noindent RULE F5: Only those expressions obtained by repeated
application of Rules F1, F2, F3, F4  are formulas. 
\vspace{2mm}
\end{minipage}
}
\end{center}


\medbreak\noindent Why is $F$ a formula according to our rules?  In
the following box we show you in detail how one can construct $F$
following the rules.

\bigbreak


\begin{center}
\fbox{
\begin{minipage}{5.4in}

\parindent=0pt

\null
\vspace{0.1mm}


\begin{center} 
{\bf AN EXAMPLE OF THE USE OF RULES F1 TO F9}
\end{center}

Let us look at the formula $F$ defined earlier. 
Recall that $F$ is

\medbreak\noindent $\exists y\,c(y,y)\vee\exists x\forall y\exists z\,
(((s(a,x)\wedge s(a,y))
\wedge((\sim x =y)\wedge(\sim a=z)))$\\
$\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$

\medbreak

Now, here is a ``derivation'' of $F$ from the rules: 

\medbreak

STEP $\,\,\,$1:$\,\,$ $y$ is a term by Rule T1.

STEP $\,\,\,$2:$\,\,$ $c(y,y)$ is an atomic formula by Rule AF1 .

STEP $\,\,\,$3:$\,\,$ $c(y,y)$ is a formula by Rule F1 .

STEP $\,\,\,$4:$\,\,$ $\exists y\,c(y,y)$ is a formula by Rule F5.

STEP $\,\,\,$5:$\,\,$ $a$ is a term by Rule T1.

STEP $\,\,\,$6:$\,\,$ $x$ is a term by Rule T1.

STEP $\,\,\,$7:$\,\,$ $s(a,x)$ is an atomic formula by Rule AF1 and Steps 5,6.

STEP $\,\,\,$8:$\,\,$ $s(a,y)$ is an atomic formula by Rule AF1 and Steps 1,5.

STEP $\,\,\,$9:$\,\,$ $s(a,x)$ is a formula by Rule F1 and Step 7.

STEP 10:$\,\,$ $s(a,y)$ is a formula by Rule F1 and Step 8.

STEP 11:$\,\,$ $(s(a,x)\wedge s(a,y))$ is a formula by Rule F4 and  Steps 9,10.

STEP 12:$\,\,$ $(x=y)$ is an atomic formula by Rule AF2 and Steps 1,6.

STEP 13:$\,\,$ $(x=y)$ is a formula by Rule AF2 and Step 12.

STEP 14:$\,\,$ $(\sim x=y)$ is a formula by Rule F2.

STEP 15:$\,\,$ $z$ is a term by Rule T1.

STEP 16:$\,\,$ $(a=z)$ is an atomic formula by Rule AF1 and Steps
5,15.

STEP 17:$\,\,$ $(a=z)$ is a formula by Rule F1 and Step 16.

STEP 18:$\,\,$ $(\sim a=z)$ is a formula by Rule F2 and Step 17.

STEP 19:$\,\,$ $((\sim x=y)
\wedge(\sim a=z))$ is a formula by Rule F3 and Steps 14, 18.

STEP 20:$\,\,$ $((s(a,x)\wedge s(a,y))\wedge((\sim x =y)\wedge(\sim a=z)))$ 
is a formula\\
\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}
by Rule F3 and Steps 11,19.

STEP 21:$\,\,$ $c(z,x)$ is a atomic formula by Rule AF1 and Steps 6,15.

STEP 22:$\,\,$ $c(z,x)$ is a formula by Rule F1 and Step 21.

STEP 23:$\,\,$ $c(z,y)$ is an atomic formula by Rule AF1 and Steps 1,15.

STEP 24:$\,\,$ $c(z,y)$ is a formula by Rule F1 and Step 23.

STEP 25:$\,\,$ $(c(z,x)\wedge c(z,y))$ is a formula by Rule F3 and Steps 
23,24.

STEP 26:$\,\,$ $b$ is a term by Rule T1.

STEP 27:$\,\,$ $c(b,x)$ is an atomic formula by Rule AF1 and Step 26.

STEP 28:$\,\,$ $((c(z,x)\wedge c(z,y))\wedge c(b,x))$ is a formula by Rule F3
and Steps 25,27.

STEP 29:$\,\,$ $(((s(a,x)\wedge s(a,y))\wedge((\sim x =y)
\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$\\ 
\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaSTEP 17:$\,\,$}is a formula by Rule F3 and Steps 20,28.

STEP 30:$\,\,$ $\exists z\,(((s(a,x)\wedge s(a,y))\wedge((\sim x =y)
\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$\\ 
\phantom{aaaaaaaaaaaaaaaaaaaaaaaaSTEP 17:$\,\,$} is a formula by Rule F4 and Step 29.

STEP 31:$\,\,$ $\forall y\exists z\,(((s(a,x)\wedge s(a,y))\wedge
((\sim x =y)\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$\\ 
\phantom{aaaaaaaaaaaaaaaaaaaaaaaaSTEP 17:$\,\,$} is a formula by Rule F4 and Step 30.

STEP 32:$\,\,$ $\exists x\forall y\exists z\,(((s(a,x)\wedge s(a,y))
\wedge((\sim x =y)\wedge(\sim a=z)))
\wedge((c(z,x)$\\
\null$\qquad\qquad\quad \wedge c(z,y))\wedge c(b,x)))$
\phantom{aTEP$\,\,\,$} is a formula by Rule F4 and Step 31.

STEP 33:$\,\,$ $\exists y\,c(y,y)\vee
\exists x\forall y\exists z\,(((s(a,x)\wedge s(a,y))
\wedge((\sim x =y)\wedge(\sim a=z)))$\\
\null$\qquad\qquad\quad \wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$
\phantom{$\,\,$} is a formula by Rule F4 and Steps 4,32.



\vspace{0.1mm}

\end{minipage}
}

\end{center}

\bigbreak
\centerline{{\bf VARIABLES AND CONSTANTS}}

\medbreak\noindent As explained before, 
the basic constituents of terms are {\sl variables} and {\sl constants}.

{\sl Constants} are symbols that stand ---permanently or temporarily--- for one
{\sl fixed}, {\sl definite} thing.  
{\sl Variables} are letters such as $x$, $y$ that are meant to stand 
for things that only become definite when you assign a specific {\sl value}
to the variable.  

In principle, any symbol could be used for a constant, and also for a
variable. What matters is that in each case it should be clear which use is
intended, e.g. by {\sl declaring} a symbol to be a constant or a
variable. Usually, letters such as $x$, $y$, $z$, are used as variables for
real numbers, letters such as $k$, $m$, $n$, $p$ are used as variables for
natural numbers, and letters such as $X$, $Y$, $Z$ are used as variables
for sets.  So, for example, we would write formulas such as 
$$
(\forall x)(\exists y)(y>x)\,,
$$
$$
(\forall m)(\forall n)(\forall k)((k|m\wedge k|n)\Rightarrow k|m+n)\,,
$$
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,.
$$
But there is nothing sacred about this particular use.   We could, for
example, write 
$$
(\forall a)(\exists b)(b>a)\,,
$$
$$
(\forall x)(\forall y)(\forall z)((z|x\wedge z|y)\Rightarrow z|x+y)\,,
$$
$$
(\forall \alpha)(\forall \beta)(\forall \gamma)((\alpha\subseteq\beta\wedge
\beta\subseteq\gamma)\Rightarrow\alpha\subseteq\gamma)\,,
$$
provided that we specify that ``$a$, $b$ are real variables, 
$x$, $y$, $z$ are natural-number variables, $\alpha,\beta,\gamma$ are set
variables.'' 

One way to make it clear that something is a variable and to specify its
range is to put the range in the quantifier.  For example, if we write
$$
(\forall a\in\rr)(\exists b\in\rr)(b>a)\,,
$$
$$
(\forall x\in\nn)(\forall y\in\nn)(\forall z\in\nn)((z|x\wedge
z|y)\Rightarrow z|x+y)\,, 
$$
then it is clear that $a$, $b$ are real variables, 
and $x$, $y$, $z$ are natural-number variables.   

For our third sentence, involving sets, the situation is a little bit more
complicated. The obvious thing to do would be to write something like
$$
(\forall X\in{\cal S})(\forall Y\in{\cal S})(\forall Z\in{\cal S})
((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,
$$
where ${\cal S}$ is ``the set of all sets.''  This, however, is {\sl
forbidden}, because the ``set of all sets'' has to be avoided, due to the
Russell Paradox (to be discussed later in the course). 

So we can get around this by writing, instead, 
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,\,\,\,\,{\rm 
where\,\,\,}X,Y,Z\,\,\,{\rm are\,\,\,set\,\,\,variables},
$$
or
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,}
(X\subseteq Y\wedge Y\subseteq Z)\Rightarrow  X\subseteq Z\,.
$$

Naturally, we could even use more informal expressions such as 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,if}
\,\,\,X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z\,\,\,{\rm it\,\,\,follows\,\,\,that\,\,\,}
\Rightarrow  X\subseteq Z\,,
$$
or 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,and\,\,\,}
X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z,\,\,\,{\rm then\,\,\,}
\Rightarrow  X\subseteq Z\,.
$$

More complicated terms can be constructed from the basic constants 
and variables by means of (i)~function symbols, and (ii)~the 
``curly bracket'' set notation.  For example, the expression 
$x+3$ is a term, obtained by combining the variable $x$ and the constant 
$3$ by means of the binary (i.e. 2-argument) function symbol $+$.
The expression $\{x\in\rr:x^2>2\}$ is a term, that stands 
for a certain set of numbers.  {\bf A more detailed description of 
how function symbols work, and how to build more complicated terms 
and work with them, will be given later. For the time being, 
let us forget about function symbols until further notice, 
and concentrate on predicates, constants, variables,
and the logical connectives.}




{convincing argument} that establishes {\bf beyond the shadow of a 
doubt} that the statement is true.  

You should think of proving statements as a {\bf game} that you are playing
against an opponent who does not believe that the statement you are proving
is true.  You make statements, your opponent may find them convincing or
not, and it is up to you to clear up all your opponent's doubts by 
arguing your case persuasively every time your opponent isn't convinced. 
(Maybe you should think that I, the instructor, am the ``opponent.''
{\bf You} have to convince {\bf me}. If you convince me, then your proof
has worked. Otherwise it has not.)

The trouble with all this is that it is not very clear what will count as a
``convincing argument.''  Suppose your opponent says, after you have given
what looks to you like a very convincing proof: ``I am not convinced; I
don't see why what you just said proves anything.''  How would you deal
with something like that?

Moreover, any argument you give will have to {\it start} somewhere.  (You
cannot prove anything starting from nothing.)  What can you do if your
opponent refuses to accept your starting point?  

\bigbreak\noindent EXAMPLE: suppose you want to prove that, 

\medbreak
\bi
\item{} {\sl If a real number is greater than or equal to $4$, then 
the square root of the number is less than or equal to half the number.}

\medbreak\noindent Equivalently ---and more concisely--- you want to show that 

\medbreak
\item{} {\sl If $x$ is any real number such
that $x\ge 4$, then $\sqrt{x}\le{x\over 2}$.}
\ei

\medbreak

\noindent Or ---even more concisely--- you want to prove that 

$$
(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)\,.
$$

\noindent To prove this, you can proceed as follows: 

\noindent 
To prove $(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)$,
we pick an arbitrary $x$ and try to prove that 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$.

\noindent 
So, let $x$ be arbitrary.  We want to prove 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$. To prove this, we assume 
$x\ge 4$ and try to prove that $\sqrt{x}\le{x\over 2}$. 

\noindent 
Now we are assuming that $x\ge 4$.  From this it follows ---multiplying
both sides by $x$--- that $x^2\ge 4x$. Then ---dividing both sides by
$4$--- we get that ${x^2\over 4}\ge x$. Then ---taking the square root of
both sides--- we get $\sqrt{{x^2\over 4}}\ge\sqrt{x}$. Since
$\sqrt{{x^2\over 4}}={x\over 2}$, we have shown that 
${x\over 2}\ge \sqrt{x}$, which is ``the same as''
$\sqrt{x}\le{x\over 2}$. {\bf END OF PROOF}

\bigbreak\noindent {\bf Is this a truly convincing argument?}

\medbreak\noindent The answer is ``yes'' if all the steps are convincing. 
But {\bf if one step is not convincing, then the whole argument fails.}
(This is why you should {\bf not} expect to get partial credit for a wrong
proof if most steps are O.K. but just a few of them are wrong.  A proof is
like a chain. And {\bf a chain is no stronger than its weakest link.}  If a
16-Ton weight is hanging right above your head, held by a chain, and the
chain breaks because of one weak link, and the weight falls on your head
and kills you, do you think the chain manufacturer should be entitled to
``partial credit'' because ``most links were O.K.; only one broke''?  If
the driver of Lady Diana's car didn't drink every night, but only drank on
the particular night of the accident, should he be entitled to partial
credit because ``after all, most of the time he didn't drink''?)

\bigbreak\noindent So, {\bf are {\mymagtwobf ALL} the steps in the above 
proof truly convincing?}

\medbreak\noindent In the proof, we have given justifications for each
step. For example, at one point we went from 

$$x\ge 4$$
to 
$$x^2\ge 4x\,,$$
and we said that this was done 
``multiplying both sides by $x$.''   Now, why is this step permissible?

Suppose I tell you that ``I don't believe that you can do this''? 
How do you persuade me?   Well, let us give it a try. 
You can say: ``because there is a general rule that says 
if you take an inequality and multiply both sides by a number, then
you still get a valid inequality.''  (Or, in more symbolic notation:
$$
(\forall a)(\forall b)(\forall c)(a\ge b\Longrightarrow ac\ge bc)\,,
$$ 
using the language of a predicate calculus that we will discuss later.)  

This, however, will not do. This ``rule'' is NOT VALID.   
(For example, if $a=5$, $b=2$ and $c=-2$, then $a\ge b$, but 
it's not true that $ac\ge bc$, since $ac=-10$ and $bc=-4$, 
so $ac<bc$.)      So WE CANNOT USE IT.   

Now, here is a second try.  You can say: ``because there is a general rule
that says if you take an inequality and multiply both sides by a
nonnegative number, then you still get a valid inequality, that is,
using symbolic notation: 
$$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)\,.$$


It turns out that this time our rule IS VALID.  But notice that, in order
to apply this rule, we must be sure that $c\ge 0$. In our case, the number
that plays the role of $c$ is $x$, and it is indeed true that $x\ge 0$,
because $x\ge 4$.  (Since this is a crucial fact for us to be able to apply
the above ``rule,'' it would have been better to include in our proof a
little remark highlighting this. For example, instead of ``multiplying
both sides by $x$,'' it would have been better to say ``multiplying both
sides by $x$, which is possible because $x\ge 0$, since we are assuming
that $x\ge 4$.'')

But we haven't solved our problem.  Remember that your proof is really a
two-person game that you are playing against me, your opponent, in which
you have to convince me. What if I say ``How do we know 
that $(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)$?   I am not really convinced 
that this is true.''?

You could try to prove for me that 
$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b\wedge c\ge 0)\Longrightarrow 
ac\ge bc\Bigr)$  is indeed true.   But I am sure that in our proof you
will use other ``rules,''  and I, your opponent, could pick any of these 
and claim, once again, that I am not convinced of its validity.  

\medbreak
{\mymagtworm 

$\,$$\,$$\,$$\,$$\,$$\,$IF I WANTED TO, \ \ $\,\,$I COULD GO ON LIKE THIS 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$FOREVER. NO MATTER WHAT JUSTIFICATION 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$YOU $\,$$\,$GIVE $\,$$\,$FOR $\,$$\,$A $\,$$\,$STEP
$\,$$\,$IN $\,$$\,$YOUR $\,$$\,$PROOF, $\,$$\,$$\,$$\,$I 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$COULD ALWAYS ASK: $\,$$\,$$\,$$\,$``WHY IS THAT
VALID?''} 

\medbreak\noindent
In other words, if your opponent is truly stubborn and uncooperative, then
there is {\it nothing} you can do. Maybe you will come up with some reason,
but this reason will be based on some other facts that look obviously true
to you.  But your opponent can still refuse to accept these other facts,
and this process can go on for ever.

To take care of this ``problem of the opponent that doesn't play fair,'' 
mathematicians have come up with the concept of a {\bf formal proof}.
We give very precise rules defining what a ``formal proof'' is, 
and if you prove something correctly, and strictly following the rules, 
then you are {\it guaranteed} that your proof will be accepted as valid. 

If you do {\it not} obey the rules, then you are on uncertain ground: you
might be able to convince an instructor or grader, but you cannot be sure,
because any time you take a step one of us might ask ``how do you 
justify that step?''

You should think of giving formal proofs as similar to playing a game
such as chess: you are given very strict, precise rules, and you are only
allowed to make moves following the rules. As long as you obey the rules,
your opponent does not have the right to reject your moves. (If your
opponent unilaterally decides that a move of yours is not acceptable,
because ``you are not allowed to move a bishop if in the previous move you
moved a knight,'' then you would say that your opponent is not playing
fair, because this ``rule'' was not in the list of rules that had been
agreed upon before the game started.) 

{\bf In Mathematics we play fair.} We tell you what the rules are, and if
you prove something correctly following the rules, then you can be sure
that it will be accepted.  You may think that having to follow the rules is
a restriction to your freedom. But this is not so.  You should think of it
as {\bf your own protection against unfair play.}  If you correctly justify
every step according to the rules, then we promise you that we will not
change the rules in the middle of the game, kick the board, or come up with
other surprises.

\medbreak\noindent REMARK. I said that you should think of giving formal
proofs of theorems as a game similar to chess, in which you are given a
small number of very strict rules telling you which moves are permitted,
and each move has to be justifiable by explaining exactly which rule you
are using and how you are using it.  However, exactly as in the case of
chess, if all you know is the rules, and you always apply the rules
carefully and scrupulously, you may still lose all the time.  If you want
to {\bf win} (that is, in the case of a proof, if you want to get to the
desired conclusion) then you have to {\bf think}. You have to plan a
strategy, make a move now because it will be useful five or six steps
later, etc.

The one big difference between chess and theorem-proving is that {\bf chess
is a zero-sum game}, but {\bf theorem-proving is a cooperative game.} 
What this means is that in chess your opponent wants you to lose,
whereas in theorem-proving you and your opponent are working together,
trying to find out whether something is true, and to understand why.

That's why in chess you don't explain openly why you make a move.  You do
{\sl not} say to your opponent things like: ``I am making this move now so
that then in the next move I can do this other thing, so that three moves
later I can do that, and then I will checkmate you.'' {\sl You don't want your
opponent to be able to figure out your strategy.}

In theorem-proving you definitely {\bf do} want your opponent to understand
your strategy.  Your opponent is going to be ruthless in criticizing you if
you don't obey the rules, but as long as you obey them you will get full
coopertaion. In fact, usually you don't have to go through the whole
business of explaining every detail of every application of every rule.
{\sl As long as you persuade your opponent that something can be done using
the rules, there is no need to go through all the steps and actually do
it.}  And, since your opponent wants to be persuaded, the more you do to 
explain why you are doing something, the better. That's why {\sl in 
theorem-proving it's always nice to explain your strategy, so that your
opponent can understand where you are headed.}    

Because theorem-proving is a cooperative game, when writing proofs it is
always nice to use the first person plural ---the so-called ``editorial
we''--- as we did in our first example of a proof (``we pick,''\ \ ``we are
assuming,''\ \ ``we have shown'').  This is a good way to convey the idea
that the writer and the reader are {\sl working together}, that {\bf the writer
is leading the reader gently along a path of discovery.} \proofend

\vfill\eject

\centerline{{\mymagtwobf 2.\ VALID INFERENCES}} 

\bigbreak\noindent{\bf Proofs} involve {\bf steps} in which you move
from some statements (the {\it premisses}) to a new statement (the
conclusion).     We also say that you {\it infer}, or {\it derive}, or {\it
deduce} the conclusion from the premisses, and the steps where you 
do that are called {\it inferences}. 

So the first thing we need to discuss is:

\bigbreak
\centerline{{\mymagonebf WHAT IS A VALID INFERENCE?}}

\medbreak\noindent An {\bf inference} is a way of passing  from
certain statements (the {\bf premisses}) to another statement (the {\bf
conclusion}).   

\medbreak

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf VALID INFERENCES}}\\
(Informal definition)
\end{center}

A {\bf valid inference} is an inference for
which it is guaranteed  that, if the premisses are true, then the
conclusion is also true.
\end{minipage}

}

\end{center}

\medbreak\noindent For example:

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (1)}

\noindent\hbox to 7cm{\hfill Socrates is mortal}

\medbreak\noindent  is an example of a valid inference known as a {\it
syllogism}.  The premisses are 
``All men are mortal'' and ``Socrates is a man,''  and the conclusion is 
``Socrates is mortal.''

\medbreak\noindent  Roughly speaking, we would like to give the following 
definition of ``logically valid inference'': 

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf LOGICALLY VALID INFERENCES}}\\
(Informal definition)
\end{center}

An inference is {\bf logically valid} if the conclusion follows
from the premisses {\bf by virtue of the ``form'' of the statements
involved.}
\end{minipage}
}
\end{center}

\medbreak\noindent  Eventually, we will say exactly that. But we are not
ready to say it yet, because we still do not know precisely
what we mean by the ``form'' of a statement. 
   
Before we talk about ``form,'' let us go on with examples to illustrate 
what we are trying to say. We would like to 
call the syllogism of (1) logically valid {\bf because in (1) it does not
matter what ``Socrates'' or ``man'' or ``mortal'' mean.}  It would be
equally valid to say

\medbreak\noindent\hbox to 7cm{\hfill All men are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (2)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All pigs are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a pig}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (3)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Donald Duck is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (4)}

\noindent\hbox to 7cm{\hfill Donald Duck  is mortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All borogoves are mimsy}

\noindent\hbox to 7cm{\hfill The Mad Hatter is a borogove}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (5)}

\noindent\hbox to 7cm{\hfill The Mad Hatter is mimsy}

\medbreak\noindent It does not matter whether the premisses are true (in
some of the above examples they are true, in others they are not) or
whether the conclusion is true.  It doesn't even matter whether you know
what a ``borogove'' is or what ``mimsy'' means.     The point is:

\bigbreak

\bi
\item{} {\it The four examples of syllogism given above all are of the
same {\bf logical form}, namely,}
\ei

\medbreak\noindent\hbox to 7cm{\hfill All $X$ are $Y$}

\noindent\hbox to 7cm{\hfill $Z$ is an $X$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (6)}

\noindent\hbox to 7cm{\hfill $Z$ is a $Y$}

\medbreak\noindent To be able to talk about an inference being ``logically 
valid'' we have to be able to talk about the {\bf logical form} (or, even
better, ``a'' logical form, rather than ``the'' logical form) of an
inference. 

For example, we could say that all five inferences (1), (2), (3), (4), (5)
are of the form (6), and any reasoning of this form is valid.  

Another example of a valid inference form would be:

\medbreak\noindent\hbox to 7cm{\hfill $X$ is  $Z$}

\noindent\hbox to 7cm{\hfill $Y$ is $Z$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (7)}

\noindent\hbox to 7cm{\hfill $X$ and $Y$ are $Z$}

\medbreak\noindent Examples of inferences of the form (7) are:

\medbreak\noindent\hbox to 7cm{\hfill John is  happy}

\noindent\hbox to 7cm{\hfill Mary is happy}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (8)}

\noindent\hbox to 7cm{\hfill John and Mary are happy}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is smart}

\noindent\hbox to 7cm{\hfill Hillary Clinton is smart}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (9)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are smart}

\medbreak\noindent  So it would seem that we have two examples of ``valid
forms'', namely, (6) and (7).  Consider, however, the following
examples: 

\medbreak\noindent\hbox to 7cm{\hfill All men are equal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (10)}

\noindent\hbox to 7cm{\hfill Socrates is equal}

\medbreak\noindent and 

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is married}

\noindent\hbox to 7cm{\hfill Hillary Clinton is married}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (11)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are married}

\medbreak\noindent Superficially, (10) is ``of the form (6),'' and
(11) is ``of the form (7).'' Something is clearly wrong here, because
in (110) the conclusion is clearly false, even though the premisses are
true, and in (10) the conclusion is meaningless. 

What is wrong here is that ``statement forms'' such as 
``All $X$ are $Y$'' or ``$X$ and $Y$ are $Z$'' are not good enough 
to capture logical distinctions that matter.   For example, the statements
``Al Gore and Hillary Clinton are married'' and
``Al Gore and Hillary Clinton are smart'' both appear to be of the same
``form,'' namely, ``$X$ and $Y$ are $Z$,'' but they really are not. 
``Smart'' is a {\bf one-variable predicate}. (If $x$ is a person, you can
say that ``$x$ is smart.'')  But ``married'' is {\it both} 
a {\it one-variable predicate} (in ``Al Gore is married'') and a
2-variable predicate (in ``Al Gore and Hillary Clinton are married''). 
In normal English discourse, when we say that ``$x$ and $y$ are married''
we do not mean that each one, separately, is married. We mean that 
``$x$ and $y$ are married to each other.''  (If you don't believe that,
ask yourself the following: when someone asks a couple ``are you two 
married?'', isn't it clear that this means ``to each other''?)

So, if we really want to talk about {\bf forms} of statements and
inferences, then we need a more sophisticated language.  We have to be able
to distinguish, for example, between one-variable predicates and two-variable
predicates. The languages that do that are the {\bf first-order predicate
calculi}. (``Calculi'' is the plural of ``calculus.'') Before we discuss
these rather complicated languages, we will say a few words about 
{\bf propositional calculi}, which are simpler. Both propositional calculi 
and first-order predicate calculi are examples of {\bf formal languages}, 
so we will start by talking about formal languages in general. 

\sepa

\begin{center}

{\mymagtwobf 5.\ WHAT ARE PROPOSITIONAL CALCULI}

{\mymagtwobf GOOD FOR?}

\end{center}

\bigbreak\noindent So far, we have just given some rules for writing
certain weird strings of symbols and calling them 
``propositional forms.''  What do we want to with all that?

The answer is: {\bf we use propositional forms to represent 
forms of propositions or statements.}  

For example, the statement 

\medbreak
\bi
\item{(S1)}  Abraham Lincoln was a great president 
and all the serious historians who have studied his career think very highly 
of him.
\ei

\medbreak\noindent could be represented by the propositional form $P\wedge 
Q$, where the 
propositional variable $P$ stands for ``Abraham Lincoln was a great
president'' and the propositional variable $Q$ stands for ``all the serious
historians who have studied Abraham Lincoln's career think very highly of
him.''

\medbreak Notice that (S1) could also be represented in the form 
$P$, where the propositional variable $P$ stands for ``Abraham Lincoln was
a great president and all the serious historians who have studied his
career think very highly of him.''

So {\bf the representation of a statement by means of a propositional form
is not unique.} There may be more than one way to represent the statement,
depending on how deeply you want to go into the details its structure.  In
our example, ``$P$'' is a {\bf coarser} propositional-form representation
of (S1) than ``$P\wedge Q$'' (and ``$P\wedge Q$'' is {\bf finer} than ``$P$'').

As explained in the book, a propositional form has a {\bf truth table}.    
The truth table tells you how to assing a ``truth value'' (True or False) 
to a propositional form if you assign truth values to the propositional
variables.        

\medbreak\noindent A {\bf tautology} is a propositional form that has the 
value True for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent A {\bf contradiction} is a propositional form that has the 
value False for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent Often, we also apply the words ``tautology'' and 
``contradiction'' to statements in ordinary English. In that case, what we
mean by a ``tautology'' is ``a statement that has a propositional form
which is a tautology.''  Also, what we mean by a ``contradiction'' is ``a
statement that has a propositional form which is a contradiction.''

For example: the statement 

\medbreak

\bi
\item{(S2)}  Hector Sussmann is a great teacher.
\ei


\medbreak\noindent is true (of course!), but it is of the form $P$, which is
not a tautology, and it has no other, finer propositional form.  So 
(S2) is not a tautology. 

\medbreak On the other hand,  the statement 

\medbreak
\bi
\item{(S3)}  Either Hector Sussmann is a great teacher
or he is not.
\ei


\medbreak\noindent has the propositional form $P$, which is
not a tautology, but is also has the form $(P\vee(\sim P))$, which is a
tautology.   So (S3) {\bf is} a tautology.  

\medbreak\noindent{\bf Remark.} The example of (S3) shows that, for
something to be a tautology, it suffices that {\bf one} of its
propositional forms be a tautology.  In the case of statement (S3), we can
find at least three different propositional form for it, namely, $P$,
$(P\vee Q)$, and $(P\vee(\sim P))$, which are progressively finer, and only
the third one is a tautology. This, however, suffices to make (S3) a 
tautology. \proofend

Let us look at another example of a formal representation. Consider the
statement (in which we assume that $f$ is a fixed function described
earlier, and $x_0$ is some given number)

\medbreak
\bi
\item{(S4)} If $f$ has a relative minimum at $x_0$ and
$f$ is differentiable at $x_0$ then $f'(x_0)=0$.  
\ei

\medbreak\noindent This is of the form $P$  (taking $P$ to stand for the
whole statement), or of the form $P\imply Q$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$ and $f$ is differentiable at $x_0$''
and $Q$ to stand for ``$f'(x_0)=0$''), 
or of the form $(P\wedge Q)\imply R$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$,'' $Q$ to stand for 
``$f$ is differentiable at $x_0$,'' and $R$ to stand for ``$f'(x_0)=0$''). 
Once again, notice that the logical form is not unique.

\sepa

\begin{center}

{\mymagtwobf 6.\ SOME TYPES OF VALID INFERENCES}

\end{center}

\bigbreak\noindent Now that we know how to represent statements, we can
start giving examples of valid inferences of the propositional calculus.

\medbreak\noindent Assume that $f$ is a given function, and $n$ is a given
natural number. Then the inferences


\medbreak\noindent\hbox to 7cm{\hfill If $f$ is differentiable then 
$f$ is continuous.}

\noindent\hbox to 7cm{\hfill $f$ is differentiable.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (12)}

\noindent\hbox to 7cm{\hfill $f$ is continuous.}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill If $n$ is not prime
then $n$ is divisible by 3.}

\noindent\hbox to 7cm{\hfill $n$ is not a prime number.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (13)}

\noindent\hbox to 7cm{\hfill $n$ is divisible by 3.}

\medbreak\noindent are both of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (14)}

\noindent\hbox to 7cm{\hfill $Q$}

\medbreak\noindent This is a {\bf valid inference form.} Why is it valid?  
Because {\it if the premisses are true then the conclusion must be true}. 
Indeed: if $P\imply Q$ is true and $P$ is true then $Q$ is true, because
if $Q$ was false then $P\imply Q$ would be false (since $P$ is true) 
because the truth table for $\imply$ gives an $F$ to $P\imply Q$ if $P$ is
$T$ and $Q$ is $F$. 

\medbreak\noindent{\bf Remark.} Notice that in (13) the first premiss is 
usually not true.  (For example, it is not true if $n=25$, because 
$25$ is not a prime number and it's not divisible by $3$. So, if 
$n=25$, the first premiss says ``If $25$ is not a prime number
then $25$ is divisible by 3.'' This is of the form $P\imply Q$, where 
$P$ stands for ``$25$ is not a prime number'' and $Q$ for 
``$25$ is divisible by 3.'' Then $P$ is true and $Q$ is false, so $P\imply
Q$ is false.)

The inference form (14) is one of the most basic ones of all logic. It is
called {\bf modus ponens.}

What do we mean, in general, by a ``valid inference form''?

First of all, an {\bf inference form} is a list of propositional forms
(the premisses), followed by a horizontal line and then a propositional
form (the conclusion).  

A {\bf valid inference form} is an inference form such that the conclusion 
is true for all assignments of truth values to the propositional variables
that render the premisses true. 

Here is another example: the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $\sim Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim P$}

\medbreak\noindent is valid. Examples of inferences of this form are:


\medbreak\noindent\hbox to 7cm{\hfill If I was rich I would own a Mercedes.}

\noindent\hbox to 7cm{\hfill I don't own a Mercedes.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill I am not rich.}

\medbreak\noindent and


\medbreak\noindent\hbox to 7cm{\hfill If $x>3$ then $x>2\,$.}

\noindent\hbox to 7cm{\hfill $\sim x>2\,$.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim x>3\,$.}


\bigbreak
\centerline{\hbox to 5cm{\hrulefill}}


\vskip 0.7cm
\centerline{{\mymagtwobf 7.\ FIRST-ORDER PREDICATE CALCULI}}

\bigbreak\noindent We now turn to the formal languages that are really
important for this course, namely, {\mymagtworm first-order predicate
calculi.} Inferences such as (1) above cannot be justified 
by just looking at their propositional form. If we look at propositional 
form, then (1) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hfill $Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $R$}

\medbreak\noindent which is not a valid inference form of the propositional
calculus. In a first-order predicate calculus, (1) will have associated
to it a much {\bf finer} form, namely

\medbreak\noindent\hbox to 7cm{\hfill $(\forall x)(p(x)\imply q(x))$}

\noindent\hbox to 7cm{\hfill $p(a)$}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (16)}

\noindent\hbox to 7cm{\hfill $q(a)$}

\medbreak\noindent (In (1), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is mortal,'' and ``$a$'' for ``Socrates.'' )

On the other hand, {\it (10) is not if the form (16)}.  In fact, to
represent (10) correctly we need a binary predicate $E(x,y)$ (meaning ``$x$
and $y$ are equal to each other''), and then {\it (10) cannot even be
represented in the language.}  Using a 1-variable predicate $M(x)$ for
``$x$ is a man'', and the symbol $a$ for ``Socrates'', we can represent the
premisses as $(\forall x)(\forall y)((M(x)\wedge M(y))\imply E(x,y))$ and
$M(a)$.  But {\it there is no way to represent the conclusion}, which is
fine because the conclusion is meaningless. In other words: {\bf by trying
to represent an inference such as (10) in a first-order predicate calculus,
we discover that it cannot be done, and this explains why it is mistaken 
to think that (10) is valid.}


Similarly, (8) and (9) are of the form

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (17)}

\noindent\hbox to 7cm{\hfill $p(x)\wedge p(y)$}

\medbreak\noindent which is valid, but (11) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (18)}

\noindent\hbox to 7cm{\hfill $q(x,y)$}

\medbreak\noindent where $q$ is a {\it different predicate} from $p$.
(In fact, $q$ is a two-variable predicate, and $p$ is a one-variable
predicate. Naturally, $p$ and $q$ are
related, because $p(x)$ ---i.e. ``$x$ is married''--- is a way of saying
$(\exists y)q(x,y)$, i.e. ``$x$ is married to somebody,'' or ``there exists
a person that $x$ is married to,'' or ``there exists a $y$ such that $x$ is
married to $y$.'')

Once again, {\bf we looked at an inference that appeared to be valid but
really should not have been.  We tried to represent it in a first-order
predicate calculus, and we discovered {\mymagtwobf} why it is not valid.
the reason is that a word that appeared in it was being used with two
different meanings, namely, the one-variable predicate ``married'' and the
two-variable predicate ``married.''}  The attempt to formalize our
inference in a first-order predicate calculus  forced us to be precise
about what kind of predicate ``married'' is, and we discovered that it 
is really two different predicates.  This solved our problem. 

\bigbreak We now turn to a more precise description of 
{\bf PREDICATE CALCULI}.

\bigbreak
\centerline{{\sl (to be continued)}}

\newpage

\bigbreak

\centerline{\hbox to 6cm{\hrulefill}}

\sepa

\begin{center}

{\mymagtwobf ON THE USE OF LETTERS}

\end{center}

\medbreak\noindent Letter variables are only used in mathematics in order
to shorten certain statements that would otherwise be too long.

\bigbreak\noindent EXAMPLE 1:  Suppose you want to say 

\medbreak
\bi
\item{} If you take two real numbers, add them, and square the sum,
then you get the same result as if you had squared each number separately,
added the squares, and added to the result twice the product of the two
numbers.
\ei

\medbreak\noindent This looks awfully complicated and ugly.
You can make it much shorter by saying 

\medbreak
\bi
\item{} For all real numbers $x$, $y$, $(x+y)^2=x^2+y^2+2xy\,$.
\ei

\medbreak\noindent (Later, we, will learn to say this even more
concisely, by writing   

$$(\forall x\in\rr)(\forall y\in\rr)\Bigl((x+y)^2=x^2+y^2+2xy\Bigr)\,.$$

\medbreak\noindent  Here ``$(\forall x\in\rr)(\forall y\in\rr)$'' is an
abbreviation of ``for all real numbers $x$, $y$.'')

\bigbreak\noindent EXAMPLE 2:  Suppose you want to say 

\medbreak
\bi
\item{} It is possible to find a real number whose cube, added to its 
square, is equal to the number itself.  
\ei

\medbreak\noindent Again, this looks awfully complicated and ugly, and you 
can make it much shorter by saying 

\medbreak
\bi
\item{} There exists a real number $x$ such that $x^2+x^3=x\,,$
\ei

\noindent or 

\ \ \ \ \ \ \ $x^2+x^3=x$ for some real number $x$.  

\medbreak\noindent (Later, we, will learn to say this thing even more
concisely, by writing   

$$(\exists x\in\rr)\Bigl(x^2+x^3=x\Bigr)\,.$$

\medbreak\noindent  Here ``$(\exists x\in\rr)$'' is an
abbreviation of ``for some real number $x$.'')

\medbreak\noindent RECOMMENDED QUESTION FOR YOU TO THINK ABOUT: 
if I were to say ``There exists a real number such that $x^2+x^3=x$.''
This would {\bf not} be acceptable. Why? 

\bigbreak\noindent EXAMPLE 3:  suppose you want to say 

\medbreak
\bi
\item{} For each integer between 2 and 25, take that integer, 
square it, add 5, and compute the multiplicative inverse of the result. 
Having done this for all the integers between 2 and 25 (including 2 and
25 themselves), add all the resulting numbers. Then the number 
you get is smaller than 20.   
\ei

\medbreak\noindent Once again, this is awful, so you can shorten the whole
thing and just say 

\medbreak
\bi
\item{} Let $N=\sum_{k=2}^{25}{1\over k^2+5}\,.$  Then $N<20$. 
\ei

\medbreak\noindent Or, if you prefer, you could just say: 

\medbreak
\bi
\item{} $\sum_{k=2}^{25}{1\over k^2+5}<20\,.$  
\ei

\medbreak\noindent{\bf RECOMMENDED QUESTIONS FOR YOU TO THINK ABOUT:}
is it true or false that $\sum_{k=2}^{25}{1\over k^2+5}<20$?  How about 
$\sum_{k=2}^{25}{1\over k^2}<20$? And $\sum_{k=2}^{25}{1}<20$? And 
what would you say if I ask you whether it's true that 
$\sum_{n=2}^{25}{1\over k^2+5}<20$? 

\bigbreak\noindent EXAMPLE 4:  Suppose you want to say

\medbreak
\bi
\item{} If you take the function that to each real number assigns its 
cube, and integrate it from $0$ to $4$, then the result is 64. 
\ei

\medbreak\noindent Again, this can be made much shorter 
by saying 

\medbreak
\bi
\item{} Let $N=\int_{0}^4x^3\,dx\,.$  Then $N=64$.
\ei

\medbreak\noindent Or you could just say: 

\medbreak
\bi
\item{} Let $\int_{0}^4x^3\,dx=64\,.$ 
\ei

\medbreak\noindent{\bf RECOMMENDED QUESTIONS FOR YOU TO THINK ABOUT:}
is it true or false that $\int_{0}^4x^3\,dx=64$? What would you say about 
$\int_{0}^4y^3\,dx=64$? 


\bigbreak\noindent EXAMPLE 5:  Suppose I want to say:  

\medbreak
	
\bi
\item{} The set of all real numbers whose square is 
larger than the cube of the number has at least two different members.
\ei

\medbreak\noindent We can make this shorter by saying

\medbreak
\bi
\item{} Let $S=\{x\in\rr:x^2>x^3\}\,.$ Then $S$ has at least 
two members.   
\ei

\medbreak\noindent Or you could just say: 

\bi
\item{} Let $S=\{x\in\rr:x^2>x^3\}\,.$ Then $(\exists x)(\exists y)
\Bigl(x\in S, y\in S, {\rm \,\,and\,\,}x\neq y\Bigr)$.
\ei

\bigbreak\noindent EXAMPLE 6:  Suppose I want to say that 
``my age is larger than 20 and less than 70, and my brother is younger than I
am, but his age is larger than half my age.''

\medbreak\noindent Here too we can be more concise, and say

\medbreak
\bi
\item{} Let $x$ be my age and let $y$ be my brother's age. Then
$20<x<70$ and ${x\over 2}<y<x$.
\ei

\bigbreak\noindent In all these cases, we have used {\bf letters} 
(namely, $x,y,k,N,S$) as labels, so that we can avoid  writing clumsy phrases 
such as ``the number that we have been talking about.'' 


\bigbreak

\centerline{\hbox to 6cm{\hrulefill}}

\bigbreak

In all the above examples, the statements we wrote using letters only
contain the letters as {\bf dummy variables}.  The meaning of this is quite
simple:   the letter is just a label used to keep track of a number 
(or set, in Example 6) that occurs several times.   

The distinctive feature of a dummy variable is that you could have used any
other symbol instead, and you would be saying exactly the same thing.  
For example, if you write 

\medbreak
\bi
\item{} For all real numbers $x$, $y$, $(x+y)^2=x^2+y^2+2xy\,$, 
\ei

\mnin then that is exactly the same thing as you had written 

\medbreak
\bi
\item{} For all real numbers $a$, $b$, $(a+b)^2=a^2+b^2+2ab\,$,
\ei

\mnin or 

\medbreak
\bi
\item{} For all real numbers $\alpha$, $\beta$, $(\alpha+\beta)^2=
\alpha^2+\beta^2+2\alpha\beta\,,$
\ei

\mnin or 

\medbreak
\bi
\item{} For all real numbers $m$, $n$, $(m+n)^2=m^2+n^2+2mn\,$,
\ei

\mnin or 

\medbreak
\bi
\item{} For all real numbers $\diamond$, $\#$,
$(\diamond+\#)^2=\diamond^2+\#^2+2\diamond\#\,$. 
\ei

\bnin Similarly, instead of 

$(\exists x\in\rr)\Bigl(x^2+x^3=x\Bigr)\,$, 

\nin you could have said 

$(\exists z\in\rr)\Bigl(z^2+z^3=z\Bigr)\,,$

\nin or 

$(\exists a\in\rr)\Bigl(a^2+a^3=a\Bigr)\,,$

\nin or 

$(\exists \omega\in\rr)\Bigl(\omega^2+\omega^3=\omega\Bigr)\,,$

\nin or 

$(\exists \aleph\in\rr)\Bigl(\aleph^2+\aleph^3=\aleph\Bigr)\,,$

\nin or 

$(\exists \diamond\in\rr)\Bigl(\diamond^2+\diamond^3=\diamond\Bigr)\,.$

\bnin Just one more example:  

\mnin Instead of 

$$\sum_{k=2}^{25}{1\over k^2+5}<20\,,$$

\nin you could have said 

$$\sum_{n=2}^{25}{1\over n^2+5}<20\,,$$

\nin or 

$$\sum_{\rho=2}^{25}{1\over \rho^2+5}<20\,.$$

\bnin  So the ``dummy variables'' in the above examples are really just
{\bf internal tags}, using  to keep track of numbers {\bf within a particular
text}.  Outside this text, they variables do not have any meaning. For
example, after you have written 

$$\sum_{k=2}^{25}{1\over k^2+5}<20\,,$$  

\mnin it would not make sense for me to ask you ``what is the value of
$k$?''   The letter $k$ is just a tag to specify a {\it range of values},
as part of a set of instructions on how to compute a sum.  Any other letter
could be used instead, so in fact ``$k$ isn't really there.''   You can
persuade yourself of this in another way, by remembering that 

$$\sum_{k=2}^{25}{1\over k^2+5}<20$$  

\mnin is just another way of saying 

\medbreak
\bi
\item{} For each integer between 2 and 25, take that integer, 
square it, add 5, and compute the multiplicative inverse of the result. 
Having done this for all the integers between 2 and 25 (including 2 and
25 themselves), add all the resulting numbers. Then the number 
you get is smaller than 20.   
\ei

\mnin As you can see, there is no $k$ there!

\bnin  This way of speaking and writing, using dummy variables, 
is very useful, especially when what we
are trying to say is quite long.  But {\bf you have to be careful} when
writing this way, and have to make sure that you are using the letters
correctly.

\vskip 0.3cm


\bigskip
\centerline{{\largerbold THE FIRST FUNDAMENTAL RULE}}

{\largerbold
\mnin A letter symbol can only be used if either its value has been
declared before or it is a dummy variable.}


\bigskip
\nin For example: if you write 

$$x^2+x=90\,$$

\nin just like that, without any words before or after explaining 
what $x$ is, or making it a dummy variable, then what you wrote is {\bf
meaningless}.  We have no way to tell if what you wrote is true or false,
because we have no idea what you are talking about.     

If you write this, then I will write in red {\largebold what is $x$?}

For comparison, here are some perfectly meaningful things you could have
said:

\medbreak
\bi
\item[1.] Let $x=9$. Then $x^2+x=90$.  

\medbreak
\item[2.] Let $x=6$. Then $x^2+x=90$.  

\medbreak
\item[3.] Let $x$ be the number of planets in the solar system. 
Then $x^2+x=90$.  

\medbreak
\item[4.] $x^2+x=90$ for every real number $x$.

\medbreak

\item[5.] $x^2+x=90$ for some real number $x$.

\medbreak
\item[6.] $(\forall x\in\rr)(x^2+x=90)$.  

\medbreak
\item[7.] $(\exists x\in\rr)(x^2+x=90)$.  

\medbreak
\item[8.] The equation $x^2+x=90$ has a solution $x\in\rr$.
(Or ``the equation $x^2+x=90$ has a real solution.'')
\ei


\bnin Some of these are true, while the others are false. (Precisely:
1,3,5,7,8 are true; 2,4,6 are false. Moreover:  6 is just another way of
saying the same thing as 4, and 7 and 8 are just alternative ways of
stating 5.) 

Notice that in 1, 2 and 3 the letter $x$ has a value that was declared 
before.    In 4, 5, 6, 7 and 8 the letter $x$ is a dummy variable. 

Again for comparison, let us now write some things that do {\bf not} have a
precise meaning:

\medbreak
\bi
\item[A.] The equation $x^2=2$ has a solution.

\medbreak
\item[B.] The equation $x^2+1=0$ has a solution.

\medbreak
\item[C.] The equation $3x=5$ has a solution.
\ei


\bnin Why are these things bad?   Because they do not specify 
the {\bf range} of values within which the solution $x$ is to be sought. 
For example, A is true if we look for solutions $x\in\rr$, but false 
if we look for solutions $x\in\qq$.  B is true if we look for solutions 
$x\in{\bf C}$ (the set of complex numbers) but false if we look for
solutions $x\in\rr$.   C is true if we look for solutions $x\in\qq$
(or $x\in\rr$) but false if we look for solutions $x\in\zz$.   


On the other hand, the following are perfectly meaningful statements: 

\medbreak
\bi
\item{A1.} The equation $x^2=2$ has a solution $x\in\rr$.  (That is:
$(\exists x\in\rr)(x^2=2)$.) 

\medbreak
\item{A2.} The equation $x^2=2$ has a solution $x\in\qq$.  (That is:
$(\exists x\in\qq)(x^2=2)$.) 

\medbreak
\item{B1.} The equation $x^2+1=0$ has a solution $x\in\zz$.  (That is:
$(\exists x\in\zz)(x^2+1=0)$.) 

\medbreak
\item{B2.} The equation $x^2+1=0$ has a solution $x\in\rr$.  (That is:
$(\exists x\in\rr)(x^2+1=0)$.) 

\medbreak
\item{B3.} The equation $x^2+1=0$ has a solution $x\in{\bf C}$.  (That is:
$(\exists x\in{\bf C})(x^2+1=0)$.) 

\medbreak
\item{C1.} The equation $3x=5$ has a solution $x\in\zz$.  (That is:
$(\exists x\in\zz)(3x=5)$.) 

\medbreak
\item{C2.} The equation $3x=5$ has a solution $x\in\qq$.  (That is:
$(\exists x\in\qq)(3x=5)$.) 

\medbreak
\item{C3.} The equation $3x=5$ has a solution $x\in\rr$.  (That is:
$(\exists x\in\rr)(3x=5)$.) 

\medbreak
\item{C4.} The equation $3x=5$ has a solution $x\in{\bf C}$.  (That is:
$(\exists x\in{\bf C})(3x=5)$.) 
\ei

\bnin (Actually, A1, B3, C2, C3 and C4 are true, whereas 
A2, B1, B2 and C1 are false. Of course, in each case you could say the same
thing in lots of different ways.  For example, instead of 
C2 you could say 
``the equation $3x=5$ has a rational solution,''   or 
``the equation $3x=5$ can be solved in the set of rational numbers,''  
or ``there is a rational number that satisfies the equation $3x=5$,''
or ``$3x=5$ is solvable in the set of rational numbers.'')

\mnin A sentence whose meaning is imprecise, because it can be interpreted
in more than one way, is said to be {\it ambiguous}.   For example, the
sentences 

\bnin\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Flying planes can be dangerous

\mnin\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ College students are turning to
vegetables

\mnin\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ The University president promised
to stop drinking on campus

\bnin are ambiguous. Why? Think of other examples of ambiguous sentences.

\mnin Similarly, 
A, B, C above are ambiguous.  Another example of ambiguous sentence is a
sentence where a letter does not have a declared value and is not a dummy
variable either. For example, if I tell you to ``read chapter $x$,'' then
that's ambiguous ---unless I have told you before what ``$x$'' stands for,
or given you clues so you can find out, or perhaps I have 
said ``read Chapter $x$ for all $x$'' (that is ``read all the chapters'')
or ``read Chapter $x$ for some $x$'' (that is ``read one chapter'')--- 
because you don't know what $x$  is supposed to be.

\bigbreak

\centerline{{\largerbold THE SECOND FUNDAMENTAL RULE}}

\bigskip\nin {\largerbold Avoid ambigous sentences!} 

\vskip 0.3cm
\bigskip
\centerline{{\largerbold THE THIRD FUNDAMENTAL RULE}}

{\largebold\bnin Formulas without words are usually meaningless and should
be avoided  like the plague.}

It's the {\bf words} that glue the formulas together and give a structure and a
meaning to mathematical writing.  If you just write a bunch of formulas
then you probably do not know what you are saying, and if you don't know
then you should not expect that the instructor or the grader will know.

Look, for example,  at Statements 1, 2, 4 and 5 above. In all cases, 
the ``formula'' $x^2+x=90$ is the same, but the statements are quite
different and have totally different meanings.   That's because the words
are different!   

\bigskip
\nin The three ``fundamental rules'' can in fact be subsumed into a 
single principle: 

\vskip 0.6cm 

{\hugeee
\centerline{BE PRECISE!}}

\vskip 0.6cm 

\centerline{\hbox to 9cm{\hrulefill}}

\bigskip
\centerline{{\largerbold AN EXAMPLE}}

\medbreak\noindent Here is an example of the kind of thing that you should
{\it not} do when you write the solution of a homework problem.

\medbreak\noindent QUESTION: describe the set 
$\{x\in\rr:1<x^2<2\}$ as a union of intervals.

\medbreak\noindent A TRULY AWFUL SOLUTION ({\it the final answer 
is correct, so this would get some partial credit, say 40{\%}}):

\centerline{$1<x^2<2$}
\centerline{$1<x<\sqrt{2}$\ \ \ or\ \ \ $-\sqrt{2}<x<-1$}
\centerline{$A=(-\sqrt{2},-1)\cup(1,\sqrt{2})$}

\medbreak\noindent WHAT MAKES THIS BAD? Well, lots of things. For
example: 

\medbreak
\bi
\item{1.} To begin with, I do not see any {\it statements} here.
Remember that a {\it statement} is supposed to be something that is
either true or false.  When the student writes the inequality
``$1<x^2<2$,'' what is the
student asserting? What statement is the student making? (I know, you are
probably thinking that what the student is saying is that ``$x^2$ is
between $1$ and $2$.''  But that does not do me any good. I still do not
know what this means, because I have no idea what $x$ is supposed to be. Do
you?  And please don't tell me ``$x$ is supposed to be a real number.''
What good does that do to me? I still don't know whether $1<x^2<2$ is true
or false, unless you tell me {\it which} number it
is.) Is this true or is it false? If you cannot tell, it means
that there is no statement there.  (There is a standard convention 
used very often in
mathematics: when you say something involving variables, and you don't
specify the values of the variables or bound the variables with a
quantifier, it means that you are saying that your formula is true for {\it
all} values of the variables.  In this case: could the student be 
trying to say
that $1<x^2<2$ is true for {\it all} values of $x$?  I sure hope not!
Because if that was it, then it would be {\bf false}! For example,
$1<x^2<2$ is not true for $x=3$.)

\item{2.} The letters $x$ and $A$ are used, but we are never told what they
are supposed to stand for.  (I know: you will tell me that $A$ is the set
$\{x\in\rr:1<x^2<2\}$. But the student did not say that anywhere!)

\item{3.} Throughout this proposed ``solution,'' there are no words
explaining what is being done. And there is not a single punctuation mark!

\item{4.} There is nothing connecting each line to the next one. 
\ei


\medbreak\noindent
Now, here is an example of a good solution of the same problem. (The words in
italics are my own  comments. They are not part of the student's 
solution.)

\medbreak\noindent A GOOD SOLUTION ({\it which gets full credit}):

\noindent
Let $A=\{x\in\rr:1<x^2<2\}$. ({\it Comment: the student is announcing
what the letter $A$ is going to mean in this solution.}) 

\noindent
Let $x\in\rr$. ({\it Comment: the student is announcing
that $x$ is going to stand for an arbitrary real number.  This means that,
from now on, whatever assertion the student makes about $x$ should be true 
for {\bf all} real numbers $x$.  For example, if in the next line the
student said ``$1<x^2<2$,'' then that would be bad, because 
``$1<x^2<2$'' is {\bf not} true for all $x$.})

\noindent
Then  ``$x\in A$'' is equivalent to ``$1<x^2<2$.'' 
({\it Comment: the student is {\bf not} saying 
that  $x\in A$, or that $1<x^2<2$. He/she is just saying that those
two statements are equivalent, that is, that they are either both true or both
false. So the statement being made here is perfectly true.}) 

\noindent
And ``$1<x^2<2$'' is equivalent to ``either $1<x<\sqrt{2}$ or
$-\sqrt{2}<x<-1$.''
({\it Comment: again, a true statement.}) 

\noindent
Therefore ``$x\in A$'' is equivalent to ``either $x\in(1,\sqrt{2})$ or
$x\in(-\sqrt{2},-1)$.''  So, finally,

\medbreak
\centerline{$A=(1,\sqrt{2})\cup(-\sqrt{2},-1)$, and this is the desired 
answer.}

\medbreak\noindent
{\it GENERAL COMMENT: notice the use of the words ``so'', ``therefore,''
``then.''  These words give continuity to the whole thing. They tell you 
how each paragraph is related to the previous one. 
English is a very rich language. Take advantage of this to write well!
Notice also the nice touch of putting the word ``finally,'' to announce
that we are getting to the end.  Why not use ``then''
all the time? {\bf Because it's nicer when you don't repeat 
the same words over and over again.} It is not nice to  give the 
impression that you have a very poor vocabulary.}

\medbreak\noindent And here is another example of a
correct solution to the same problem.

\medbreak\noindent ANOTHER PERFECTLY FINE SOLUTION ({\it which also gets
full credit}):

\medbreak\noindent A real number belongs to the set under discussion if and
only if the square of this number is larger than one and less than two.
This will happen if either the number itself is strictly between one and
the square root of two ({\it Comment: why ``strictly''? Because I am saying
that it cannot be equal to one or square root of two.}), or its negative is
strictly between minus the square root of two and minus one.  So our set is
$(-\sqrt{2},-1)\cup(1,\sqrt{2})$.  {\it General comment: notice that the
student did everything with words, without using any letter variables. Is
that bad?  Not at all! There is nothing wrong
with using words rather than math symbols.  The reason that we usually
don't do it is that math symbols make it {\bf easier} to say things,
as we saw before.}

\medbreak\noindent{\bf A FINAL COMMENT.} The main point of
mathematical writing is not the use of weird symbols.  It is the use of
{\bf precise language.} You can be precise with words or with symbols.  But
symbols are just a way to say more concisely certain things that would take
too long to say with words.  As a rule, {\bf if you say something with 
mathematical symbols, then you ought to know  how to say exactly the same
thing using only words.}

\bigskip
\centerline{{\largerbold ANOTHER  EXAMPLE}}

\mnin
Let us look at the definition of ``interval,'' and let us compare 
some correct ways of stating it and some incorrect ones. 

\bigbreak\noindent The definition says: 

\medbreak
\bi	
\item{(D1)} an {\it interval} is a set $I$ of real numbers such that, 
whenever $a$, $b\in I$ and $c\in\rr$ are such that $a<c<b$, it follows
that $c\in I$.
\ei

\medbreak\noindent
Again, as in our previous examples, all this can be said 
without letters. For example, here is one possible way:

\medbreak\noindent
\bi
\item{(D2)} an {\it interval} is a set of real numbers with the property
that any time you take two elements of the set then every real number
lying between those two elements is also in the set.
\ei

\medbreak\noindent  It should be clear to you that (D2) and (D1) say 
exactly the same thing.  (If it isn't clear, then you should go on thinking
and rethinking this whole thing until it becomes clear.) 

\medbreak\centerline{\hbox to4cm{\hrulefill}}

\medbreak\noindent
Now let us look at some bad definitions of ``interval'':

\medbreak\noindent EXAMPLE B1: trying to define ``interval,'' a student
wrote:  

\medbreak\noindent
\bi
\item{(W1)} for a set to be an interval $I$, 
$(\forall a\in\rr),(\forall b\in\rr),(\forall c\in\rr)$ If $(a\in I),(b\in
I)$, there exists a $c$ such that $a\le c\le b$, If $c\in I$, then
the set $I$ is an interval.  
\ei

\medbreak\noindent (NOTE: {\it This a real example.  I actually took it from a
midterm exam written by a Math 300 student.}) Can you figure out what this
means?  Can you say what this means without letters?  I cannot figure it
out. When I grade, this definition of interval is worth $0$. (Can you make
sense of things like ``for all $c$ there exists a $c$''?  I cannot.)

\medbreak\centerline{\hbox to4cm{\hrulefill}}

\medbreak\noindent EXAMPLE B2: also trying to define ``interval,'' another
student writes:  

\medbreak\noindent
\bi
\item{(W2)} an {\it interval}, $I$, exists when real numbers $a$, $b$ 
are in $I$, and there is a real number $c$ in the {\it interval}
$I$ if and only if $a<c<b$.   
\ei

\medbreak\noindent Again, see if you can figure out what this means.
As before, ask yourself: how would you say this without letters?  I cannot
figure it out. When I grade, this definition of interval is also worth $0$.

\medbreak\centerline{\hbox to4cm{\hrulefill}}

\medbreak\noindent EXAMPLE B3: here is another answer:  

\medbreak\noindent
\bi
\item{(W3)} an {\it interval} $I$ is defined by the following rule
where for all real numbers $a$ and $b$ such that $a\in I$ and $b\in I$,
there exists a real number $c$ such that $a<c<b$.
\ei

\medbreak\noindent Again, let's see if we can figure out what this means.
How can we say this without letters?  Well, in this case I am afraid we
{\it can} say it.  (There are some very wrong things that the student 
is doing here, like for instance using the expression ``the following rule
where ...'' which is completely ungrammatical English. Perhaps ``namely''
would have been better than ``where''?  Also, the student 
never said that an interval is supposed to be a set of real numbers, so for
all we know the set of all apples could be an interval. 
But let's ignore all this.)

What this student is saying is this: {\sl for a set to be an interval, what
has to happen is that whenever you pick two members of this set then there
has to be a third number strictly between these two members.} (Why do I say
``strictly''?  Because the student wrote $a<c<b$, so $c$ cannot be equal to
one of the two numbers you picked.) 

We could also say what the student seems to be saying using the symbols 
that we introduced earlier. An interval, according to the student, is a set
$I$ such that 
$$(\forall a\in I)(\forall b\in I)(\exists c\in\rr)(a<c<b)\,.$$
(Look at this formula, think it over, and see whether or not it is
an adequate translation of what the student wrote.  If you {\it don't}
agree, tell us why and we can discuss this in class.)

Now, let us see what sets turn out to be ``intervals'' according to this
definition.  Suppose $I$ is a set of real numbers.  Suppose $I$ is not the
empty set. So $I$ has at least one member.  Now, nothing prevents me from
picking $a$ to be one of the members of $I$, and picking $b$ to be exactly
that very same element.  (The student said ``for all $a\in I$ and $b\in
I$'' so this certainly includes the possibility of taking $a$ and $b$ to be
equal.) But if I pick $a$ and $b$ this way, then of course there cannot
exist a $c$ such that $a<c<b$.  So the definition of interval is not
satisfied.  So: {\bf according to this student's definition of
``interval,'' the only set of real numbers that is a true interval is the
empty set!}

So this definition of ``interval'' also gets a $0$, but this time my
reason is different.  This time the definition does say something. But what
it says is wrong! (Probably, the student did not ask himself/herself 
the question: ``what am I saying here? What does this say in plain
English?''  You should always ask yourself that question!)

I am sure the student does not believe that the only set of real numbers
that is an interval is the empty set.  So what we have got here is not
failure to know, but {\bf failure to communicate}.  This needs serious
work, and it's you, the students, who have to do the work.  I cannot teach
you how to express your thoughts.  I can help you, but most of the work has
to be done by you.  And it will require {\bf hard work} on your part.

\begin{center}

{\large{\bf WRITING MATHEMATICS}}

\end{center}
%\phantom{9/19/94}}

\bigskip\noindent
You should write so that you can be easily understood by a properly trained
English-speaking individual. In particular, this means that you must 

\medbreak
{\parindent=1.71828\parindent

\bi
\item{$\bullet$} Use {\it complete English sentences}, that make 
clearly identifiable {\it statements}.  (For example, if you tell me that
``she is very smart,'' but you haven't told me who ``she'' is, then I don't
know what you are talking about, so you haven't made a statement.
Similarly, if you say ``$x>0$,'' but you haven't told me who ``$x$'' is,
then I don't know what you are talking about, so you haven't made a
statement.)

\item{$\bullet$} Avoid exaggerated or incorrect use of cryptic
mathematical notation. 

\item{$\bullet$} Explain what you are doing.

\item{$\bullet$} Make sure that letter ``variables'' are only used if
either: (i)~it has been said before what these letters stand for, or
(ii)~they are ``closed variables'' (or ``dummy variables,'' or ``bound
variables'') in the sense that will be discussed in detail in class, and 
also further below in this handout.

\item{$\bullet$} Provide proper connectives between equations as well as
between ideas. 

\item{$\bullet$} Make sure that all the rules of English grammar (including
those of spelling and punctuation) are strictly obeyed.  

\item{$\bullet$} Try to say things correctly,  following all the rules,
but {\it in your own words.} Please {\it no rote learning.} If you have to
memorize a definition or a statement, this is an indication of lack of
understanding.

\item{$\bullet$} {\bf Pay attention to presentation}:    

\bi

\item{{\bf -}} If you have to cross out
lots of words, then you should rewrite the whole thing anew, {\bf cleanly and
neatly}.  If you are not willing to spend some of your time doing this; if
what you hand in shows that you were in a hurry and that you did not make
the effort to write things neatly and properly, then there is no reason for
me or the grader to spend any of our time reading what you wrote, and we
will not do it.  

\item{{\bf -}} Make sure that {\bf your name appears in every sheet 
of paper you hand in}, and that if you are handing in more that one sheet
then the sheets are {\bf stapled} and the pages are {\bf numbered}.

\item{{\bf -}} {\bf Homework assignments that do not meet these
requirements will be marked ``unacceptable'' and returned unread.}

\ei\ei


}

\medbreak\noindent
Do not assume that the person reading the paper can read your mind. Do
assume that the person reading the paper is intelligent, but also assume
that the person reading the paper is busy, and cannot and will not spend an
excessive amount of time puzzling out your meaning.  Communication is
difficult, and written technical communication is close to an art.

{\bf Effective written exposition will be worth at least 50\% of your grade,
and, conversely, lousy exposition may be penalized as much as 50\%
of the grade or even more.}

Please proofread carefully what you hand in. Ideally, you should read and
reread and revise almost any formal communication.  {\bf Neatness and
clarity count}, as you well know if you've tried to read any complicated
document.

The best reference known to me on effective writing is {\sl
The Elements of Style} by Strunk and White, a very thin paperback
published by Macmillan. It isn't expensive, and it is easy to read.  I
recommend it.

\bigbreak



\sepa

\vspace{0.7cm}

\begin{center}
{\mymagtwobf 1.\ FIRST-ORDER PREDICATE CALCULI}

\end{center}

\medbreak\noindent We have been talking about ``proofs'' from the
beginning of the course, but we have not yet given a precise explanation of
what a proof is.   The goal of this discussion will be to show how the 
notion of ``proof'' can be made precise.

We ``prove'' statements. A ``proof'' will consist of writing 
several statements, one after the other, in such a way that every time 
we write a new statement we can fully justify putting  it there, by
invoking the {\bf rules of logic}.  

The rules of logic tell us when we can go from some statements (the
premisses) to another statement (the conclusion) just because of 
the {\bf form} if the statements involved, irrespective of the {\bf
meaning}, and even irrespective of whether we know what the statements
mean.  For example, suppose that I give you as ``axioms''
the following two statements:

\medbreak\noindent Axiom A: $\quad(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$, 

\medbreak\noindent Axiom B:.$\quad G(a,b)$.

\medbreak\noindent(An ``axiom'' is something that we accept as true. 
This means that in a proof you are always allowed to bring it in.) 

Suppose that I then write 

\begin{center}
(P)$\quad\quad$
\fbox{
\begin{minipage}{11cm}

\medbreak\noindent 1.$\quad(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$, 

\medbreak\noindent 2.$\quad G(a,b)$, 

\medbreak\noindent 3.$\quad G(a,b)\imply(a\diamond b)\oplus a=47$, 

\medbreak\noindent 4.$\quad(a\diamond b)\oplus a=47$.

\end{minipage}
}
\end{center}

\medbreak\noindent Then the four steps in (P) can be justified as
follows:

\bi
\item{} Step 1 is valid because $(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$ is an axiom.

\item{}  Step 2 is valid because $G(a,b)$ is an axiom. 

\item{}  Step 3 is valid because $G(a,b)\imply(a\diamond
b)\oplus a=47$ follows from the formula of Step 1 by ``specialization''
(that is, pluggin in particular values for the variables).

\item{}  Step 4 is valid because $(a\diamond b)\oplus a=47$
follows from the formulas of Steps 2 and 3 by ``Modus Ponens'' (that is, 
the rule that allows you to go to $Q$ whenever you have $P\imply Q$ and
$P$). 

\ei

\medbreak So $\quad$\fbox{{\bf (P) is a proof}}.

\medbreak Notice that we did not say anything about the meaning of the
symbols $\diamond$ and $\oplus$.  You probably don't know what they mean
and---in case you wonder---I haven't the faintest idea either. And neither
you nor I know what $a$ and $b$ are. But {\bf it does not matter.} 
The validity of (P) as a proof depends only on the {\bf form} of the
sentences involved, and you don't need to know what they mean.  
(We already discussed examples of this situation earlier. I you know that 
``all borogoves are mimsy,'' and ``Mr. T is a borogove,'' then 
you can conclude that ``Mr. T is mimsy.'' It doesn't matter what ``mimsy''
means, or what a ``borogove'' is. (Nor does it matter who Mr. T is.) 

\bigbreak To make the notion of {\sl proof} precise we have to start
by making the notion of {\sl (logical) form} precise. The things that have a 
logical form are {\sl statements}, so we have to start by giving a precise 
definition of {\sl statement}.  

At this point we have a choice: we can either

\bi
\item{} try to define precisely what a ``statement'' in the English
language is, 
\ei
or
\bi
\item{} create a language of our own, for which it will be easier to 
define precisely what a ``statement'' is.
\ei
The first option is basically hopeless. We have already discussed, eariler
in the course, the notion of statement. We said things like:

\begin{center}
\begin{minipage}{11cm}
A {\bf sentence} is a complete unit of thought. 

Sentences can be declarative, interrogative, exclamatory or imperative.

A {\bf statement} is a declarative sentence. 

A {\bf statement} is a declarative sentence that has a definite truth value. 
\end{minipage}
\end{center}

\medbreak\noindent Those are impressive words, but {\bf do they really add
up to a {\large truly precise} definition of ``statement''?}

For example: {\bf would it be possible to write a computer program to check
whether a string of English words is a statement?}  How would you tell the
computer to check whether a string of words is a statement, using the above
explanation?  

In case you believe that this can be done, here are some observations 
that ought to change your mind.  First of all, ask yourself whether you
would accept 

\begin{center}
\begin{minipage}{11cm}
This book are very interesting.
\end{minipage}
\end{center}

\medbreak\noindent as a statement. I am sure your answer is ``no,'' because
``this book are very interesting'' is {\sl grammatically incorrect.}  The
verb (``are'') ought to agree in number with the subject (``this book'')
but in this example it does not. (Had we said ``this book is very
interesting,'' or ``these books are very interesting,'' then we would have
had real statements.)

So you probably agree by now that, if our computer program is going to
check whether or not something is a statement, then it should first be 
able to check if it is grammatically correct.  

And we have opened a can of worms! First of all, the notion of {\sl
grammatical correctness} is itself fuzzy. (Are split infinitives O.K.?) 
Second, ask yourself whether you 
would accept 

\begin{center}
(S)$\quad\quad\quad\quad$
\begin{minipage}{10cm}
I don't know if George Eliot was a good writer because I have not read any
of her books.
\end{minipage}
\end{center}

\medbreak\noindent as a statement. Yes or no? ``George'' is a man's name,
so talking about ``her'' books seems to be grammatically wrong. However, if
you know that {\bf George Eliot was a woman}, then you realize that there
is no problem.

So, for our imaginary computer program to be able to tell whether (S) is a
statement, it would have to know, for example, that George Eliot was a
woman. This kind of knowledge goes far beyond knowing a few rules, and
requires a sophisticated and extensive knowledge of the full context in
which our language operates, of history and of literature.

\medbreak In any case, if you are not convinced that one cannot give 
a precise definition of ``statement'' in the English language---a
definition so precise that one can actually translate it into a 
computer program for checking whather something is a statement---then try to
come up with one.      I doubt that you will be able to. 

\medbreak English is an example of a {\bf natural language}, that is, one
of that many languages that evolved in different parts of the world over
thousands of years. Nobody directed this evolution with the idea of making
those languages amenable to the precise study that is required in logic. 

So mathematicians and logicians have chosen a more realistic solution: 
we create our own languages, and do it in such a way that we can 
define in a completely precise way the notions that interest us, such as 
``statement'' and ``proof.'' 

The situation here is very much the same as with computer programming.  Ask
yourself: why is it that when I want to write a computer program I have to
use a special programming language? Why can't I just write programs in
ordinary English?  The answer is that ordinary English is too fuzzy, too
imprecise for a computer to be able to deal with it and understand it.

\sepa

The {\bf formal languages} that we will be using here are 
{\bf predicate calculi}. (``Calculi'' is the plural of
``calculus.'') In these languages, 
we have letters such as $x$, $y$, $a$, $b$, etc., that stand for 
specfic ``things,'' or are ``variables'' (that is, slots where you can plug
in things). These letters can be combined by means of {\bf function
symbols} to write {\bf terms}, which are more complicated expressions 
that also stand for things.  (For example, $6.(3+5)$ is a term.)
Finally, we will have {\bf predicate symbols} that will enable 
us to make assertions about the terms.   (For example, using the predicate 
symbol $>$ we can combine the terms $6.(3+5)$ and $7$ and make the 
assertion that $6.(3+5)>7$.) The simplest such assertions are 
the {\bf atomic formulas}. These can be combined by means of {\bf logical
connectives} to build even more complicated formulas. (For example, 
$6.(3+5)>7$ and $7>5$ are atomic formulas. We can combine them to get 
the formula $6.(3+5)>7\wedge 7>5$, which is no longer atomic. The symbol
$\wedge$ is therefore used to {\sl connect} two formulas, which is why it
is called a {\sl connective.} Another example:  $x=3+z$ and $y<5$ are
atomic formulas, and we can combine them into 
$(\forall x)(x=3+z\wedge y<5)$, which is a formula obtained in two steps,
by first forming  $x=3+z\wedge y<5$---that is, using the logical connective
$\wedge$---and then forming  $(\forall x)(x=3+z\wedge y<5)$---that is, 
using the logical connective $\forall$, known as the {\bf universal
quantifier}---to produce a {\bf quantified version} of 
$x=3+z\wedge y<5$.)

To make all this precise, we will run into a lot of trouble because of the
parentheses. To see why, look at the formula 
\begin{equation}\label{flaab}
(\forall x)(x=3+z\wedge y<5)\,\,.
\end{equation}
Is (\ref{flaa}) obtained by 

\bi
\item[(A)] ``quantifying'' the formula 
\fbox{$x=3+z\wedge y<5$}$\,$?  
\ei
or by
\bi
\item[(B)] ``quantifying'' the formula \fbox{$(x=3+z\wedge y<5)$}$\,$?
\ei

Suppose we choose the possibility (A).  This would mean that we are
counting the expression $x=3+z\wedge y<5$ as a formula, and surrounding it
with parentheses is part of the quantification process.  So the 
``universal quantification rule'' would have to say:

\begin{center}
(UQR-Version A)$\quad$
\fbox{
\begin{minipage}{9.5cm}
If $F$ is a formula and $v$ is a variable then 
$(\forall v)(F)$ is a formula.  
\end{minipage} 
}
\end{center} 
This would have some awful consequences.  Just to give you an example, let
us look at what would happen when you want to quantify two or three times.
Suppose, for example, you want to state the associative property of
addition of real numbers, which says that ``for all $x$, $y$, $z$, the
equality $(x+y)+z=x+(y+z)$ holds.''  When you quantify, each time you add a
new quantifier you would have to surround the formula being quantified
with parentheses.  So you would get 
$(\forall z)((x+y)+z=x+(y+z))$ in the first step, then 
$(\forall y)((\forall z)((x+y)+z=x+(y+z)))$ in the second step,
and after the third and last step 
you would end up with the ugly expression
$$
(\forall x)((\forall y)((\forall z)((x+y)+z=x+(y+z))))\,\,.
$$

So it seems that choosing (A), and using  (UQR-Version A) for the 
universal quantification ruleis not a very good idea. So let try to follow
(B) instead. This means that we need the rule

\begin{center}
(UQR-Version B)$\quad$
\fbox{
\begin{minipage}{9.5cm}
If $F$ is a formula and $v$ is a variable then 
$(\forall v)F$ is a formula,  
\end{minipage} 
}
\end{center} 
which could then be applied, for example, to the 
formula $((x+y)+z=x+(y+z))$. Now we will get 
$(\forall z)((x+y)+z=x+(y+z))$ in the first step, then 
$(\forall y)(\forall z)((x+y)+z=x+(y+z))$ in the second step,
and after the third and last step 
we would end up with the much nicer expression
$$
(\forall x)(\forall y)(\forall z)((x+y)+z=x+(y+z))\,\,.
$$
This looks fine, but we have to pay a price, namely, we have to call
$((x+y)+z=x+(y+z))$ a formula but we are not allowed to call
$(x+y)+z=x+(y+z)$ a formula.   This is also quite unpleasant. 

We are going to take care of this problem by insisting that the first time
we quantify we have to put parentheses, but then when we quantify again, we
do not.  This will require that we first define the notion of a
``quantified version'' of a formula before we go on.  


\sepa

We now do the (boring) details and define 
precisely what a {\bf predicate calculus} is.


First of all, a predicate calculus involves a list of 
{\bf basic symbols} (or ``primitive symbols''). 

\begin{center}
\fbox{
\begin{minipage}{11cm}
\begin{center}
{\large{\bf THE BASIC SYMBOLS\\
OF A FIRST-ORDER PREDICATE CALCULUS}}
\end{center} 
A first-order predicate calculus uses {\bf five} kinds of basic symbols: 
\bi
\item[1.] individual symbols\footnote{WARNING: the phrase ``individual
symbol'' is not meant to refer to ``a signle symbol,'' as in the 
statement ``if you only look at the individual symbols then you might not 
understand the words.''  ``Individual symbol'' here means 
``symbol for an individual,'' in the same way as further down 
we use ``predicate symbol'' in the sense of ``symbol for a predicate.'' 
In other words, ``individual'' here is a {\bf noun}, not 
an {\bf adjective.}}$\,\,$;
\vspace{-2.5mm}
\item[2.] function symbols$\,\,$;
\vspace{-2.5mm}
\item[3.] predicate symbols$\,\,$;
\vspace{-2.5mm}
\item[4.] the logical connectives $\forall$, $\exists$, $\wedge$, $\vee$,
$\Rightarrow$, $\Longleftrightarrow$, $\sim$$\,\,$;
\vspace{-2.5mm}
\item[5.] the left and right parentheses, ``$($'' and ``$)$,'' and the
comma ``,''$\,\,$.  
\ei
\end{minipage} 
}
\end{center} 

\medbreak\noindent {\bf INDIVIDUAL SYMBOLS} are usually letters such as
$x$, $y$, $z$, $u$, $r$, $a$, $b$, $c$, $n$, $p$, $q$, 
$\ldots$. In principle, you can use {\sl any symbol} to be an individual,
even if it is not a Roman letter. For example, you can use Greek
letters ($\alpha$, $\beta$, $\varepsilon$, $\pi$, etc.), Hebrew letters
(e.~g. $\aleph$), other symbols such as 
$\diamondsuit$, $\clubsuit$, or $\#$, 
and even longer and more complicated expressions such as
$GNU$, $grrr$,  $oink$, $my$-$uncle$-$Jimmy$, etc. 

An individual symbol is either a {\it constant} or a {\it variable}, and
can never be both at the same time.  Whether a symbol is a {\it constant}
or a {\it variable} depends on the context. There are various ways to {\it
declare} something to be a constant or a variable:

\bi
\item[1.] If within a proof you say ``Let $\#$ be $\ldots$''  then you are
declaring  $\#$ to be a constant until the end of the proof. (For example: 
once you say ``Let $a=3$'' then from that moment on $a$ is a constant. 
If you say ``Let $\varepsilon$ be arbitrary'' then from then on 
$\varepsilon$ is a constant. (The same applies to ``choose'' or ``pick.'') 

\item[2.] Any symbol that appears right next to a quantifier 
is a variable. For example, if you see the formula $(\forall x)(\exists
y)(b(x,y))$, then you know that $x$ and $y$ are variables. 
If I write $(\forall x)(x^2\ge 0)$, then 
$x$ {\sl has} to be a variable, because only variables can appear in
quantifiers such as $(\forall x)(x^2\ge 0)$. 

\item[3.] Some symbols are {\it reserved constants}, that is, declared to be
constants once and for all, and intended to have a specific value that is
fixed for ever. For example, in real-number
arithmetic the symbols $0$, $1$, $2$, $\pi$, $e$, 
are fixed constants. In set theory, the symbol $\emptyset$ is a fixed
constant. 
\ei

\medbreak\noindent At each point, it is important to know  whether a symbol
is a constant or a variable, because this helps you know what you can and
what you cannot do with it.  

\medbreak For example, after you have declared
``$a$'' to be a constant by saying, for example, ``let $a$ be an arbitrary
real number,'' {\sl you cannot use quantifiers such as $(\forall a)$},
because $a$ is now fixed, not variable.

\medbreak Similarly, if a symbol is a variable, then there 
are things you cannot do with it.  For example, after 
you have said in a proof $(\forall x)(\forall y)(x+y=y+x)$, 
you cannot write in your proof something like $x+y=y+x$, 
because {\sl this is not a statement}.  (This means that, if $x$ and $y$
are variables, then ``$x+y=y+x$'' does not have  
a definite truth value. It doesn't make sense to say that 
$x+y=y+x$'' is true or false unless $x$ and $y$ are given specific 
values.    And, if you give $x$ and $y$ specific, definite values, then
they become constants.) 

\medbreak\noindent{\bf FUNCTION SYMBOLS} are letters such as $f$, $g$,
$F$, $G$, or other symbols such as $+$, $\cdot$, $\cap$, $\cup$,
$\in $, $\sim$. Each function symbol admits a {\sl fixed} number of arguments.
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} symbol. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} symbol.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} symbol. And so on.)   If you write 
a function symbol, followed by a ``$($,'' then as many arguments
as the function symbol admits, separated by commas, and then 
``$)$,'' you get a {\sl term}.  The strings that can be used 
as arguments are variables, or constants, or more general terms.

In other words: 

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``TERM''}}
\end{center}

\medbreak\noindent RULE T1: constants and variables are terms;

\medbreak\noindent RULE T2: if $t_1,\ldots,t_k$ are terms 
and $F$ is a $k$-variable function symbol, then $F(t_1,\ldots,t_k)$ is a
term. (For example: if $F$ is a one-variable function symbol and $t$ is a
term, then $F(t)$ is a term; if $F$ is a two-variable function symbol and
$t$, $s$ are terms, then $F(t,s)$ is a term; 
if $F$ is a three-variable function symbol and
$t$, $s$, $u$ are terms, then $F(t,s,u)$ is a term.) 

\medbreak\noindent RULE T3: only those expressions obtained by repeated
application of Rules T1 and T2 are terms. 
\vspace{2mm}
\end{minipage}
}
\end{center}

\medbreak\noindent Later, when we discuss how to {\it interpret} 
(that is, assign a meaning), function symbols will correspond to {\it
functions}, that is, expressions that have a {\bf definite value}
once you plug in values for its arguments.  For example, if we 
use a binary function symbol $plus$ to stand for ``sum of two numbers,''
then $plus(3,5)$ will have the value $8$, $plus(x,3)$ will have a value 
for each value of $x$ (so $plus(x,3)$ will be $7$ if $x$ is $4$, 
$2$ if $x$ is $-1$, and so on) and $plus(x,y)$ will have a value 
for each choice of values of $x$ and $y$ 
(so $plus(x,y)$ will be $7$ if $x$ is $4$ and $y$ is $3$,  
$2$ if $x$ is $-4$ and $y$ is $6$, and so on). 

\medbreak\noindent {\bf EXAMPLE 1.} Let us start setting  up a 
predicate calculus to talk about {\bf real number arithmetic}.  We
introduce the two-variable function symbols $plus$, $times$, $minus$,
$max$, $min$, and two one-variable function symbols $negative$-$of$ and
$abs$. (``$max$'' is meant to stand for ``the largest of,'' so for example
$max(5,3)=5$ and $max(-7,2)=2$. Similarly,
``$min$'' is meant to stand for ``the smallest of,'' so for example
$min(5,3)=3$ and $min(-7,2)=-7$.)

In addition, we will have {\bf reserved constants}
for the usual names of the nonnegative integers,
in decimal notation (for example: $0$, $1$, $2$, $3$, $4$, etc). 

Then the following are terms:

\bi
\item[1.] $plus(x,y)$, 

\item[2.] $minus(plus(x,y),z)$, 

\item[3.] $abs(x)$, 

\item[4.] $plus(min(x,1),abs(plus(x,y)))$, 

\item[5.] $min(plus(x,y),abs(y))$, 

\item[6.] $min(plus(x,abs(y)),negative$-$of(y))$, 

\item[7.] $plus(max(x,27),min(plus(x,y),abs(y)))$, 

\item[8.] $plus(abs(min(plus(x,y),abs(y))),max(plus(x,y),abs(y)))$.
\ei

\medbreak\noindent It is common in mathematics to use $+$ rather than
$plus$, and write $t+s$ rather than $plus(t,s)$.  Similarly, one uses  
$-$ rather than $minus$ and $.$ rather than $times$,  
so we write $t-s$ rather than $minus(t,s)$ and 
$t.s$ rather than $times(t,s)$.  Also, for the one-variable symbol
$abs$, we use $|\ldots|$ instead, and write $|x|$ instead of 
$abs(x)$, and for the one-variable symbol
$negative$-$of$, we use $-$ instead, and write $-x$ instead of 
$negative$-$of(x)$. 

\begin{center}
\fbox{
\begin{minipage}{11cm}
{\bf The same symbol $-$ is used 
for two different things: the one-variable function $negative$-$of$, 
and the two-variable function $minus$.  For example, when we write
$5-8=-3$, the first ``$-$'' is the two-variable function symbol, and the
second ``$-$'' is the one-variable function symbol.} 
\end{minipage}
}
\end{center}

With these conventions, we would write

\medbreak

\pltw{1}{$x+y$}{$plus(x,y)$,} 

\pltw{2}{$(x+y)-z$}{$minus(plus(x,y),z)$,} 

\pltw{3}{$|x|$}{$abs(x)$,} 

\pltw{4}{$min(x,1)+|x+y|$}{$plus(min(x,1),abs(plus(x,y)))$,}

\pltw{5}{$min(x+y,|y|)$}{$min(plus(x,y),abs(y))$,}

\pltw{6}{$min(x+|y|,-y)$}{$min(plus(x,abs(y)),negative$-$of(y))$,} 

\pltw{7}{$max(x,27)+min(x+y,|y|)$}{$plus(max(x,27),min(plus(x,y),abs(y)))$,}

\pltw{8}{$|min(x+y,|y|)|+max(x+y,|y|))$}{$\,\,$}

$plus(abs(min(plus(x,y),abs(y))),min(plus(x,y),abs(y)))$.

\medbreak\noindent If you want to, you could get the terms to look even nicer
by also agreeing to use $\wedge$ rather than $min$ and $\vee$ rather than
$max$, so you would write $t\wedge s$ instead of $min(t,s)$ and $t\vee s$
instead of $max(t,s)$.  With these new symbols, one would write

\pltw{a}{$(x\wedge 1)+|x+y|$}{$plus(min(x,1),abs(plus(x,y)))$,}

\pltw{b}{$(x+y)\wedge |y|)$}{$min(plus(x,y),abs(y))$,}

\pltw{c}{$(x\vee 27)+((x+y)\wedge |y|)$}{$plus(max(x,27),min(plus(x,y),abs(y)))$,}

\pltw{d}{$|((x+y)\wedge|y|)|+((x+y)\vee|y|)$}{$\,\,$}

$plus(abs(min(plus(x,y),abs(y))),min(plus(x,y),abs(y)))$.


\medbreak\noindent {\bf EXAMPLE 2.} Let us start setting  up a 
predicate calculus to talk about {\bf natural number arithmetic}.  We
introduce the two-variable function symbols $plus$, $times$, 
$max$, $min$, $GCD$, $LCM$. (``$GCD$'' and ``$LCM$'' are meant to stand for
``greatest common divisor'' and ``least common multiple,'' so for example
$GCD(12,20)=4$ and $LCM(12,20)=60$.) 

In addition, we will have {\bf reserved constants} for the usual names of
the natural numbers, in decimal notation (for example: $1$, $2$, $3$, $4$,
etc). ({\sl Remember that $0$ is {\bf not} a natural number}.)

Then the following are terms:

\bi
\item[1.] $plus(x,y)$, 

\item[2.] $GCD(plus(x,5),z)$, 

\item[3.] $plus(min(x,1),LCM(plus(x,y),83)))$. 
\ei

\medbreak\noindent As in Example 1, we use $+$ rather than
$plus$, and write $t+s$ rather than $plus(t,s)$.  Similarly, we use  
$.$ rather than $times$,  
so we write $t.s$ rather than $times(t,s)$.  

With these conventions, we would write

\medbreak

\pltw{1}{$x+y$}{$plus(x,y)$,} 

\pltw{2}{$GCD(x+5,z)$}{$GCD(plus(x,5),z)$,}

\pltw{3}{$min(x,1)+LCM(x+y,83)$}{$plus(min(x,1),LCM(plus(x,y),83)))$.}

\medbreak\noindent You could get the terms to look even nicer
by also agreeing to use $\wedge$ rather than $min$ and $\vee$ rather than
$max$, so you would write $t\wedge s$ instead of $min(t,s)$ and $t\vee s$
instead of $max(t,s)$.  With these new symbols, you could write, for
example,

\medbreak

$x\wedge 1+LCM(x+y,83)\quad$ insteand of $\quad
plus(min(x,1),LCM(plus(x,y),83)))$.

\bigbreak\noindent {\bf EXAMPLE 3.} Let us start setting  up a 
predicate calculus to talk about {\bf kinship}.  We introduce two 
one-variable function symbols $f$, $m$, that will be thought of as standing
for ``father of'' and ``mother of.'' Suppose we have constants such as 
$Bill$, $John$ and $Mary$, and variables $x$, $y$, $z$. 
Then, for example, the following are terms:

\bi
\item[1.] $f(x)$, 

\item[2.] $m(y)$, 

\item[3.] $f(m(Bill))$, 

\item[4.] $f(m(x))$,

\item[5.] $f(f(f(f(f(f(Mary))))))$, 

\item[6.] $m(m(m(m(m(m(Mary))))))$, 

\item[7.] $m(f(m(f(f(m(Mary))))))$.
\ei

\medbreak\noindent The term $f(Bill)$ would have a definite value, namely, 
Bill's father.  The term $f(x)$ does not have a definite value 
unless you are told what the value of $x$ is.  If $x$ is Bill then 
the value of $f(x)$ is Bill's father. If $x$ is Mary then 
the value of $f(x)$ is Mary's father. 

\medbreak\noindent QUESTION: could we introduce function symbols to stand 
for ``son,'' ``daughter,'' ``brother'' or ``sister''?  The answer is {\bf
NO!} Why? Because {\bf ``son,'' ``daughter,'' ``brother'' and ``sister''
are not functions.} They do not have a definite value for each value of the
argument. For example,  $son(x)$ might fail to have a value 
if $x$ is Bill, because it could happen that Bill does not have a son, and
if this is the case then you don't know what $son(Bill)$ means.
Also, $son(x)$ might fail to have a {\it definite} value 
if $x$ is Bill, because it could happen that Bill has more than one son,
and then once again you would not know what $son(Bill)$ means.

As we will see later, functions that sometimes fail to have a value are
called {\bf partial functions}, and do not pose much of a problem.  For
example, in arithmetic we use two-variable partial functions such as
$over$, and write $\frac{x}{y}$ rather than $over(x,y)$.  As you know, the
expression $\frac{x}{y}$ does not have a value for every $x$ and every
$y$. (For example, $\frac{1}{0}$ does not have a value.)  Another example
is the one-variable partial function $square$-$root$ (for which we write
$\sqrt{x}$ rather than $square$-$root(x)$). Again, $\sqrt{x}$ does not have
a value for all real numbers $x$. (For example, $\sqrt{-1}$ does not have a
value.)  When we are dealing with a partial fraction, we just have to be
careful and make sure that we never plug in the ``wrong'' values of the
arguments.

On the other hand, ``functions'' that have more than one value pose a very
serious problem, and for that reason we do not want to allow them at all. 
For example, suppose we defined $\frac{x}{y}=z$ to mean 
the same as $z.y=x$, without any extra restrictions. Then we would 
have, for example, $1=\frac{0}{0}$---because $1.0=0$---and 
$0=\frac{0}{0}$---because $0.0=0$---so we would be able to conclude that 
$1=0$, which is very bad.    So what we will do instead is define 
``$\frac{x}{y}=z$'' to mean the same as ``$z.y=x\wedge y\neq 0$.''
This will take care of our problem, because now if 
$\frac{x}{y}=z$ and $\frac{x}{y}=w$ it will follow that 
$z=w$, so we will not get into any trouble.

So, returning to our kinship example, the real trouble with ``functions'' 
such as $brother$ or $sister$ is that people often have more than one
brother or sister.  So, for example, if Bill has two sisters, Jane and
Clare, and I was allowed to write things like
$Jane=sister(Bill)$ and $Clare=sister(Bill)$, I would be able to conclude
that $Jane=Clare$, which is not right.  \proofend

\begin{center}
\fbox{
\begin{minipage}{13cm}
\begin{center}
{\large {\bf Are $\frac{1}{0}$ and $\frac{0}{0}$ ``undefined''?}}
\end{center}

\medbreak\noindent Students often say things like 
``$\frac{1}{0}$ is undefined,'' or ``$\frac{0}{0}$ is undefined,''
or ``$\sqrt{-1}$ is undefined.''   Let us think this over. To ``define''
something means ``to say what it means.''  And I know perfectly well 
what $\sqrt{-1}$ means. In general, if $x$ is any real number, 
``$\sqrt{x}$'' means ``the number $y$ such that $y\ge 0$ and $y.y=x$.''  
So ``$\sqrt{-1}$'' means ``the number $y$ such that $y\ge 0$ and $y.y=-1$.''  
This is perfectly clear to me, and should be perfectly clear to you. 
So don't go on saying that ``$\sqrt{-1}$ is undefined.''   The trouble
with $\sqrt{-1}$ is {\bf not} that it is ``undefined,'' but that it does
not exist.  

Similarly, ``$\frac{x}{y}$'' means ``the number $z$ such that 
$z.y=x\wedge y\neq 0$.''    So ``$\frac{1}{0}$''  
means ``the number $z$ such that $z.0=1\wedge 0\neq 0$.'' 
And ``$\frac{0}{0}$''  
means ``the number $z$ such that $z.0=0\wedge 0\neq 0$.'' 
In both cases, the requirements on $z$ are very clear, so you cannot say
that $\frac{1}{0}$ or $\frac{0}{0}$ are ``undefined.''  What you {\bf
should} say  is 

\medbreak
\begin{center}
{\bf $\frac{1}{0}$ and $\frac{0}{0}$ do not exist.}
\end{center}
\medbreak
$\,\,$
\end{minipage}
}
\end{center}


\begin{center}
\fbox{
\begin{minipage}{12cm}

\noindent {\bf Six-legged chickens are not ``undefined.'' You now very well
what ``six-legged chicken'' means!  The problem with six-legged chickens is
not that they are ``undefined.'' It's that there aren't any.  
SIX-LEGGED CHICEKNS DO NOT EXIST!}

{\bf Unicorns are not ``undefined.'' I know perfectly well what a unicorn
is. There just aren't any. UNICORNS DO NOT EXIST!}

\end{minipage}
}
\end{center}

\medbreak
\centerline{\hbox to 5cm{\hrulefill}}

\medbreak\noindent In order ot be able to talk about about 
sons and daughters and brothers and sisters we need 
{\sl PREDICATE SYMBOLS}.

\medbreak\noindent {\bf PREDICATE SYMBOLS} are letters such as $p$, $q$,
$P$, $Q$, or other symbols such as $=$, $>$, $<$, $\le$,
$\ge$, $\epsilon$, $\subseteq$, $tall$, $stupid$, $smarter$-$than$. 
Each predicate symbol admits a {\sl fixed}
number of arguments.    
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} predicate. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} predicate.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} predicate.  And so on.)   If you write 
a predicate symbol, followed by a ``$($,'' then as many arguments
as the predicate symbol admits, separated by commas, and then 
``$)$,'' you get an {\sl atomic formula}.  The strings that can be used 
as arguments are {\sl terms}.

More formally, then:


\begin{center}
\fbox{
\begin{minipage}{13cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``ATOMIC FORMULA''}}
\end{center}

\medbreak\noindent RULE A: If $L$ is a $k$-argument predicate symbol,
and $t_1,\ldots,t_k$ are terms, then $L(t_1,\ldots,t_k)$ is an atomic
formula. (For example: if $p$ is a one-variable predicate symbol and $t$ is
a term, then $p(t)$ is an atomic formula; if $p$ is a two-variable
predicate symbol and $t$, $s$ are terms, then $p(t,s)$ is an atomic
formula; if $p$ is a three-variable predicate symbol and $t$, $s$, $u$ are
terms, then $p(t,s,u)$ is aa atomic formula.)
\vspace{2mm}
\end{minipage}
}
\end{center}

\medbreak\noindent Later, when we discuss how to {\it interpret} 
our formulas (that is, how to assign {\sl meanings} to them), predicate symbols
will correspond to {\it predicates}, that is, expressions that have a {\bf
definite truth value} once you plug in values for its arguments.  For
example, if we use a binary predicate symbol $>$ to stand for ``larger
than,'' and write $t>s$ rather than $>(t,s)$, then $3>5$ will have the
truth value $False$, $x>3$ will have a truth value for each value of $x$
(so $x>3$ will be $True$ if $x$ is $4$, $False$ if $x$ is $-1$, and so on)
and $x>y$ will have a value for each choice of values of $x$ and $y$ (so
$x>y$ will be $True$ if $x$ is $4$ and $y$ is $3$, $False$ if $x$ is $-1$
and $y$ is $0$, and so on).

\medbreak\noindent {\bf EXAMPLE 1, continued.} Let us go on 
with our language for {\bf real-number arithmetic}. Now that we know all
about predicate symbols, we introduce symbols for the basic binary
predicates ``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' which are {\sl
binary}. Strictly speaking, we should write $=(x,y)$ to mean ``$x$ is equal
to $y$,'' and $>(3,5)$ to mean ``$3$ is larger than $5$'' (which is false,
of course, but is perfectly meaningful).  But we do what we did for the
function symbols.  Instead of writing $>(3,5)$ we write $3>5$, instead of
writing $=(x,y)$ we write $x=y$, and so on.

Then the following are examples of atomic formulas: 

\bi
\item[1.] $x+y=7$, 

\item[2.] $3+7=x$, 

\item[3.] $x+y+|x+3|=7-y+z$, 

\item[4.] $x+y+|x+3|>7-y+z$,

\item[5.] $|x|=x\wedge(-x)$,

\item[6.] $|x|+5=|((x+y)\wedge|y|)|+((x+y)\vee|y|)$.

\ei
 
\medbreak\noindent {\bf EXAMPLE 2, continued.} Let us go on 
with our language for {\bf natural-number arithmetic}. Now that we know all
about predicate symbols, we introduce symbols for the basic binary
predicates ``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' ``$|$'' which
are {\sl binary}. (Note: ``$|$'' stands for ``divides'', so ``$x|y$'' means
``$x$ divides $y$,'' that is $(\exists z)(z.x=y)$.)  Strictly speaking, we
should write $=(x,y)$ to mean ``$x$ is equal to $y$,'' and $>(3,5)$ to mean
``$3$ is larger than $5$'' (which is false, of course, but is perfectly
meaningful).  But we do what we did for the function symbols.  Instead of
writing $>(3,5)$ we write $3>5$, instead of writing $=(x,y)$ we write
$x=y$, and so on.

Then the following are examples of atomic formulas: 

\bi
\item[1.] $x+y=7$, 

\item[2.] $3+7=x$, 

\item[3.] $7|3.4$,

\item[4.] $x+y+GCD(x,8)>7-y+z$,

\item[5.] $x=x\wedge(-x)$,

\ei
 
\medbreak\noindent {\bf EXAMPLE 3, continued.} Now let us go on with 
the predicate calculus that enables us to talk about kinship.
Now that we have predicate symbols, we could introduce one-variable
predicates $\mu$, $\varphi$---so that 
``$\mu(x)$'' would stand for ``$x$ is male''
and ``$\varphi(x)$'' for ``$x$ is female''---
and lots of two-variable predicates  such as 
$s$, $d$, $c$, $b$, $sis$, $u$, $a$m   so that 
``$s(x,y)$'' would stand for ``$x$ is $y$'s son'',
``$d(x,y)$'' would stand for ``$x$ is $y$'s daughter'',
``$c(x,y)$'' would stand for ``$x$ is $y$'s child,'' 
``$b(x,y)$'' would stand for ``$x$ is $y$'s brother,'' 
``$sis(x,y)$'' would stand for ``$x$ is $y$'s sister,'' 
``$u(x,y)$'' would stand for ``$x$ is $y$'s uncle,'' 
and ``$a(x,y)$'' would stand for ``$x$ is $y$'s aunt.'' \proofend

\sepa

\medbreak\noindent Atomic formulas can be combined with the logical
connectives to yield {\bf formulas}.  For example, let us look at the
following fairly long string of 82 characters, in a predicate calculus 
with three binary predicate symbols, namely, $=$, $s$ and $c$: 
$$
(\exists y)(c(y,y))\vee
(\exists x)(\forall y)(\exists z) (((s(a,x)\wedge s(a,y))\wedge((\sim
\!x\!=\!y)\wedge(\sim\! a\!=\!z))) \wedge ((c(z,x)\wedge c(z,y))\wedge c(b,z)))\,.
$$
Let us call this expression $F$. Then $F$ is {\sl not} an atomic formula,
but it is a formula.  This means that $F$ is obtained from atomic formulas
by combining them via the connectives $\wedge$, $\sim$ and $\exists$.  In
our case, the atomic formulas useud to produce $F$ are: $c(y,y)$,
$s(a,x)$, $s(a,y)$,
$x=y$, $a=z$, $c(z,x)$, $c(z,y)$, and $c(b,z)$.

We now describe precisely the rules that allow us to combine 
atomic formulas in order to make formulas.

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``FORMULA''}}
\end{center}

\medbreak\noindent RULE F1: Atomic formulas are a-formulas.

\medbreak\noindent RULE F2: If $F$ is an a-formula or a q-formula then
$\sim F$ is a p-formula. 

\medbreak\noindent RULE F3: If $F$ is a p-formula
$\sim(F)$ is a p-formula.

\medbreak\noindent RULE F4: If $F$ is an a-formula or a q-formula 
and $G$ is an a-formula or a q-formula then 
$F\wedge G$, $F\vee G$, $F\imply G$, and $F\lto G$ are p-formulas.

\medbreak\noindent RULE F5: If $F$ is an a-formula or a q-formula 
and $G$ is a p-formula 
then $F\wedge (G)$, $F\vee (G)$, $F\imply (G)$, $F\lto (G)$, 
$(G)\wedge F$, $(G)\vee F$, $(G)\imply F$, and $(G)\lto F$
are p-formulas.

\medbreak\noindent RULE F6: If $F$ and $G$ are p-formulas, 
then $(F)\wedge(G)$, $(F)\vee (G)$, $(F)\imply(G)$, and $(F)\lto (G)$
are p-formulas.

\medbreak\noindent RULE F7: If $F$ is a q-formula,
and $v$ is a variable, then $(\forall v)F$ and  $(\exists v)F$ are
q-formulas.

\medbreak\noindent RULE F8: If $F$ is an a-formula or a p-formula and
$v$ is a variable, then $(\forall v)(F)$ and  $(\exists v)(F)$ are
q-formulas.

\medbreak\noindent RULE F9: Only those expressions obtained by repeated
application of Rules F1, F2, F3, F4, F5, F6, F7, and F8  are formulas. 
\vspace{2mm}
\end{minipage}
}
\end{center}


\medbreak\noindent Why is $F$ a formula according to our rules? 
In Box GNU we show you in detail how one can construct $F$ following the
rules. 

\bigbreak\noindent Notice how the rules are carefully designed so that you
don't have to put too many parentheses.  For example, the rules allow tou
say $x<y$ and $x<y\vee x>y$ (without parentheses) but then you are required
to put parenthesis to say $(\exists x)(x<y)$ and $\exists x)(x<y\vee
x>y)$. This would appear to mean that we have a rule that says that ``if
$G$ is a formula and $v$ is a variable then $(\exists v)(G)$ and $(\forall
v)(G)$ are formulas.  In particular, if you want to quantify $(\exists
x)(x<y)$ again---universally, say---then you would have to write 
$(\forall y)((\exists x)(x<y))$, with an extra pair of parentheses. 
Our rules do not require the extra parentheses, because once you have
quantified once, then the formula yo get becomes a q-formula, and then when
you want to quantify a second time you apply Rule F7 rather than Rule F8.


(Then 
$c(a,b)\lto(s(a,b)\vee d(a,b))$. Make sure you understand why!)
We could also have the binary predicate ``$=$'' with its usual meaning
(that is, ``$a=b$'' means that $a$ and $b$ are the same person). 
With these symbols, we could say lots of more complicated
things. For example, suppose I want to say ``$a$ is $b$'s uncle.''
I could say this by writing 
$$(\exists x)(\exists y)(\exists z)
\Bigl(s(a,x)\wedge s(a,y)\wedge(\sim x=y)\wedge(\sim a=z)
\wedge c(z,x)\wedge c(z,y)\wedge c(b,z)\Bigr)\,.$$

\bigbreak
\centerline{{\bf TERMS VS. FORMULAS}}

\medbreak\noindent{\bf Terms} are supposed to be 
names of ``things.''  {\bf Formulas} are
expressions that ---when we plug in values for the
open variables--- become sentences or statements.

For example, the following are terms: 

\medbreak
\noindent\hbox to 3.5cm{\hfill $3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to 3.5cm{\hfill $x$}\hbox to 3.5cm{\hfill $5xyz$}

\noindent\hbox to 3.5cm{\hfill $x+3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to
3.5cm{\hfill $x^2+3y+7$}\hbox to 3.5cm{\hfill $\{x:x^2<3y\}$}\ ,

\noindent and the following are formulas:

\medbreak
\noindent\hbox to 3.5cm{\hfill $3<5$}\hbox to 3.5cm{\hfill $3+5=8$}\hbox to 3.5cm{\hfill $x+3=7y$}\hbox to 3.5cm{\hfill $5xyz<x+z$}

\noindent\hbox to 3.5cm{\hfill $(\forall x)(x+3>x)$}\hbox to 3.5cm{\hfill $(\forall x)(x+y>x)$}\hbox to 3.5cm{\hfill $(\exists x)(x+y>xz)$}\hbox to 3.5cm{\hfill $(\forall x)(x\!>\!y\imply y\!=\!z)$}\ .

\medbreak\noindent If you want to see a completely precise definition of
``term''  and ``formula'' then you have to consult a mathematical logic
textbook. In this course, we will take these concepts for granted.

How can you tell whether something is a term or a formula? 

\medbreak\noindent TERMS.  A {\bf term}
stands for a {\bf definite thing} (it may be a number, or a set, or a line,
or a function, or some other entity such a person or an animal), or for a
``variable thing'' that becomes definite once you plug in values for the
variables.  (For example, ``$3+5$'' denotes a definite number, whereas
``$3+x$'' denotes a ``variable number'' that becomes definite once you plug
in values for $x$: when you plug in $x=5$, $3+x$ has the value $8$; when
you plug in $x=-3$, $3+x$ has the value $0$; when you plug in $x=-20$,
$3+x$ has the value $-17$; and so on.)  For a more complicated example 
of a term, consider the expression ``$f(x)$''.   This is a term 
that contains two letter variables, namely, $f$ and $x$, standing,
respectively, for a function and a number. So, when you plug in a specific
function for $f$, and a specific number for $x$, you get a number. 
(For example, if $f$ is the ``squaring'' function, i.e. the function such
that $f(y)=y^2$ for all $y$, and $x=3$, then $f(x)$ has the value $9$. 
If $f$ is the ``exponential'' function, i.e. the function such
that $f(y)=e^y$ for all $y$, and $x=6$, then $f(x)$ has the value $e^6$. 
If $f$ is the function such
that $f(y)=y^2+3y+7$ for all $y$, and $x=6$, then $f(x)$ has the value 
$61$.) 


\medbreak\noindent FORMULAS.  A {\bf formula} makes an {\bf assertion},
which may be 
true or false.  So a formula has a {\bf definite truth value} (``true'' or
``false,'' often abbreviated $T$ or $F$) or a ``variable truth value'' that
becomes definite once you plug in values for the variables.  (For example,
``$3+5=8$'' has a definite truth value, namely, T, and ``$3+5=6$'' has a
definite truth value, namely, F.  On the other hand, ``$3+x=y$'' has a
variable truth value, which becomes definite once you plug in values for
the variables $x$ and $y$: when you plug in $x=5$ and $y=8$, $3+x=y$ is T;
when you plug in $x=2$ and $y=8$, $3+x=y$ is F; when you plug in $x=2$ and
$y=5$, $3+x=y$ is T; and so on.)

\bigbreak
\centerline{{\bf VARIABLES AND CONSTANTS}}

\medbreak\noindent As explained before, 
the basic constituents of terms are {\sl variables} and {\sl constants}.

{\sl Constants} are symbols that stand ---permanently or temporarily--- for one
{\sl fixed}, {\sl definite} thing.  
{\sl Variables} are letters such as $x$, $y$ that are meant to stand 
for things that only become definite when you assign a specific {\sl value}
to the variable.  

In principle, any symbol could be used for a constant, and also for a
variable. What matters is that in each case it should be clear which use is
intended, e.g. by {\sl declaring} a symbol to be a constant or a
variable. Usually, letters such as $x$, $y$, $z$, are used as variables for
real numbers, letters such as $k$, $m$, $n$, $p$ are used as variables for
natural numbers, and letters such as $X$, $Y$, $Z$ are used as variables
for sets.  So, for example, we would write formulas such as 
$$
(\forall x)(\exists y)(y>x)\,,
$$
$$
(\forall m)(\forall n)(\forall k)((k|m\wedge k|n)\Rightarrow k|m+n)\,,
$$
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,.
$$
But there is nothing sacred about this particular use.   We could, for
example, write 
$$
(\forall a)(\exists b)(b>a)\,,
$$
$$
(\forall x)(\forall y)(\forall z)((z|x\wedge z|y)\Rightarrow z|x+y)\,,
$$
$$
(\forall \alpha)(\forall \beta)(\forall \gamma)((\alpha\subseteq\beta\wedge
\beta\subseteq\gamma)\Rightarrow\alpha\subseteq\gamma)\,,
$$
provided that we specify that ``$a$, $b$ are real variables, 
$x$, $y$, $z$ are natural-number variables, $\alpha,\beta,\gamma$ are set
variables.'' 

One way to make it clear that something is a variable and to specify its
range is to put the range in the quantifier.  For example, if we write
$$
(\forall a\in\rr)(\exists b\in\rr)(b>a)\,,
$$
$$
(\forall x\in\nn)(\forall y\in\nn)(\forall z\in\nn)((z|x\wedge
z|y)\Rightarrow z|x+y)\,, 
$$
then it is clear that $a$, $b$ are real variables, 
and $x$, $y$, $z$ are natural-number variables.   

For our third sentence, involving sets, the situation is a little bit more
complicated. The obvious thing to do would be to write something like
$$
(\forall X\in{\cal S})(\forall Y\in{\cal S})(\forall Z\in{\cal S})
((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,
$$
where ${\cal S}$ is ``the set of all sets.''  This, however, is {\sl
forbidden}, because the ``set of all sets'' has to be avoided, due to the
Russell Paradox (to be discussed later in the course). 

So we can get around this by writing, instead, 
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,\,\,\,\,{\rm 
where\,\,\,}X,Y,Z\,\,\,{\rm are\,\,\,set\,\,\,variables},
$$
or
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,}
(X\subseteq Y\wedge Y\subseteq Z)\Rightarrow  X\subseteq Z\,.
$$

Naturally, we could even use more informal expressions such as 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,if}
\,\,\,X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z\,\,\,{\rm it\,\,\,follows\,\,\,that\,\,\,}
\Rightarrow  X\subseteq Z\,,
$$
or 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,and\,\,\,}
X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z,\,\,\,{\rm then\,\,\,}
\Rightarrow  X\subseteq Z\,.
$$

More complicated terms can be constructed from the basic constants 
and variables by means of (i)~function symbols, and (ii)~the 
``curly bracket'' set notation.  For example, the expression 
$x+3$ is a term, obtained by combining the variable $x$ and the constant 
$3$ by means of the binary (i.e. 2-argument) function symbol $+$.
The expression $\{x\in\rr:x^2>2\}$ is a term, that stands 
for a certain set of numbers.  {\bf A more detailed description of 
how function symbols work, and how to build more complicated terms 
and work with them, will be given later. For the time being, 
let us forget about function symbols until further notice, 
and concentrate on predicates, constants, variables,
and the logical connectives.}




{convincing argument} that establishes {\bf beyond the shadow of a 
doubt} that the statement is true.  

You should think of proving statements as a {\bf game} that you are playing
against an opponent who does not believe that the statement you are proving
is true.  You make statements, your opponent may find them convincing or
not, and it is up to you to clear up all your opponent's doubts by 
arguing your case persuasively every time your opponent isn't convinced. 
(Maybe you should think that I, the instructor, am the ``opponent.''
{\bf You} have to convince {\bf me}. If you convince me, then your proof
has worked. Otherwise it has not.)

The trouble with all this is that it is not very clear what will count as a
``convincing argument.''  Suppose your opponent says, after you have given
what looks to you like a very convincing proof: ``I am not convinced; I
don't see why what you just said proves anything.''  How would you deal
with something like that?

Moreover, any argument you give will have to {\it start} somewhere.  (You
cannot prove anything starting from nothing.)  What can you do if your
opponent refuses to accept your starting point?  

\bigbreak\noindent EXAMPLE: suppose you want to prove that, 

\medbreak
\bi
\item{} {\sl If a real number is greater than or equal to $4$, then 
the square root of the number is less than or equal to half the number.}

\medbreak\noindent Equivalently ---and more concisely--- you want to show that 

\medbreak
\item{} {\sl If $x$ is any real number such
that $x\ge 4$, then $\sqrt{x}\le{x\over 2}$.}
\ei

\medbreak

\noindent Or ---even more concisely--- you want to prove that 

$$
(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)\,.
$$

\noindent To prove this, you can proceed as follows: 

\noindent 
To prove $(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)$,
we pick an arbitrary $x$ and try to prove that 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$.

\noindent 
So, let $x$ be arbitrary.  We want to prove 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$. To prove this, we assume 
$x\ge 4$ and try to prove that $\sqrt{x}\le{x\over 2}$. 

\noindent 
Now we are assuming that $x\ge 4$.  From this it follows ---multiplying
both sides by $x$--- that $x^2\ge 4x$. Then ---dividing both sides by
$4$--- we get that ${x^2\over 4}\ge x$. Then ---taking the square root of
both sides--- we get $\sqrt{{x^2\over 4}}\ge\sqrt{x}$. Since
$\sqrt{{x^2\over 4}}={x\over 2}$, we have shown that 
${x\over 2}\ge \sqrt{x}$, which is ``the same as''
$\sqrt{x}\le{x\over 2}$. {\bf END OF PROOF}

\bigbreak\noindent {\bf Is this a truly convincing argument?}

\medbreak\noindent The answer is ``yes'' if all the steps are convincing. 
But {\bf if one step is not convincing, then the whole argument fails.}
(This is why you should {\bf not} expect to get partial credit for a wrong
proof if most steps are O.K. but just a few of them are wrong.  A proof is
like a chain. And {\bf a chain is no stronger than its weakest link.}  If a
16-Ton weight is hanging right above your head, held by a chain, and the
chain breaks because of one weak link, and the weight falls on your head
and kills you, do you think the chain manufacturer should be entitled to
``partial credit'' because ``most links were O.K.; only one broke''?  If
the driver of Lady Diana's car didn't drink every night, but only drank on
the particular night of the accident, should he be entitled to partial
credit because ``after all, most of the time he didn't drink''?)

\bigbreak\noindent So, {\bf are {\mymagtwobf ALL} the steps in the above 
proof truly convincing?}

\medbreak\noindent In the proof, we have given justifications for each
step. For example, at one point we went from 

$$x\ge 4$$
to 
$$x^2\ge 4x\,,$$
and we said that this was done 
``multiplying both sides by $x$.''   Now, why is this step permissible?

Suppose I tell you that ``I don't believe that you can do this''? 
How do you persuade me?   Well, let us give it a try. 
You can say: ``because there is a general rule that says 
if you take an inequality and multiply both sides by a number, then
you still get a valid inequality.''  (Or, in more symbolic notation:
$$
(\forall a)(\forall b)(\forall c)(a\ge b\Longrightarrow ac\ge bc)\,,
$$
using the language of a predicate calculus that we will discuss later.)  

This, however, will not do. This ``rule'' is NOT VALID.   
(For example, if $a=5$, $b=2$ and $c=-2$, then $a\ge b$, but 
it's not true that $ac\ge bc$, since $ac=-10$ and $bc=-4$, 
so $ac<bc$.)      So WE CANNOT USE IT.   

Now, here is a second try.  You can say: ``because there is a general rule
that says if you take an inequality and multiply both sides by a
nonnegative number, then you still get a valid inequality, that is,
using symbolic notation: 
$$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)\,.$$


It turns out that this time our rule IS VALID.  But notice that, in order
to apply this rule, we must be sure that $c\ge 0$. In our case, the number
that plays the role of $c$ is $x$, and it is indeed true that $x\ge 0$,
because $x\ge 4$.  (Since this is a crucial fact for us to be able to apply
the above ``rule,'' it would have been better to include in our proof a
little remark highlighting this. For example, instead of ``multiplying
both sides by $x$,'' it would have been better to say ``multiplying both
sides by $x$, which is possible because $x\ge 0$, since we are assuming
that $x\ge 4$.'')

But we haven't solved our problem.  Remember that your proof is really a
two-person game that you are playing against me, your opponent, in which
you have to convince me. What if I say ``How do we know 
that $(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)$?   I am not really convinced 
that this is true.''?

You could try to prove for me that 
$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b\wedge c\ge 0)\Longrightarrow 
ac\ge bc\Bigr)$  is indeed true.   But I am sure that in our proof you
will use other ``rules,''  and I, your opponent, could pick any of these 
and claim, once again, that I am not convinced of its validity.  

\medbreak
{\mymagtworm 

$\,$$\,$$\,$$\,$$\,$$\,$IF I WANTED TO, \ \ $\,\,$I COULD GO ON LIKE THIS 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$FOREVER. NO MATTER WHAT JUSTIFICATION 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$YOU $\,$$\,$GIVE $\,$$\,$FOR $\,$$\,$A $\,$$\,$STEP
$\,$$\,$IN $\,$$\,$YOUR $\,$$\,$PROOF, $\,$$\,$$\,$$\,$I 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$COULD ALWAYS ASK: $\,$$\,$$\,$$\,$``WHY IS THAT
VALID?''} 

\medbreak\noindent
In other words, if your opponent is truly stubborn and uncooperative, then
there is {\it nothing} you can do. Maybe you will come up with some reason,
but this reason will be based on some other facts that look obviously true
to you.  But your opponent can still refuse to accept these other facts,
and this process can go on for ever.

To take care of this ``problem of the opponent that doesn't play fair,'' 
mathematicians have come up with the concept of a {\bf formal proof}.
We give very precise rules defining what a ``formal proof'' is, 
and if you prove something correctly, and strictly following the rules, 
then you are {\it guaranteed} that your proof will be accepted as valid. 

If you do {\it not} obey the rules, then you are on uncertain ground: you
might be able to convince an instructor or grader, but you cannot be sure,
because any time you take a step one of us might ask ``how do you 
justify that step?''

You should think of giving formal proofs as similar to playing a game
such as chess: you are given very strict, precise rules, and you are only
allowed to make moves following the rules. As long as you obey the rules,
your opponent does not have the right to reject your moves. (If your
opponent unilaterally decides that a move of yours is not acceptable,
because ``you are not allowed to move a bishop if in the previous move you
moved a knight,'' then you would say that your opponent is not playing
fair, because this ``rule'' was not in the list of rules that had been
agreed upon before the game started.) 

{\bf In Mathematics we play fair.} We tell you what the rules are, and if
you prove something correctly following the rules, then you can be sure
that it will be accepted.  You may think that having to follow the rules is
a restriction to your freedom. But this is not so.  You should think of it
as {\bf your own protection against unfair play.}  If you correctly justify
every step according to the rules, then we promise you that we will not
change the rules in the middle of the game, kick the board, or come up with
other surprises.

\medbreak\noindent REMARK. I said that you should think of giving formal
proofs of theorems as a game similar to chess, in which you are given a
small number of very strict rules telling you which moves are permitted,
and each move has to be justifiable by explaining exactly which rule you
are using and how you are using it.  However, exactly as in the case of
chess, if all you know is the rules, and you always apply the rules
carefully and scrupulously, you may still lose all the time.  If you want
to {\bf win} (that is, in the case of a proof, if you want to get to the
desired conclusion) then you have to {\bf think}. You have to plan a
strategy, make a move now because it will be useful five or six steps
later, etc.

The one big difference between chess and theorem-proving is that {\bf chess
is a zero-sum game}, but {\bf theorem-proving is a cooperative game.} 
What this means is that in chess your opponent wants you to lose,
whereas in theorem-proving you and your opponent are working together,
trying to find out whether something is true, and to understand why.

That's why in chess you don't explain openly why you make a move.  You do
{\sl not} say to your opponent things like: ``I am making this move now so
that then in the next move I can do this other thing, so that three moves
later I can do that, and then I will checkmate you.'' {\sl You don't want your
opponent to be able to figure out your strategy.}

In theorem-proving you definitely {\bf do} want your opponent to understand
your strategy.  Your opponent is going to be ruthless in criticizing you if
you don't obey the rules, but as long as you obey them you will get full
coopertaion. In fact, usually you don't have to go through the whole
business of explaining every detail of every application of every rule.
{\sl As long as you persuade your opponent that something can be done using
the rules, there is no need to go through all the steps and actually do
it.}  And, since your opponent wants to be persuaded, the more you do to 
explain why you are doing something, the better. That's why {\sl in 
theorem-proving it's always nice to explain your strategy, so that your
opponent can understand where you are headed.}    

Because theorem-proving is a cooperative game, when writing proofs it is
always nice to use the first person plural ---the so-called ``editorial
we''--- as we did in our first example of a proof (``we pick,''\ \ ``we are
assuming,''\ \ ``we have shown'').  This is a good way to convey the idea
that the writer and the reader are {\sl working together}, that {\bf the writer
is leading the reader gently along a path of discovery.} \proofend

\vfill\eject

\centerline{{\mymagtwobf 2.\ VALID INFERENCES}} 

\bigbreak\noindent{\bf Proofs} involve {\bf steps} in which you move
from some statements (the {\it premisses}) to a new statement (the
conclusion).     We also say that you {\it infer}, or {\it derive}, or {\it
deduce} the conclusion from the premisses, and the steps where you 
do that are called {\it inferences}. 

So the first thing we need to discuss is:

\bigbreak
\centerline{{\mymagonebf WHAT IS A VALID INFERENCE?}}

\medbreak\noindent An {\bf inference} is a way of passing  from
certain statements (the {\bf premisses}) to another statement (the {\bf
conclusion}).   

\medbreak

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf VALID INFERENCES}}\\
(Informal definition)
\end{center}

A {\bf valid inference} is an inference for
which it is guaranteed  that, if the premisses are true, then the
conclusion is also true.
\end{minipage}

}

\end{center}

\medbreak\noindent For example:

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (1)}

\noindent\hbox to 7cm{\hfill Socrates is mortal}

\medbreak\noindent  is an example of a valid inference known as a {\it
syllogism}.  The premisses are 
``All men are mortal'' and ``Socrates is a man,''  and the conclusion is 
``Socrates is mortal.''

\medbreak\noindent  Roughly speaking, we would like to give the following 
definition of ``logically valid inference'': 

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf LOGICALLY VALID INFERENCES}}\\
(Informal definition)
\end{center}

An inference is {\bf logically valid} if the conclusion follows
from the premisses {\bf by virtue of the ``form'' of the statements
involved.}
\end{minipage}
}
\end{center}

\medbreak\noindent  Eventually, we will say exactly that. But we are not
ready to say it yet, because we still do not know precisely
what we mean by the ``form'' of a statement. 
   
Before we talk about ``form,'' let us go on with examples to illustrate 
what we are trying to say. We would like to 
call the syllogism of (1) logically valid {\bf because in (1) it does not
matter what ``Socrates'' or ``man'' or ``mortal'' mean.}  It would be
equally valid to say

\medbreak\noindent\hbox to 7cm{\hfill All men are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (2)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All pigs are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a pig}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (3)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Donald Duck is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (4)}

\noindent\hbox to 7cm{\hfill Donald Duck  is mortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All borogoves are mimsy}

\noindent\hbox to 7cm{\hfill The Mad Hatter is a borogove}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (5)}

\noindent\hbox to 7cm{\hfill The Mad Hatter is mimsy}

\medbreak\noindent It does not matter whether the premisses are true (in
some of the above examples they are true, in others they are not) or
whether the conclusion is true.  It doesn't even matter whether you know
what a ``borogove'' is or what ``mimsy'' means.     The point is:

\bigbreak

\bi
\item{} {\it The four examples of syllogism given above all are of the
same {\bf logical form}, namely,}
\ei

\medbreak\noindent\hbox to 7cm{\hfill All $X$ are $Y$}

\noindent\hbox to 7cm{\hfill $Z$ is an $X$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (6)}

\noindent\hbox to 7cm{\hfill $Z$ is a $Y$}

\medbreak\noindent To be able to talk about an inference being ``logically 
valid'' we have to be able to talk about the {\bf logical form} (or, even
better, ``a'' logical form, rather than ``the'' logical form) of an
inference. 

For example, we could say that all five inferences (1), (2), (3), (4), (5)
are of the form (6), and any reasoning of this form is valid.  

Another example of a valid inference form would be:

\medbreak\noindent\hbox to 7cm{\hfill $X$ is  $Z$}

\noindent\hbox to 7cm{\hfill $Y$ is $Z$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (7)}

\noindent\hbox to 7cm{\hfill $X$ and $Y$ are $Z$}

\medbreak\noindent Examples of inferences of the form (7) are:

\medbreak\noindent\hbox to 7cm{\hfill John is  happy}

\noindent\hbox to 7cm{\hfill Mary is happy}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (8)}

\noindent\hbox to 7cm{\hfill John and Mary are happy}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is smart}

\noindent\hbox to 7cm{\hfill Hillary Clinton is smart}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (9)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are smart}

\medbreak\noindent  So it would seem that we have two examples of ``valid
forms'', namely, (6) and (7).  Consider, however, the following
examples: 

\medbreak\noindent\hbox to 7cm{\hfill All men are equal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (10)}

\noindent\hbox to 7cm{\hfill Socrates is equal}

\medbreak\noindent and 

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is married}

\noindent\hbox to 7cm{\hfill Hillary Clinton is married}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (11)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are married}

\medbreak\noindent Superficially, (10) is ``of the form (6),'' and
(11) is ``of the form (7).'' Something is clearly wrong here, because
in (110) the conclusion is clearly false, even though the premisses are
true, and in (10) the conclusion is meaningless. 

What is wrong here is that ``statement forms'' such as 
``All $X$ are $Y$'' or ``$X$ and $Y$ are $Z$'' are not good enough 
to capture logical distinctions that matter.   For example, the statements
``Al Gore and Hillary Clinton are married'' and
``Al Gore and Hillary Clinton are smart'' both appear to be of the same
``form,'' namely, ``$X$ and $Y$ are $Z$,'' but they really are not. 
``Smart'' is a {\bf one-variable predicate}. (If $x$ is a person, you can
say that ``$x$ is smart.'')  But ``married'' is {\it both} 
a {\it one-variable predicate} (in ``Al Gore is married'') and a
2-variable predicate (in ``Al Gore and Hillary Clinton are married''). 
In normal English discourse, when we say that ``$x$ and $y$ are married''
we do not mean that each one, separately, is married. We mean that 
``$x$ and $y$ are married to each other.''  (If you don't believe that,
ask yourself the following: when someone asks a couple ``are you two 
married?'', isn't it clear that this means ``to each other''?)

So, if we really want to talk about {\bf forms} of statements and
inferences, then we need a more sophisticated language.  We have to be able
to distinguish, for example, between one-variable predicates and two-variable
predicates. The languages that do that are the {\bf first-order predicate
calculi}. (``Calculi'' is the plural of ``calculus.'') Before we discuss
these rather complicated languages, we will say a few words about 
{\bf propositional calculi}, which are simpler. Both propositional calculi 
and first-order predicate calculi are examples of {\bf formal languages}, 
so we will start by talking about formal languages in general. 

\bigbreak
\centerline{\hbox to 5cm{\hrulefill}}



\vskip 0.7cm
\centerline{{\mymagtwobf 5.\ WHAT ARE PROPOSITIONAL CALCULI}}

\centerline{{\mymagtwobf GOOD FOR?}}

\bigbreak\noindent So far, we have just given some rules for writing
certain weird strings of symbols and calling them 
``propositional forms.''  What do we want to with all that?

The answer is: {\bf we use propositional forms to represent 
forms of propositions or statements.}  

For example, the statement 

\medbreak
\bi
\item{(S1)}  Abraham Lincoln was a great president 
and all the serious historians who have studied his career think very highly 
of him.
\ei

\medbreak\noindent could be represented by the propositional form $P\wedge 
Q$, where the 
propositional variable $P$ stands for ``Abraham Lincoln was a great
president'' and the propositional variable $Q$ stands for ``all the serious
historians who have studied Abraham Lincoln's career think very highly of
him.''

\medbreak Notice that (S1) could also be represented in the form 
$P$, where the propositional variable $P$ stands for ``Abraham Lincoln was
a great president and all the serious historians who have studied his
career think very highly of him.''

So {\bf the representation of a statement by means of a propositional form
is not unique.} There may be more than one way to represent the statement,
depending on how deeply you want to go into the details its structure.  In
our example, ``$P$'' is a {\bf coarser} propositional-form representation
of (S1) than ``$P\wedge Q$'' (and ``$P\wedge Q$'' is {\bf finer} than ``$P$'').

As explained in the book, a propositional form has a {\bf truth table}.    
The truth table tells you how to assing a ``truth value'' (True or False) 
to a propositional form if you assign truth values to the propositional
variables.        

\medbreak\noindent A {\bf tautology} is a propositional form that has the 
value True for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent A {\bf contradiction} is a propositional form that has the 
value False for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent Often, we also apply the words ``tautology'' and 
``contradiction'' to statements in ordinary English. In that case, what we
mean by a ``tautology'' is ``a statement that has a propositional form
which is a tautology.''  Also, what we mean by a ``contradiction'' is ``a
statement that has a propositional form which is a contradiction.''

For example: the statement 

\medbreak

\bi
\item{(S2)}  Hector Sussmann is a great teacher.
\ei


\medbreak\noindent is true (of course!), but it is of the form $P$, which is
not a tautology, and it has no other, finer propositional form.  So 
(S2) is not a tautology. 

\medbreak On the other hand,  the statement 

\medbreak
\bi
\item{(S3)}  Either Hector Sussmann is a great teacher
or he is not.
\ei


\medbreak\noindent has the propositional form $P$, which is
not a tautology, but is also has the form $(P\vee(\sim P))$, which is a
tautology.   So (S3) {\bf is} a tautology.  

\medbreak\noindent{\bf Remark.} The example of (S3) shows that, for
something to be a tautology, it suffices that {\bf one} of its
propositional forms be a tautology.  In the case of statement (S3), we can
find at least three different propositional form for it, namely, $P$,
$(P\vee Q)$, and $(P\vee(\sim P))$, which are progressively finer, and only
the third one is a tautology. This, however, suffices to make (S3) a 
tautology. \proofend

Let us look at another example of a formal representation. Consider the
statement (in which we assume that $f$ is a fixed function described
earlier, and $x_0$ is some given number)

\medbreak
\bi
\item{(S4)} If $f$ has a relative minimum at $x_0$ and
$f$ is differentiable at $x_0$ then $f'(x_0)=0$.  
\ei

\medbreak\noindent This is of the form $P$  (taking $P$ to stand for the
whole statement), or of the form $P\imply Q$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$ and $f$ is differentiable at $x_0$''
and $Q$ to stand for ``$f'(x_0)=0$''), 
or of the form $(P\wedge Q)\imply R$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$,'' $Q$ to stand for 
``$f$ is differentiable at $x_0$,'' and $R$ to stand for ``$f'(x_0)=0$''). 
Once again, notice that the logical form is not unique.


\bigbreak
%\centerline{\hbox to 5cm{\hrulefill}}


\vskip 0.7cm
\vbox{

\centerline{{\mymagtwobf 6.\ SOME TYPES OF VALID INFERENCES}}

\bigbreak\noindent Now that we know how to represent statements, we can
start giving examples of valid inferences of the propositional calculus.

\medbreak\noindent Assume that $f$ is a given function, and $n$ is a given
natural number. Then the inferences
}

\medbreak\noindent\hbox to 7cm{\hfill If $f$ is differentiable then 
$f$ is continuous.}

\noindent\hbox to 7cm{\hfill $f$ is differentiable.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (12)}

\noindent\hbox to 7cm{\hfill $f$ is continuous.}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill If $n$ is not prime
then $n$ is divisible by 3.}

\noindent\hbox to 7cm{\hfill $n$ is not a prime number.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (13)}

\noindent\hbox to 7cm{\hfill $n$ is divisible by 3.}

\medbreak\noindent are both of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (14)}

\noindent\hbox to 7cm{\hfill $Q$}

\medbreak\noindent This is a {\bf valid inference form.} Why is it valid?  
Because {\it if the premisses are true then the conclusion must be true}. 
Indeed: if $P\imply Q$ is true and $P$ is true then $Q$ is true, because
if $Q$ was false then $P\imply Q$ would be false (since $P$ is true) 
because the truth table for $\imply$ gives an $F$ to $P\imply Q$ if $P$ is
$T$ and $Q$ is $F$. 

\medbreak\noindent{\bf Remark.} Notice that in (13) the first premiss is 
usually not true.  (For example, it is not true if $n=25$, because 
$25$ is not a prime number and it's not divisible by $3$. So, if 
$n=25$, the first premiss says ``If $25$ is not a prime number
then $25$ is divisible by 3.'' This is of the form $P\imply Q$, where 
$P$ stands for ``$25$ is not a prime number'' and $Q$ for 
``$25$ is divisible by 3.'' Then $P$ is true and $Q$ is false, so $P\imply
Q$ is false.)

The inference form (14) is one of the most basic ones of all logic. It is
called {\bf modus ponens.}

What do we mean, in general, by a ``valid inference form''?

First of all, an {\bf inference form} is a list of propositional forms
(the premisses), followed by a horizontal line and then a propositional
form (the conclusion).  

A {\bf valid inference form} is an inference form such that the conclusion 
is true for all assignments of truth values to the propositional variables
that render the premisses true. 

Here is another example: the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $\sim Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim P$}

\medbreak\noindent is valid. Examples of inferences of this form are:


\medbreak\noindent\hbox to 7cm{\hfill If I was rich I would own a Mercedes.}

\noindent\hbox to 7cm{\hfill I don't own a Mercedes.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill I am not rich.}

\medbreak\noindent and


\medbreak\noindent\hbox to 7cm{\hfill If $x>3$ then $x>2\,$.}

\noindent\hbox to 7cm{\hfill $\sim x>2\,$.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim x>3\,$.}


\bigbreak
\centerline{\hbox to 5cm{\hrulefill}}


\vskip 0.7cm
\centerline{{\mymagtwobf 7.\ FIRST-ORDER PREDICATE CALCULI}}

\bigbreak\noindent We now turn to the formal languages that are really
important for this course, namely, {\mymagtworm first-order predicate
calculi.} Inferences such as (1) above cannot be justified 
by just looking at their propositional form. If we look at propositional 
form, then (1) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hfill $Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $R$}

\medbreak\noindent which is not a valid inference form of the propositional
calculus. In a first-order predicate calculus, (1) will have associated
to it a much {\bf finer} form, namely

\medbreak\noindent\hbox to 7cm{\hfill $(\forall x)(p(x)\imply q(x))$}

\noindent\hbox to 7cm{\hfill $p(a)$}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (16)}

\noindent\hbox to 7cm{\hfill $q(a)$}

\medbreak\noindent (In (1), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is mortal,'' and ``$a$'' for ``Socrates.'' )

On the other hand, {\it (10) is not if the form (16)}.  In fact, to
represent (10) correctly we need a binary predicate $E(x,y)$ (meaning ``$x$
and $y$ are equal to each other''), and then {\it (10) cannot even be
represented in the language.}  Using a 1-variable predicate $M(x)$ for
``$x$ is a man'', and the symbol $a$ for ``Socrates'', we can represent the
premisses as $(\forall x)(\forall y)((M(x)\wedge M(y))\imply E(x,y))$ and
$M(a)$.  But {\it there is no way to represent the conclusion}, which is
fine because the conclusion is meaningless. In other words: {\bf by trying
to represent an inference such as (10) in a first-order predicate calculus,
we discover that it cannot be done, and this explains why it is mistaken 
to think that (10) is valid.}


Similarly, (8) and (9) are of the form

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (17)}

\noindent\hbox to 7cm{\hfill $p(x)\wedge p(y)$}

\medbreak\noindent which is valid, but (11) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (18)}

\noindent\hbox to 7cm{\hfill $q(x,y)$}

\medbreak\noindent where $q$ is a {\it different predicate} from $p$.
(In fact, $q$ is a two-variable predicate, and $p$ is a one-variable
predicate. Naturally, $p$ and $q$ are
related, because $p(x)$ ---i.e. ``$x$ is married''--- is a way of saying
$(\exists y)q(x,y)$, i.e. ``$x$ is married to somebody,'' or ``there exists
a person that $x$ is married to,'' or ``there exists a $y$ such that $x$ is
married to $y$.'')

Once again, {\bf we looked at an inference that appeared to be valid but
really should not have been.  We tried to represent it in a first-order
predicate calculus, and we discovered {\mymagtwobf} why it is not valid.
the reason is that a word that appeared in it was being used with two
different meanings, namely, the one-variable predicate ``married'' and the
two-variable predicate ``married.''}  The attempt to formalize our
inference in a first-order predicate calculus  forced us to be precise
about what kind of predicate ``married'' is, and we discovered that it 
is really two different predicates.  This solved our problem. 

\bigbreak We now turn to a more precise description of 
{\bf PREDICATE CALCULI}.

\bigbreak
\centerline{{\sl (to be continued)}}

\newpage


\begin{center}
{\mymagtwobf 1.\ \ NATURAL LANGUAGES VS. FORMAL LANGUAGES}
\end{center}

Written English is a {\bf natural language}, that is, a language 
that has evolved over centuries GNUGNU.

The most basic units of Written English are the {\bf characters}
such as letters, the blank space, and punctuation marks.

There are of course 26 letters:
\begin{center} 
a$\,\,\,\,\,$b$\,\,\,\,\,$c$\,\,\,\,\,$d$\,\,\,\,\,$e$\,\,\,\,\,$f$\,\,\,\,\,$g$\,\,\,\,\,$h$\,\,\,\,\,$i$\,\,\,\,\,$j$\,\,\,\,\,$k$\,\,\,\,\,$l$\,\,\,\,\,$m$\,\,\,\,\,$n$\,\,\,\,\,$o$\,\,\,\,\,$p$\,\,\,\,\,$q$\,\,\,\,\,$r$\,\,\,\,\,$s$\,\,\,\,\,$t$\,\,\,\,\,$u$\,\,\,\,\,$v$\,\,\,\,\,$w$\,\,\,\,\,$x$\,\,\,\,\,$y$\,\,\,\,\,$z
\end{center}
as well as the 10 punctuation marks 
\begin{center} 
,$\,\,\,\,\,$;$\,\,\,\,\,$:$\,\,\,\,\,$.$\,\,\,\,\,$!$\,\,\,\,\,$?$\,\,\,\,\,$($\,\,\,\,\,$)$\,\,\,\,\,$`$\,\,\,\,\,$'
\end{center}
and the digits 
\begin{center}
$0\,\,\,\,\,\,1\,\,\,\,\,\,2\,\,\,\,\,\,3\,\,\,\,\,\,4\,\,\,\,\,\,5\,\,\,\,\,\,6\,\,\,\,\,\,7\,\,\,\,\,\,8\,\,\,\,\,\,9$
\end{center}
that we use to represent numbers. Moreover, there is also the blank space,
which is very hard to display on the page but is definitely there. 

So you might think that Written English has 26+10+10+1=47 basic symbols.
Things are not so clear, however. First of all, there are also the capital
letters: 
\begin{center} 
A$\,\,\,\,\,$B$\,\,\,\,\,$C$\,\,\,\,\,$D$\,\,\,\,\,$E$\,\,\,\,\,$F$\,\,\,\,\,$G$\,\,\,\,\,$H$\,\,\,\,\,$I$\,\,\,\,\,$J$\,\,\,\,\,$K$\,\,\,\,\,$L$\,\,\,\,\,$M$\,\,\,\,\,$N$\,\,\,\,\,$O$\,\,\,\,\,$P$\,\,\,\,\,$Q$\,\,\,\,\,$R$\,\,\,\,\,$S$\,\,\,\,\,$T$\,\,\,\,\,$U$\,\,\,\,\,$V$\,\,\,\,\,$W$\,\,\,\,\,$X$\,\,\,\,\,$Y$\,\,\,\,\,$Z
\end{center}
and there are other widely used symbols such as: 
\begin{center} 
@$\,\,\,\,\,${\$}$\,\,\,\,\,${\%}$\,\,\,\,\,${\&}$\,\,\,\,\,${*}
$\,\,\,\,\,${$\#$}
\end{center}
that you will find on the  keybord of your
typewriter\footnote{{\sl Typewriters} are machines with keys for producing
printlike characters one at a time on paper inserted round a roller.  They
were once upon a time widely used to type letters and papers, but are
rapidly becoming extinct.  Some specimens can still be found in museums and
in the homes of people who were born before the advent of computers.}
or computer.

Should we also count these symbols? Should we count the capitals as
different symbols?   

In addition to all this, you can use different fonts,
such as {\bf boldface} or {\it italic type} or {\sl slant}.  Should we
count a boldface ${\bf a}$ as different from an ordinary a? 

As you must be realizing by now, there is an enormous number of
possibilities. So {\bf it is not at all clear exactly what are the 
``basic characters'' or ``basic symbols'' of Written English.}

\medbreak In any case, letters are put together to form {\bf words}. 
So we might now ask:
\begin{center}
\fbox{\large{{\bf Exactly what is a ``word''?}}}
\end{center}
Undoubtedly, words are strings of letters. But {\bf not every string of
letters is a word}. For example: 

\begin{center}
tiger$\,\,\,\,\,$house$\,\,\,\,\,$number
\end{center}
are English words, but 
\begin{center}
qwertyuiop$\,\,\,\,\,$adfgofger$\,\,\,\,\,$grrooafaf$\,\,\,\,\,$bbbbb$\,\,\,\,\,$auauau
\end{center}
are not. 

Now, look at the following list, and try to decide for each item 
whether or not it is an English word:

\begin{minipage}{4.5cm}

\parindent=0pt
\null
\vspace{1mm}
linguini

chutzpah

shlemiel

aloha

luau

moo

oink
\vspace{1mm}
\end{minipage}
\begin{minipage}{4.5cm}

\parindent=0pt
\null
\vspace{1mm}
brillig

tove

gyre 

gimble

wabe

mimsy

borogove
\vspace{1mm}
\end{minipage}
\begin{minipage}{4.5cm}

\parindent=0pt
\null
\vspace{1mm}
deoxyribonucleic 

hypercontractivity

subriemannian

neoclintonite

ultrareaganite

gingrichian

interpersonality 

\vspace{1mm}
\end{minipage}

You are not sure in all cases, right?  So this makes my point: 
{\bf the concept of ``word'' in Written English is fuzzy, not precise.}
In many cases, it is not clear whether something counts as a word or not. 
Reasonable people could disagree, and they would have no way to settle
their difference. 

You might think that one can solve these problems by resorting to the
dictionary. But, if you have ever played {\it Scrabble}, you know that
sometimes the dictionary will not really answer your questions. 
For example, the dictionary does not have all possible inflections of a
word, so for example you will find ``like'' but you will not find
``likes.''  You have to {\sl reason} and use your knowledge of grammar to
know that ``likes'' is the third person singular of the present tense of
the verb ``to like,'' and is a perfectly fine word.

Should I talk about spell-checkers here? 

\medbreak Now suppose we look more complex items, such as {\bf sentences.}


\vskip 0.7cm
\vbox{

\centerline{{\mymagtwobf 4.\ PROPOSITIONAL CALCULI}}

\bigbreak\noindent In a propositional calculus, the {\bf symbols} are:

\medbreak
\bi
\item{1.} the {\it propositional variables} (usually letters such as $P$,
$Q$, $R$, $S$),

\item{2.} the {\it logical connectives} $\vee$, $\wedge$, $\imply$, $\sim$,

\item{3.} the {\it parentheses} $\,\,\,(\,\,\,$ and $\,\,\,)\,\,\,$.
\ei

}

\medbreak\noindent The well-formed formulas are also called 
{\bf propositional forms.} The {\bf rules} determining which strings are
well-formed formulas (i.~e. propositional forms) are:

\medbreak\noindent RULE 1: One-letter strings consisting of a propositional
variable are well-formed formulas. 

\medbreak\noindent RULE 2: If $A$ is a well-formed formula then 
$(\sim A)$ is a well-formed formula.

\medbreak\noindent RULE 3: If $A$ and $B$ are well-formed formulas then 
$(A\wedge B)$, $(A\vee B)$, and $(A\imply B)$ are  well-formed formulas.

\medbreak\noindent RULE 4: Only those strings obtained by applying 
Rules 1, 2 and 3 are well-formed formulas. 

\medbreak\noindent For example: 

$(((P\vee Q)\wedge(\sim Q))\imply(P\vee Q))$ is well formed, 
because one can arrive at it by applying the rules as follows:

\medbreak\noindent STEP 1: $P$ and $Q$ are well formed, by Rule 1.

\medbreak\noindent STEP 2: Applying Rule 2 with $Q$ in the role of $A$,
we see that $(\sim Q)$ is well formed. 

\medbreak\noindent STEP 3: Applying Rule 3 with $P$ in the role of $A$ and
$Q$ in the role of $B$, we see that $(P\vee Q)$ is well formed. 

\medbreak\noindent STEP 4: Applying Rule 3 with $(P\vee Q)$ in the role of
$A$ and $(\sim Q)$ in the role of $B$, we see that $((P\vee Q)\wedge(\sim Q))$
is well formed. 

\medbreak\noindent STEP 5: Applying Rule 3 with $((P\vee Q)\wedge(\sim Q))$
in the role of $A$ and $(P\vee Q)$ in the role of $B$, we 
we see that $(((P\vee Q)\wedge(\sim Q))\imply(P\vee Q))$ is well formed.

\medbreak\noindent On the other hand, I claim that 
$((P\wedge Q)\imply Q$ is {\it not} well formed. Once again, as in the
previous section, it is a good exercise to try to understand {\it why}.
First of all, you can try to produce $((P\wedge Q)\imply Q$ using the three
rules, and no matter how hard you try you will not be able to.  But,
{\mymagonerm just because you have not been able to do something it does
not follow that it cannot be done.}  So, how do we know that $((P\wedge
Q)\imply Q$ {\bf cannot} be obtained using the rules?  We need an {\it
argument}. We need {\it reasoning}.  We need a {\bf proof.}

Here is a proof. First, we prove a small theorem.

\medbreak\noindent THEOREM 1. {\sl In every well-formed formula the number
of left parentheses is equal to the number of right parentheses.}

\medbreak\noindent PROOF.  Call a string ``balanced'' if it has the same
number of left and right parentheses. Any string obtained using Rule 1 is
balanced, because it has zero left parentheses and 
zero right parentheses. Any application of Rule 2 will add an extra left
parenthesis and an extra right parenthesis, so when you apply Rule 
2 to a balanced string you get a balanced string.  
In any application of Rule 3, the number of left
parentheses in the result is that of $A$, plus that of $B$, plus one, 
and the same is true for the number of right parentheses. So when you apply
Rule  3 to two balanced strings you get a balanced string.   So the
application of Rules 1, 2, 3 can never create unbalanced strings. END OF
PROOF. 

\medbreak\noindent{\bf Remark.}  Theorem 1 gives an example 
of a {\bf necessary condition} for a string to be well formed.  (We say
that $A$ is a {\it necessary condition} for $B$ if $A$ happens whenever $B$
does, that is, if $B$ implies $A$.  For example: ``he took the final exam''
is a necessary condition for ``he passed the course,'' because if he
passed the course then he must have taken the final exam. But is of course
not a ``sufficient condition.'' Just taking the final exam is not enough to
pass the course.)  The theorem says that if a string is well formed then it
has to have the same number of left and right parentheses. This condition
is not sufficient:  there are many strings that have 
the same number of left and right parentheses but are not well formed.  
(For example, the string ``$P)($''.)   How can we tell if a string is well
formed? Even 
more precisely, can one give a {\bf computer algorithm} to check?  
The answer is ``yes,'' but this is a hard thing to do, and 
if you are interested you should look it up in a mathematical logic 
book. \proofend



\medbreak\noindent PROBLEM 4. 
Is $(((P\vee Q)\wedge\sim Q)\imply(P\vee Q))$ well formed?  Justify your
answer. 

\medbreak\noindent PROBLEM 5. 
Is $(((P\vee Q)\wedge(\sim Q))(\imply(P\vee Q)))$ well formed?  Justify your
answer. 

\bigbreak\noindent When we work with a propositional calculus, we usually
do not want to have to write too many parentheses. So we add the

\medbreak\noindent PARENTHESIS DELETION OPTION:   If there is only one way
to put the parentheses back in to produce a well-formed formula, then you
are allowed to omit them. 

\medbreak\noindent For example, you can write 
$P\wedge Q$ rather than $(P\wedge Q)$, or $\sim P$ rather than $(\sim P)$. 

\medbreak\noindent{\bf Remark}. The book, in the bottom half of page 6, 
gives you extra leeway to omit parentheses. For example, the book gives you
{\bf extra rules} so that, if you write $\sim P\wedge Q$, it is understood
that this stands for $((\sim P)\wedge Q))$. (In principle, it could also
have stood for $(\sim (P\wedge Q))$.  The rules given in the book exclude
this second possibility, because ``$\sim$ applies to the smallest
proposition following it.'')  

Notice that {\bf if you want to make the formulas simpler by allowing
deletion of lots of parentheses, as the book does, then the rules become
more complicated.} In my own personal view, it is better to keep the rules
simple, even at the cost of writing formulas with more parentheses. {\bf
You are free to do it my way, or the book's way, as you prefer, long as you
do it correctly.} \proofend

\sepa

\vspace{0.7cm}

\begin{center}
{\mymagtwobf 1.\ FIRST-ORDER PREDICATE CALCULI}

\end{center}

\medbreak\noindent We have been talking about ``proofs'' from the
beginning of the course, but we have not yet given a precise explanation of
what a proof is.   The goal of this discussion will be to show how the 
notion of ``proof'' can be made precise.

We ``prove'' statements. A ``proof'' will consist of writing 
several statements, one after the other, in such a way that every time 
we write a new statement we can fully justify putting  it there, by
invoking the {\bf rules of logic}.  

The rules of logic tell us when we can go from some statements (the
premisses) to another statement (the conclusion) just because of 
the {\bf form} if the statements involved, irrespective of the {\bf
meaning}, and even irrespective of whether we know what the statements
mean.  For example, suppose that I give you as ``axioms''
the following two statements:

\medbreak\noindent Axiom A: $\quad(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$, 

\medbreak\noindent Axiom B:.$\quad G(a,b)$.

\medbreak\noindent(An ``axiom'' is something that we accept as true. 
This means that in a proof you are always allowed to bring it in.) 

Suppose that I then write 

\begin{center}
(P)$\quad\quad$
\fbox{
\begin{minipage}{11cm}

\medbreak\noindent 1.$\quad(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$, 

\medbreak\noindent 2.$\quad G(a,b)$, 

\medbreak\noindent 3.$\quad G(a,b)\imply(a\diamond b)\oplus a=47$, 

\medbreak\noindent 4.$\quad(a\diamond b)\oplus a=47$.

\end{minipage}
}
\end{center}

\medbreak\noindent Then the four steps in (P) can be justified as
follows:

\bi
\item{} Step 1 is valid because $(\forall x)(\forall y)
(G(x,y)\imply(x\diamond y)\oplus x=47)$ is an axiom.

\item{}  Step 2 is valid because $G(a,b)$ is an axiom. 

\item{}  Step 3 is valid because $G(a,b)\imply(a\diamond
b)\oplus a=47$ follows from the formula of Step 1 by ``specialization''
(that is, pluggin in particular values for the variables).

\item{}  Step 4 is valid because $(a\diamond b)\oplus a=47$
follows from the formulas of Steps 2 and 3 by ``Modus Ponens'' (that is, 
the rule that allows you to go to $Q$ whenever you have $P\imply Q$ and
$P$). 

\ei

\medbreak So $\quad$\fbox{{\bf (P) is a proof}}.

\medbreak Notice that we did not say anything about the meaning of the
symbols $\diamond$ and $\oplus$.  You probably don't know what they mean
and---in case you wonder---I haven't the faintest idea either. And neither
you nor I know what $a$ and $b$ are. But {\bf it does not matter.} 
The validity of (P) as a proof depends only on the {\bf form} of the
sentences involved, and you don't need to know what they mean.  
(We already discussed examples of this situation earlier. I you know that 
``all borogoves are mimsy,'' and ``Mr. T is a borogove,'' then 
you can conclude that ``Mr. T is mimsy.'' It doesn't matter what ``mimsy''
means, or what a ``borogove'' is. (Nor does it matter who Mr. T is.) 

\bigbreak To make the notion of {\sl proof} precise we have to start
by making the notion of {\sl (logical) form} precise. The things that have a 
logical form are {\sl statements}, so we have to start by giving a precise 
definition of {\sl statement}.  

At this point we have a choice: we can either

\bi
\item{} try to define precisely what a ``statement'' in the English
language is, 
\ei
or
\bi
\item{} create a language of our own, for which it will be easier to 
define precisely what a ``statement'' is.
\ei
The first option is basically hopeless. We have already discussed, eariler
in the course, the notion of statement. We said things like:

\begin{center}
\begin{minipage}{11cm}
A {\bf sentence} is a complete unit of thought. 

Sentences can be declarative, interrogative, exclamatory or imperative.

A {\bf statement} is a declarative sentence. 

A {\bf statement} is a declarative sentence that has a definite truth value. 
\end{minipage}
\end{center}

\medbreak\noindent Those are impressive words, but {\bf do they really add
up to a {\large truly precise} definition of ``statement''?}

For example: {\bf would it be possible to write a computer program to check
whether a string of English words is a statement?}  How would you tell the
computer to check whether a string of words is a statement, using the above
explanation?  

In case you believe that this can be done, here are some observations 
that ought to change your mind.  First of all, ask yourself whether you
would accept 

\begin{center}
\begin{minipage}{11cm}
This book are very interesting.
\end{minipage}
\end{center}

\medbreak\noindent as a statement. I am sure your answer is ``no,'' because
``this book are very interesting'' is {\sl grammatically incorrect.}  The
verb (``are'') ought to agree in number with the subject (``this book'')
but in this example it does not. (Had we said ``this book is very
interesting,'' or ``these books are very interesting,'' then we would have
had real statements.)

So you probably agree by now that, if our computer program is going to
check whether or not something is a statement, then it should first be 
able to check if it is grammatically correct.  

And we have opened a can of worms! First of all, the notion of {\sl
grammatical correctness} is itself fuzzy. (Are split infinitives O.K.?) 
Second, ask yourself whether you 
would accept 

\begin{center}
(S)$\quad\quad\quad\quad$
\begin{minipage}{10cm}
I don't know if George Eliot was a good writer because I have not read any
of her books.
\end{minipage}
\end{center}

\medbreak\noindent as a statement. Yes or no? ``George'' is a man's name,
so talking about ``her'' books seems to be grammatically wrong. However, if
you know that {\bf George Eliot was a woman}, then you realize that there
is no problem.

So, for our imaginary computer program to be able to tell whether (S) is a
statement, it would have to know, for example, that George Eliot was a
woman. This kind of knowledge goes far beyond knowing a few rules, and
requires a sophisticated and extensive knowledge of the full context in
which our language operates, of history and of literature.

\medbreak In any case, if you are not convinced that one cannot give 
a precise definition of ``statement'' in the English language---a
definition so precise that one can actually translate it into a 
computer program for checking whather something is a statement---then try to
come up with one.      I doubt that you will be able to. 

\medbreak English is an example of a {\bf natural language}, that is, one
of that many languages that evolved in different parts of the world over
thousands of years. Nobody directed this evolution with the idea of making
those languages amenable to the precise study that is required in logic. 

So mathematicians and logicians have chosen a more realistic solution: 
we create our own languages, and do it in such a way that we can 
define in a completely precise way the notions that interest us, such as 
``statement'' and ``proof.'' 

The situation here is very much the same as with computer programming.  Ask
yourself: why is it that when I want to write a computer program I have to
use a special programming language? Why can't I just write programs in
ordinary English?  The answer is that ordinary English is too fuzzy, too
imprecise for a computer to be able to deal with it and understand it.

\sepa

The {\bf formal languages} that we will be using here are 
{\bf predicate calculi}. (``Calculi'' is the plural of
``calculus.'') In these languages, 
we have letters such as $x$, $y$, $a$, $b$, etc., that stand for 
specfic ``things,'' or are ``variables'' (that is, slots where you can plug
in things). These letters can be combined by means of {\bf function
symbols} to write {\bf terms}, which are more complicated expressions 
that also stand for things.  (For example, $6.(3+5)$ is a term.)
Finally, we will have {\bf predicate symbols} that will enable 
us to make assertions about the terms.   (For example, using the predicate 
symbol $>$ we can combine the terms $6.(3+5)$ and $7$ and make the 
assertion that $6.(3+5)>7$.) The simplest such assertions are 
the {\bf atomic formulas}. These can be combined by means of {\bf logical
connectives} to build even more complicated formulas. (For example, 
$6.(3+5)>7$ and $7>5$ are atomic formulas. We can combine them to get 
the formula $6.(3+5)>7\wedge 7>5$, which is no longer atomic. The symbol
$\wedge$ is therefore used to {\sl connect} two formulas, which is why it
is called a {\sl connective.} Another example:  $x=3+z$ and $y<5$ are
atomic formulas, and we can combine them into 
$(\forall x)(x=3+z\wedge y<5)$, which is a formula obtained in two steps,
by first forming  $x=3+z\wedge y<5$---that is, using the logical connective
$\wedge$---and then forming  $(\forall x)(x=3+z\wedge y<5)$---that is, 
using the logical connective $\forall$, known as the {\bf universal
quantifier}---to produce a {\bf quantified version} of 
$x=3+z\wedge y<5$.)

To make all this precise, we will run into a lot of trouble because of the
parentheses. To see why, look at the formula 
\begin{equation}\label{flaac}
(\forall x)(x=3+z\wedge y<5)\,\,.
\end{equation}
Is (\ref{flaa}) obtained by 

\bi
\item[(A)] ``quantifying'' the formula 
\fbox{$x=3+z\wedge y<5$}$\,$?  
\ei
or by
\bi
\item[(B)] ``quantifying'' the formula \fbox{$(x=3+z\wedge y<5)$}$\,$?
\ei

Suppose we choose the possibility (A).  This would mean that we are
counting the expression $x=3+z\wedge y<5$ as a formula, and surrounding it
with parentheses is part of the quantification process.  So the 
``universal quantification rule'' would have to say:

\begin{center}
(UQR-Version A)$\quad$
\fbox{
\begin{minipage}{9.5cm}
If $F$ is a formula and $v$ is a variable then 
$(\forall v)(F)$ is a formula.  
\end{minipage} 
}
\end{center} 
This would have some awful consequences.  Just to give you an example, let
us look at what would happen when you want to quantify two or three times.
Suppose, for example, you want to state the associative property of
addition of real numbers, which says that ``for all $x$, $y$, $z$, the
equality $(x+y)+z=x+(y+z)$ holds.''  When you quantify, each time you add a
new quantifier you would have to surround the formula being quantified
with parentheses.  So you would get 
$(\forall z)((x+y)+z=x+(y+z))$ in the first step, then 
$(\forall y)((\forall z)((x+y)+z=x+(y+z)))$ in the second step,
and after the third and last step 
you would end up with the ugly expression
$$
(\forall x)((\forall y)((\forall z)((x+y)+z=x+(y+z))))\,\,.
$$

So it seems that choosing (A), and using  (UQR-Version A) for the 
universal quantification ruleis not a very good idea. So let try to follow
(B) instead. This means that we need the rule

\begin{center}
(UQR-Version B)$\quad$
\fbox{
\begin{minipage}{9.5cm}
If $F$ is a formula and $v$ is a variable then 
$(\forall v)F$ is a formula,  
\end{minipage} 
}
\end{center} 
which could then be applied, for example, to the 
formula $((x+y)+z=x+(y+z))$. Now we will get 
$(\forall z)((x+y)+z=x+(y+z))$ in the first step, then 
$(\forall y)(\forall z)((x+y)+z=x+(y+z))$ in the second step,
and after the third and last step 
we would end up with the much nicer expression
$$
(\forall x)(\forall y)(\forall z)((x+y)+z=x+(y+z))\,\,.
$$
This looks fine, but we have to pay a price, namely, we have to call
$((x+y)+z=x+(y+z))$ a formula but we are not allowed to call
$(x+y)+z=x+(y+z)$ a formula.   This is also quite unpleasant. 

We are going to take care of this problem by insisting that the first time
we quantify we have to put parentheses, but then when we quantify again, we
do not.  This will require that we first define the notion of a
``quantified version'' of a formula before we go on.  


\sepa

We now do the (boring) details and define 
precisely what a {\bf predicate calculus} is.


First of all, a predicate calculus involves a list of 
{\bf basic symbols} (or ``primitive symbols''). 

\begin{center}
\fbox{
\begin{minipage}{11cm}
\begin{center}
{\large{\bf THE BASIC SYMBOLS\\
OF A FIRST-ORDER PREDICATE CALCULUS}}
\end{center} 
A first-order predicate calculus uses {\bf five} kinds of basic symbols: 
\bi
\item[1.] individual symbols\footnote{WARNING: the phrase ``individual
symbol'' is not meant to refer to ``a signle symbol,'' as in the 
statement ``if you only look at the individual symbols then you might not 
understand the words.''  ``Individual symbol'' here means 
``symbol for an individual,'' in the same way as further down 
we use ``predicate symbol'' in the sense of ``symbol for a predicate.'' 
In other words, ``individual'' here is a {\bf noun}, not 
an {\bf adjective.}}$\,\,$;
\vspace{-2.5mm}
\item[2.] function symbols$\,\,$;
\vspace{-2.5mm}
\item[3.] predicate symbols$\,\,$;
\vspace{-2.5mm}
\item[4.] the logical connectives $\forall$, $\exists$, $\wedge$, $\vee$,
$\Rightarrow$, $\Longleftrightarrow$, $\sim$$\,\,$;
\vspace{-2.5mm}
\item[5.] the left and right parentheses, ``$($'' and ``$)$,'' and the
comma ``,''$\,\,$.  
\ei
\end{minipage} 
}
\end{center} 

\medbreak\noindent {\bf INDIVIDUAL SYMBOLS} are usually letters such as
$x$, $y$, $z$, $u$, $r$, $a$, $b$, $c$, $n$, $p$, $q$, 
$\ldots$. In principle, you can use {\sl any symbol} to be an individual,
even if it is not a Roman letter. For example, you can use Greek
letters ($\alpha$, $\beta$, $\varepsilon$, $\pi$, etc.), Hebrew letters
(e.~g. $\aleph$), other symbols such as 
$\diamondsuit$, $\clubsuit$, or $\#$, 
and even longer and more complicated expressions such as
$GNU$, $grrr$,  $oink$, $my$-$uncle$-$Jimmy$, etc. 

An individual symbol is either a {\it constant} or a {\it variable}, and
can never be both at the same time.  Whether a symbol is a {\it constant}
or a {\it variable} depends on the context. There are various ways to {\it
declare} something to be a constant or a variable:

\bi
\item[1.] If within a proof you say ``Let $\#$ be $\ldots$''  then you are
declaring  $\#$ to be a constant until the end of the proof. (For example: 
once you say ``Let $a=3$'' then from that moment on $a$ is a constant. 
If you say ``Let $\varepsilon$ be arbitrary'' then from then on 
$\varepsilon$ is a constant. (The same applies to ``choose'' or ``pick.'') 

\item[2.] Any symbol that appears right next to a quantifier 
is a variable. For example, if you see the formula $(\forall x)(\exists
y)(b(x,y))$, then you know that $x$ and $y$ are variables. 
If I write $(\forall x)(x^2\ge 0)$, then 
$x$ {\sl has} to be a variable, because only variables can appear in
quantifiers such as $(\forall x)(x^2\ge 0)$. 

\item[3.] Some symbols are {\it reserved constants}, that is, declared to be
constants once and for all, and intended to have a specific value that is
fixed for ever. For example, in real-number
arithmetic the symbols $0$, $1$, $2$, $\pi$, $e$, 
are fixed constants. In set theory, the symbol $\emptyset$ is a fixed
constant. 
\ei

\medbreak\noindent At each point, it is important to know  whether a symbol
is a constant or a variable, because this helps you know what you can and
what you cannot do with it.  

\medbreak For example, after you have declared
``$a$'' to be a constant by saying, for example, ``let $a$ be an arbitrary
real number,'' {\sl you cannot use quantifiers such as $(\forall a)$},
because $a$ is now fixed, not variable.

\medbreak Similarly, if a symbol is a variable, then there 
are things you cannot do with it.  For example, after 
you have said in a proof $(\forall x)(\forall y)(x+y=y+x)$, 
you cannot write in your proof something like $x+y=y+x$, 
because {\sl this is not a statement}.  (This means that, if $x$ and $y$
are variables, then ``$x+y=y+x$'' does not have  
a definite truth value. It doesn't make sense to say that 
$x+y=y+x$'' is true or false unless $x$ and $y$ are given specific 
values.    And, if you give $x$ and $y$ specific, definite values, then
they become constants.) 

\medbreak\noindent{\bf FUNCTION SYMBOLS} are letters such as $f$, $g$,
$F$, $G$, or other symbols such as $+$, $\cdot$, $\cap$, $\cup$,
$\in $, $\sim$. Each function symbol admits a {\sl fixed} number of arguments.
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} symbol. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} symbol.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} symbol. And so on.)   If you write 
a function symbol, followed by a ``$($,'' then as many arguments
as the function symbol admits, separated by commas, and then 
``$)$,'' you get a {\sl term}.  The strings that can be used 
as arguments are variables, or constants, or more general terms.

In other words: 

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``TERM''}}
\end{center}

\medbreak\noindent RULE T1: constants and variables are terms;

\medbreak\noindent RULE T2: if $t_1,\ldots,t_k$ are terms 
and $F$ is a $k$-variable function symbol, then $F(t_1,\ldots,t_k)$ is a
term. (For example: if $F$ is a one-variable function symbol and $t$ is a
term, then $F(t)$ is a term; if $F$ is a two-variable function symbol and
$t$, $s$ are terms, then $F(t,s)$ is a term; 
if $F$ is a three-variable function symbol and
$t$, $s$, $u$ are terms, then $F(t,s,u)$ is a term.) 

\medbreak\noindent RULE T3: only those expressions obtained by repeated
application of Rules T1 and T2 are terms. 
\vspace{2mm}
\end{minipage}
}
\end{center}

\medbreak\noindent Later, when we discuss how to {\it interpret} 
(that is, assign a meaning), function symbols will correspond to {\it
functions}, that is, expressions that have a {\bf definite value}
once you plug in values for its arguments.  For example, if we 
use a binary function symbol $plus$ to stand for ``sum of two numbers,''
then $plus(3,5)$ will have the value $8$, $plus(x,3)$ will have a value 
for each value of $x$ (so $plus(x,3)$ will be $7$ if $x$ is $4$, 
$2$ if $x$ is $-1$, and so on) and $plus(x,y)$ will have a value 
for each choice of values of $x$ and $y$ 
(so $plus(x,y)$ will be $7$ if $x$ is $4$ and $y$ is $3$,  
$2$ if $x$ is $-4$ and $y$ is $6$, and so on). 

\medbreak\noindent {\bf EXAMPLE 1.} Let us start setting  up a 
predicate calculus to talk about {\bf real number arithmetic}.  We
introduce the two-variable function symbols $plus$, $times$, $minus$,
$max$, $min$, and two one-variable function symbols $negative$-$of$ and
$abs$. (``$max$'' is meant to stand for ``the largest of,'' so for example
$max(5,3)=5$ and $max(-7,2)=2$. Similarly,
``$min$'' is meant to stand for ``the smallest of,'' so for example
$min(5,3)=3$ and $min(-7,2)=-7$.)

In addition, we will have {\bf reserved constants}
for the usual names of the nonnegative integers,
in decimal notation (for example: $0$, $1$, $2$, $3$, $4$, etc). 

Then the following are terms:

\bi
\item[1.] $plus(x,y)$, 

\item[2.] $minus(plus(x,y),z)$, 

\item[3.] $abs(x)$, 

\item[4.] $plus(min(x,1),abs(plus(x,y)))$, 

\item[5.] $min(plus(x,y),abs(y))$, 

\item[6.] $min(plus(x,abs(y)),negative$-$of(y))$, 

\item[7.] $plus(max(x,27),min(plus(x,y),abs(y)))$, 

\item[8.] $plus(abs(min(plus(x,y),abs(y))),max(plus(x,y),abs(y)))$.
\ei

\medbreak\noindent It is common in mathematics to use $+$ rather than
$plus$, and write $t+s$ rather than $plus(t,s)$.  Similarly, one uses  
$-$ rather than $minus$ and $.$ rather than $times$,  
so we write $t-s$ rather than $minus(t,s)$ and 
$t.s$ rather than $times(t,s)$.  Also, for the one-variable symbol
$abs$, we use $|\ldots|$ instead, and write $|x|$ instead of 
$abs(x)$, and for the one-variable symbol
$negative$-$of$, we use $-$ instead, and write $-x$ instead of 
$negative$-$of(x)$. 

\begin{center}
\fbox{
\begin{minipage}{11cm}
{\bf The same symbol $-$ is used 
for two different things: the one-variable function $negative$-$of$, 
and the two-variable function $minus$.  For example, when we write
$5-8=-3$, the first ``$-$'' is the two-variable function symbol, and the
second ``$-$'' is the one-variable function symbol.} 
\end{minipage}
}
\end{center}

With these conventions, we would write

\medbreak

\pltw{1}{$x+y$}{$plus(x,y)$,} 

\pltw{2}{$(x+y)-z$}{$minus(plus(x,y),z)$,} 

\pltw{3}{$|x|$}{$abs(x)$,} 

\pltw{4}{$min(x,1)+|x+y|$}{$plus(min(x,1),abs(plus(x,y)))$,}

\pltw{5}{$min(x+y,|y|)$}{$min(plus(x,y),abs(y))$,}

\pltw{6}{$min(x+|y|,-y)$}{$min(plus(x,abs(y)),negative$-$of(y))$,} 

\pltw{7}{$max(x,27)+min(x+y,|y|)$}{$plus(max(x,27),min(plus(x,y),abs(y)))$,}

\pltw{8}{$|min(x+y,|y|)|+max(x+y,|y|))$}{$\,\,$}

$plus(abs(min(plus(x,y),abs(y))),min(plus(x,y),abs(y)))$.

\medbreak\noindent If you want to, you could get the terms to look even nicer
by also agreeing to use $\wedge$ rather than $min$ and $\vee$ rather than
$max$, so you would write $t\wedge s$ instead of $min(t,s)$ and $t\vee s$
instead of $max(t,s)$.  With these new symbols, one would write

\pltw{a}{$(x\wedge 1)+|x+y|$}{$plus(min(x,1),abs(plus(x,y)))$,}

\pltw{b}{$(x+y)\wedge |y|)$}{$min(plus(x,y),abs(y))$,}

\pltw{c}{$(x\vee 27)+((x+y)\wedge |y|)$}{$plus(max(x,27),min(plus(x,y),abs(y)))$,}

\pltw{d}{$|((x+y)\wedge|y|)|+((x+y)\vee|y|)$}{$\,\,$}

$plus(abs(min(plus(x,y),abs(y))),min(plus(x,y),abs(y)))$.


\medbreak\noindent {\bf EXAMPLE 2.} Let us start setting  up a 
predicate calculus to talk about {\bf natural number arithmetic}.  We
introduce the two-variable function symbols $plus$, $times$, 
$max$, $min$, $GCD$, $LCM$. (``$GCD$'' and ``$LCM$'' are meant to stand for
``greatest common divisor'' and ``least common multiple,'' so for example
$GCD(12,20)=4$ and $LCM(12,20)=60$.) 

In addition, we will have {\bf reserved constants} for the usual names of
the natural numbers, in decimal notation (for example: $1$, $2$, $3$, $4$,
etc). ({\sl Remember that $0$ is {\bf not} a natural number}.)

Then the following are terms:

\def\pltw#1#2#3{\noindent#1.$\,\,\,$\hbox to 5cm{#2\hfill}$\qquad$instead
of\hbox to 6.5cm{$\quad$#3\hfill}} 

\bi
\item[1.] $plus(x,y)$, 

\item[2.] $GCD(plus(x,5),z)$, 

\item[3.] $plus(min(x,1),LCM(plus(x,y),83)))$. 
\ei

\medbreak\noindent As in Example 1, we use $+$ rather than
$plus$, and write $t+s$ rather than $plus(t,s)$.  Similarly, we use  
$.$ rather than $times$,  
so we write $t.s$ rather than $times(t,s)$.  

With these conventions, we would write

\medbreak

\pltw{1}{$x+y$}{$plus(x,y)$,} 

\pltw{2}{$GCD(x+5,z)$}{$GCD(plus(x,5),z)$,}

\pltw{3}{$min(x,1)+LCM(x+y,83)$}{$plus(min(x,1),LCM(plus(x,y),83)))$.}

\medbreak\noindent You could get the terms to look even nicer
by also agreeing to use $\wedge$ rather than $min$ and $\vee$ rather than
$max$, so you would write $t\wedge s$ instead of $min(t,s)$ and $t\vee s$
instead of $max(t,s)$.  With these new symbols, you could write, for
example,

\medbreak

$\quad x\wedge 1+LCM(x+y,83)\quad$ instead of $\quad
plus(min(x,1),LCM(plus(x,y),83)))$. 

\bigbreak\noindent {\bf EXAMPLE 3.} Let us start setting  up a 
predicate calculus to talk about {\bf kinship}.  We introduce two 
one-variable function symbols $f$, $m$, that will be thought of as standing
for ``father of'' and ``mother of.'' Suppose we have constants such as 
$Bill$, $John$ and $Mary$, and variables $x$, $y$, $z$. 
Then, for example, the following are terms:

\bi
\item[1.] $f(x)$, 

\item[2.] $m(y)$, 

\item[3.] $f(m(Bill))$, 

\item[4.] $f(m(x))$,

\item[5.] $f(f(f(f(f(f(Mary))))))$, 

\item[6.] $m(m(m(m(m(m(Mary))))))$, 

\item[7.] $m(f(m(f(f(m(Mary))))))$.
\ei

\medbreak\noindent The term $f(Bill)$ would have a definite value, namely, 
Bill's father.  The term $f(x)$ does not have a definite value 
unless you are told what the value of $x$ is.  If $x$ is Bill then 
the value of $f(x)$ is Bill's father. If $x$ is Mary then 
the value of $f(x)$ is Mary's father. 

\medbreak\noindent QUESTION: could we introduce function symbols to stand 
for ``son,'' ``daughter,'' ``brother'' or ``sister''?  The answer is {\bf
NO!} Why? Because {\bf ``son,'' ``daughter,'' ``brother'' and ``sister''
are not functions.} They do not have a definite value for each value of the
argument. For example,  $son(x)$ might fail to have a value 
if $x$ is Bill, because it could happen that Bill does not have a son, and
if this is the case then you don't know what $son(Bill)$ means.
Also, $son(x)$ might fail to have a {\it definite} value 
if $x$ is Bill, because it could happen that Bill has more than one son,
and then once again you would not know what $son(Bill)$ means.

As we will see later, functions that sometimes fail to have a value are
called {\bf partial functions}, and do not pose much of a problem.  For
example, in arithmetic we use two-variable partial functions such as
$over$, and write $\frac{x}{y}$ rather than $over(x,y)$.  As you know, the
expression $\frac{x}{y}$ does not have a value for every $x$ and every
$y$. (For example, $\frac{1}{0}$ does not have a value.)  Another example
is the one-variable partial function $square$-$root$ (for which we write
$\sqrt{x}$ rather than $square$-$root(x)$). Again, $\sqrt{x}$ does not have
a value for all real numbers $x$. (For example, $\sqrt{-1}$ does not have a
value.)  When we are dealing with a partial fraction, we just have to be
careful and make sure that we never plug in the ``wrong'' values of the
arguments.

On the other hand, ``functions'' that have more than one value pose a very
serious problem, and for that reason we do not want to allow them at all. 
For example, suppose we defined $\frac{x}{y}=z$ to mean 
the same as $z.y=x$, without any extra restrictions. Then we would 
have, for example, $1=\frac{0}{0}$---because $1.0=0$---and 
$0=\frac{0}{0}$---because $0.0=0$---so we would be able to conclude that 
$1=0$, which is very bad.    So what we will do instead is define 
``$\frac{x}{y}=z$'' to mean the same as ``$z.y=x\wedge y\neq 0$.''
This will take care of our problem, because now if 
$\frac{x}{y}=z$ and $\frac{x}{y}=w$ it will follow that 
$z=w$, so we will not get into any trouble.

So, returning to our kinship example, the real trouble with ``functions'' 
such as $brother$ or $sister$ is that people often have more than one
brother or sister.  So, for example, if Bill has two sisters, Jane and
Clare, and I was allowed to write things like
$Jane=sister(Bill)$ and $Clare=sister(Bill)$, I would be able to conclude
that $Jane=Clare$, which is not right.  \proofend

\begin{center}
\fbox{
\begin{minipage}{13cm}
\begin{center}
{\large {\bf Are $\frac{1}{0}$ and $\frac{0}{0}$ ``undefined''?}}
\end{center}

\medbreak\noindent Students often say things like 
``$\frac{1}{0}$ is undefined,'' or ``$\frac{0}{0}$ is undefined,''
or ``$\sqrt{-1}$ is undefined.''   Let us think this over. To ``define''
something means ``to say what it means.''  And I know perfectly well 
what $\sqrt{-1}$ means. In general, if $x$ is any real number, 
``$\sqrt{x}$'' means ``the number $y$ such that $y\ge 0$ and $y.y=x$.''  
So ``$\sqrt{-1}$'' means ``the number $y$ such that $y\ge 0$ and $y.y=-1$.''  
This is perfectly clear to me, and should be perfectly clear to you. 
So don't go on saying that ``$\sqrt{-1}$ is undefined.''   The trouble
with $\sqrt{-1}$ is {\bf not} that it is ``undefined,'' but that it does
not exist.  

Similarly, ``$\frac{x}{y}$'' means ``the number $z$ such that 
$z.y=x\wedge y\neq 0$.''    So ``$\frac{1}{0}$''  
means ``the number $z$ such that $z.0=1\wedge 0\neq 0$.'' 
And ``$\frac{0}{0}$''  
means ``the number $z$ such that $z.0=0\wedge 0\neq 0$.'' 
In both cases, the requirements on $z$ are very clear, so you cannot say
that $\frac{1}{0}$ or $\frac{0}{0}$ are ``undefined.''  What you {\bf
should} say  is 

\medbreak
\begin{center}
{\bf $\frac{1}{0}$ and $\frac{0}{0}$ do not exist.}
\end{center}
\medbreak
$\,\,$
\end{minipage}
}
\end{center}


\begin{center}
\fbox{
\begin{minipage}{12cm}

\noindent {\bf Six-legged chickens are not ``undefined.'' You now very well
what ``six-legged chicken'' means!  The problem with six-legged chickens is
not that they are ``undefined.'' It's that there aren't any.  
SIX-LEGGED CHICEKNS DO NOT EXIST!}

{\bf Unicorns are not ``undefined.'' I know perfectly well what a unicorn
is. There just aren't any. UNICORNS DO NOT EXIST!}

\end{minipage}
}
\end{center}

\medbreak
\centerline{\hbox to 5cm{\hrulefill}}

\medbreak\noindent In order ot be able to talk about about 
sons and daughters and brothers and sisters we need 
{\sl PREDICATE SYMBOLS}.

\medbreak\noindent {\bf PREDICATE SYMBOLS} are letters such as $p$, $q$,
$P$, $Q$, or other symbols such as $=$, $>$, $<$, $\le$,
$\ge$, $\epsilon$, $\subseteq$, $tall$, $stupid$, $smarter$-$than$. 
Each predicate symbol admits a {\sl fixed}
number of arguments.    
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} predicate. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} predicate.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} predicate.  And so on.)   If you write 
a predicate symbol, followed by a ``$($,'' then as many arguments
as the predicate symbol admits, separated by commas, and then 
``$)$,'' you get an {\sl atomic formula}.  The strings that can be used 
as arguments are {\sl terms}.

More formally, then:

\begin{center}
\fbox{
\begin{minipage}{13cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``ATOMIC FORMULA''}}
\end{center}

\medbreak\noindent RULE A: If $L$ is a $k$-argument predicate symbol,
and $t_1,\ldots,t_k$ are terms, then $L(t_1,\ldots,t_k)$ is an atomic
formula. (For example: if $p$ is a one-variable predicate symbol and $t$ is
a term, then $p(t)$ is an atomic formula; if $p$ is a two-variable
predicate symbol and $t$, $s$ are terms, then $p(t,s)$ is an atomic
formula; if $p$ is a three-variable predicate symbol and $t$, $s$, $u$ are
terms, then $p(t,s,u)$ is aa atomic formula.)
\vspace{2mm}
\end{minipage}
}
\end{center}

\medbreak\noindent Later, when we discuss how to {\it interpret} 
our formulas (that is, how to assign {\sl meanings} to them), predicate symbols
will correspond to {\it predicates}, that is, expressions that have a {\bf
definite truth value} once you plug in values for its arguments.  For
example, if we use a binary predicate symbol $>$ to stand for ``larger
than,'' and write $t>s$ rather than $>(t,s)$, then $3>5$ will have the
truth value $False$, $x>3$ will have a truth value for each value of $x$
(so $x>3$ will be $True$ if $x$ is $4$, $False$ if $x$ is $-1$, and so on)
and $x>y$ will have a value for each choice of values of $x$ and $y$ (so
$x>y$ will be $True$ if $x$ is $4$ and $y$ is $3$, $False$ if $x$ is $-1$
and $y$ is $0$, and so on).

\medbreak\noindent {\bf EXAMPLE 1, continued.} Let us go on 
with our language for {\bf real-number arithmetic}. Now that we know all
about predicate symbols, we introduce symbols for the basic binary
predicates ``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' which are {\sl
binary}. Strictly speaking, we should write $=(x,y)$ to mean ``$x$ is equal
to $y$,'' and $>(3,5)$ to mean ``$3$ is larger than $5$'' (which is false,
of course, but is perfectly meaningful).  But we do what we did for the
function symbols.  Instead of writing $>(3,5)$ we write $3>5$, instead of
writing $=(x,y)$ we write $x=y$, and so on.

Then the following are examples of atomic formulas: 

\bi
\item[1.] $x+y=7$, 

\item[2.] $3+7=x$, 

\item[3.] $x+y+|x+3|=7-y+z$, 

\item[4.] $x+y+|x+3|>7-y+z$,

\item[5.] $|x|=x\wedge(-x)$,

\item[6.] $|x|+5=|((x+y)\wedge|y|)|+((x+y)\vee|y|)$.

\ei
 
\medbreak\noindent {\bf EXAMPLE 2, continued.} Let us go on 
with our language for {\bf natural-number arithmetic}. Now that we know all
about predicate symbols, we introduce symbols for the basic binary
predicates ``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' ``$|$'' which
are {\sl binary}. (Note: ``$|$'' stands for ``divides'', so ``$x|y$'' means
``$x$ divides $y$,'' that is $(\exists z)(z.x=y)$.)  Strictly speaking, we
should write $=(x,y)$ to mean ``$x$ is equal to $y$,'' and $>(3,5)$ to mean
``$3$ is larger than $5$'' (which is false, of course, but is perfectly
meaningful).  But we do what we did for the function symbols.  Instead of
writing $>(3,5)$ we write $3>5$, instead of writing $=(x,y)$ we write
$x=y$, and so on.

Then the following are examples of atomic formulas: 

\bi
\item[1.] $x+y=7$, 

\item[2.] $3+7=x$, 

\item[3.] $7|3.4$,

\item[4.] $x+y+GCD(x,8)>7-y+z$,

\item[5.] $x=x\wedge(-x)$,

\ei
 
\medbreak\noindent {\bf EXAMPLE 3, continued.} Now let us go on with 
the predicate calculus that enables us to talk about kinship.
Now that we have predicate symbols, we could introduce one-variable
predicates $\mu$, $\varphi$---so that 
``$\mu(x)$'' would stand for ``$x$ is male''
and ``$\varphi(x)$'' for ``$x$ is female''---
and lots of two-variable predicates  such as 
$s$, $d$, $c$, $b$, $sis$, $u$, $a$m   so that 
``$s(x,y)$'' would stand for ``$x$ is $y$'s son'',
``$d(x,y)$'' would stand for ``$x$ is $y$'s daughter'',
``$c(x,y)$'' would stand for ``$x$ is $y$'s child,'' 
``$b(x,y)$'' would stand for ``$x$ is $y$'s brother,'' 
``$sis(x,y)$'' would stand for ``$x$ is $y$'s sister,'' 
``$u(x,y)$'' would stand for ``$x$ is $y$'s uncle,'' 
and ``$a(x,y)$'' would stand for ``$x$ is $y$'s aunt.'' \proofend

\sepa

\medbreak\noindent Atomic formulas can be combined with the logical
connectives to yield {\bf formulas}.  For example, let us look at the
following fairly long string of 82 characters, in a predicate calculus 
with three binary predicate symbols, namely, $=$, $s$ and $c$: 
$$
(\exists y)(c(y,y))\vee
(\exists x)(\forall y)(\exists z) (((s(a,x)\wedge s(a,y))\wedge((\sim
\!x\!=\!y)\wedge(\sim\! a\!=\!z))) \wedge ((c(z,x)\wedge c(z,y))\wedge c(b,z)))\,.
$$
Let us call this expression $F$. Then $F$ is {\sl not} an atomic formula,
but it is a formula.  This means that $F$ is obtained from atomic formulas
by combining them via the connectives $\wedge$, $\sim$ and $\exists$.  In
our case, the atomic formulas useud to produce $F$ are: $c(y,y)$,
$s(a,x)$, $s(a,y)$,
$x=y$, $a=z$, $c(z,x)$, $c(z,y)$, and $c(b,z)$.

We now describe precisely the rules that allow us to combine 
atomic formulas in order to make formulas.

\begin{center}
\fbox{
\begin{minipage}{12cm}
\null\vspace{1mm}
\begin{center}
{\large{\bf PRECISE DEFITION OF ``FORMULA''}}
\end{center}

\medbreak\noindent RULE F1: Atomic formulas are a-formulas.

\medbreak\noindent RULE F2: If $F$ is an a-formula or a q-formula then
$\sim F$ is a p-formula. 

\medbreak\noindent RULE F3: If $F$ is a p-formula
$\sim(F)$ is a p-formula.

\medbreak\noindent RULE F4: If $F$ is an a-formula or a q-formula 
and $G$ is an a-formula or a q-formula then 
$F\wedge G$, $F\vee G$, $F\imply G$, and $F\lto G$ are p-formulas.

\medbreak\noindent RULE F5: If $F$ is an a-formula or a q-formula 
and $G$ is a p-formula 
then $F\wedge (G)$, $F\vee (G)$, $F\imply (G)$, $F\lto (G)$, 
$(G)\wedge F$, $(G)\vee F$, $(G)\imply F$, and $(G)\lto F$
are p-formulas.

\medbreak\noindent RULE F6: If $F$ and $G$ are p-formulas, 
then $(F)\wedge(G)$, $(F)\vee (G)$, $(F)\imply(G)$, and $(F)\lto (G)$
are p-formulas.

\medbreak\noindent RULE F7: If $F$ is a q-formula,
and $v$ is a variable, then $(\forall v)F$ and  $(\exists v)F$ are
q-formulas.

\medbreak\noindent RULE F8: If $F$ is an a-formula or a p-formula and
$v$ is a variable, then $(\forall v)(F)$ and  $(\exists v)(F)$ are
q-formulas.

\medbreak\noindent RULE F9: Only those expressions obtained by repeated
application of Rules F1, F2, F3, F4, F5, F6, F7, and F8  are formulas. 
\vspace{2mm}
\end{minipage}
}
\end{center}


\medbreak\noindent Why is $F$ a formula according to our rules? 
In Box GNU we show you in detail how one can construct $F$ following the
rules. 

\bigbreak\noindent Notice how the rules are carefully designed so that you
don't have to put too many parentheses.  For example, the rules allow tou
say $x<y$ and $x<y\vee x>y$ (without parentheses) but then you are required
to put parenthesis to say $(\exists x)(x<y)$ and $\exists x)(x<y\vee
x>y)$. This would appear to mean that we have a rule that says that ``if
$G$ is a formula and $v$ is a variable then $(\exists v)(G)$ and $(\forall
v)(G)$ are formulas.  In particular, if you want to quantify $(\exists
x)(x<y)$ again---universally, say---then you would have to write 
$(\forall y)((\exists x)(x<y))$, with an extra pair of parentheses. 
Our rules do not require the extra parentheses, because once you have
quantified once, then the formula yo get becomes a q-formula, and then when
you want to quantify a second time you apply Rule F7 rather than Rule F8.


(Then 
$c(a,b)\lto(s(a,b)\vee d(a,b))$. Make sure you understand why!)
We could also have the binary predicate ``$=$'' with its usual meaning
(that is, ``$a=b$'' means that $a$ and $b$ are the same person). 
With these symbols, we could say lots of more complicated
things. For example, suppose I want to say ``$a$ is $b$'s uncle.''
I could say this by writing 
$$(\exists x)(\exists y)(\exists z)
\Bigl(s(a,x)\wedge s(a,y)\wedge(\sim x=y)\wedge(\sim a=z)
\wedge c(z,x)\wedge c(z,y)\wedge c(b,z)\Bigr)\,.$$

\bigbreak
\centerline{{\bf TERMS VS. FORMULAS}}

\medbreak\noindent{\bf Terms} are supposed to be 
names of ``things.''  {\bf Formulas} are
expressions that ---when we plug in values for the
open variables--- become sentences or statements.

For example, the following are terms: 

\medbreak
\noindent\hbox to 3.5cm{\hfill $3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to 3.5cm{\hfill $x$}\hbox to 3.5cm{\hfill $5xyz$}

\noindent\hbox to 3.5cm{\hfill $x+3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to
3.5cm{\hfill $x^2+3y+7$}\hbox to 3.5cm{\hfill $\{x:x^2<3y\}$}\ ,

\noindent and the following are formulas:

\medbreak
\noindent\hbox to 3.5cm{\hfill $3<5$}\hbox to 3.5cm{\hfill $3+5=8$}\hbox to 3.5cm{\hfill $x+3=7y$}\hbox to 3.5cm{\hfill $5xyz<x+z$}

\noindent\hbox to 3.5cm{\hfill $(\forall x)(x+3>x)$}\hbox to 3.5cm{\hfill $(\forall x)(x+y>x)$}\hbox to 3.5cm{\hfill $(\exists x)(x+y>xz)$}\hbox to 3.5cm{\hfill $(\forall x)(x\!>\!y\imply y\!=\!z)$}\ .

\medbreak\noindent If you want to see a completely precise definition of
``term''  and ``formula'' then you have to consult a mathematical logic
textbook. In this course, we will take these concepts for granted.

How can you tell whether something is a term or a formula? 

\medbreak\noindent TERMS.  A {\bf term}
stands for a {\bf definite thing} (it may be a number, or a set, or a line,
or a function, or some other entity such a person or an animal), or for a
``variable thing'' that becomes definite once you plug in values for the
variables.  (For example, ``$3+5$'' denotes a definite number, whereas
``$3+x$'' denotes a ``variable number'' that becomes definite once you plug
in values for $x$: when you plug in $x=5$, $3+x$ has the value $8$; when
you plug in $x=-3$, $3+x$ has the value $0$; when you plug in $x=-20$,
$3+x$ has the value $-17$; and so on.)  For a more complicated example 
of a term, consider the expression ``$f(x)$''.   This is a term 
that contains two letter variables, namely, $f$ and $x$, standing,
respectively, for a function and a number. So, when you plug in a specific
function for $f$, and a specific number for $x$, you get a number. 
(For example, if $f$ is the ``squaring'' function, i.e. the function such
that $f(y)=y^2$ for all $y$, and $x=3$, then $f(x)$ has the value $9$. 
If $f$ is the ``exponential'' function, i.e. the function such
that $f(y)=e^y$ for all $y$, and $x=6$, then $f(x)$ has the value $e^6$. 
If $f$ is the function such
that $f(y)=y^2+3y+7$ for all $y$, and $x=6$, then $f(x)$ has the value 
$61$.) 


\medbreak\noindent FORMULAS.  A {\bf formula} makes an {\bf assertion},
which may be 
true or false.  So a formula has a {\bf definite truth value} (``true'' or
``false,'' often abbreviated $T$ or $F$) or a ``variable truth value'' that
becomes definite once you plug in values for the variables.  (For example,
``$3+5=8$'' has a definite truth value, namely, T, and ``$3+5=6$'' has a
definite truth value, namely, F.  On the other hand, ``$3+x=y$'' has a
variable truth value, which becomes definite once you plug in values for
the variables $x$ and $y$: when you plug in $x=5$ and $y=8$, $3+x=y$ is T;
when you plug in $x=2$ and $y=8$, $3+x=y$ is F; when you plug in $x=2$ and
$y=5$, $3+x=y$ is T; and so on.)

\bigbreak
\centerline{{\bf VARIABLES AND CONSTANTS}}

\medbreak\noindent As explained before, 
the basic constituents of terms are {\sl variables} and {\sl constants}.

{\sl Constants} are symbols that stand ---permanently or temporarily--- for one
{\sl fixed}, {\sl definite} thing.  
{\sl Variables} are letters such as $x$, $y$ that are meant to stand 
for things that only become definite when you assign a specific {\sl value}
to the variable.  

In principle, any symbol could be used for a constant, and also for a
variable. What matters is that in each case it should be clear which use is
intended, e.g. by {\sl declaring} a symbol to be a constant or a
variable. Usually, letters such as $x$, $y$, $z$, are used as variables for
real numbers, letters such as $k$, $m$, $n$, $p$ are used as variables for
natural numbers, and letters such as $X$, $Y$, $Z$ are used as variables
for sets.  So, for example, we would write formulas such as 
$$
(\forall x)(\exists y)(y>x)\,,
$$
$$
(\forall m)(\forall n)(\forall k)((k|m\wedge k|n)\Rightarrow k|m+n)\,,
$$
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,.
$$
But there is nothing sacred about this particular use.   We could, for
example, write 
$$
(\forall a)(\exists b)(b>a)\,,
$$
$$
(\forall x)(\forall y)(\forall z)((z|x\wedge z|y)\Rightarrow z|x+y)\,,
$$
$$
(\forall \alpha)(\forall \beta)(\forall \gamma)((\alpha\subseteq\beta\wedge
\beta\subseteq\gamma)\Rightarrow\alpha\subseteq\gamma)\,,
$$
provided that we specify that ``$a$, $b$ are real variables, 
$x$, $y$, $z$ are natural-number variables, $\alpha,\beta,\gamma$ are set
variables.'' 

One way to make it clear that something is a variable and to specify its
range is to put the range in the quantifier.  For example, if we write
$$
(\forall a\in\rr)(\exists b\in\rr)(b>a)\,,
$$
$$
(\forall x\in\nn)(\forall y\in\nn)(\forall z\in\nn)((z|x\wedge
z|y)\Rightarrow z|x+y)\,, 
$$
then it is clear that $a$, $b$ are real variables, 
and $x$, $y$, $z$ are natural-number variables.   

For our third sentence, involving sets, the situation is a little bit more
complicated. The obvious thing to do would be to write something like
$$
(\forall X\in{\cal S})(\forall Y\in{\cal S})(\forall Z\in{\cal S})
((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,
$$
where ${\cal S}$ is ``the set of all sets.''  This, however, is {\sl
forbidden}, because the ``set of all sets'' has to be avoided, due to the
Russell Paradox (to be discussed later in the course). 

So we can get around this by writing, instead, 
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,\,\,\,\,{\rm 
where\,\,\,}X,Y,Z\,\,\,{\rm are\,\,\,set\,\,\,variables},
$$
or
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,}
(X\subseteq Y\wedge Y\subseteq Z)\Rightarrow  X\subseteq Z\,.
$$

Naturally, we could even use more informal expressions such as 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,if}
\,\,\,X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z\,\,\,{\rm it\,\,\,follows\,\,\,that\,\,\,}
\Rightarrow  X\subseteq Z\,,
$$
or 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,and\,\,\,}
X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z,\,\,\,{\rm then\,\,\,}
\Rightarrow  X\subseteq Z\,.
$$

More complicated terms can be constructed from the basic constants 
and variables by means of (i)~function symbols, and (ii)~the 
``curly bracket'' set notation.  For example, the expression 
$x+3$ is a term, obtained by combining the variable $x$ and the constant 
$3$ by means of the binary (i.e. 2-argument) function symbol $+$.
The expression $\{x\in\rr:x^2>2\}$ is a term, that stands 
for a certain set of numbers.  {\bf A more detailed description of 
how function symbols work, and how to build more complicated terms 
and work with them, will be given later. For the time being, 
let us forget about function symbols until further notice, 
and concentrate on predicates, constants, variables,
and the logical connectives.}




{convincing argument} that establishes {\bf beyond the shadow of a 
doubt} that the statement is true.  

You should think of proving statements as a {\bf game} that you are playing
against an opponent who does not believe that the statement you are proving
is true.  You make statements, your opponent may find them convincing or
not, and it is up to you to clear up all your opponent's doubts by 
arguing your case persuasively every time your opponent isn't convinced. 
(Maybe you should think that I, the instructor, am the ``opponent.''
{\bf You} have to convince {\bf me}. If you convince me, then your proof
has worked. Otherwise it has not.)

The trouble with all this is that it is not very clear what will count as a
``convincing argument.''  Suppose your opponent says, after you have given
what looks to you like a very convincing proof: ``I am not convinced; I
don't see why what you just said proves anything.''  How would you deal
with something like that?

Moreover, any argument you give will have to {\it start} somewhere.  (You
cannot prove anything starting from nothing.)  What can you do if your
opponent refuses to accept your starting point?  

\bigbreak\noindent EXAMPLE: suppose you want to prove that, 

\medbreak
\bi
\item{} {\sl If a real number is greater than or equal to $4$, then 
the square root of the number is less than or equal to half the number.}

\medbreak\noindent Equivalently ---and more concisely--- you want to show that 

\medbreak
\item{} {\sl If $x$ is any real number such
that $x\ge 4$, then $\sqrt{x}\le{x\over 2}$.}
\ei

\medbreak

\noindent Or ---even more concisely--- you want to prove that 

$$
(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)\,.
$$

\noindent To prove this, you can proceed as follows: 

\noindent 
To prove $(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)$,
we pick an arbitrary $x$ and try to prove that 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$.

\noindent 
So, let $x$ be arbitrary.  We want to prove 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$. To prove this, we assume 
$x\ge 4$ and try to prove that $\sqrt{x}\le{x\over 2}$. 

\noindent 
Now we are assuming that $x\ge 4$.  From this it follows ---multiplying
both sides by $x$--- that $x^2\ge 4x$. Then ---dividing both sides by
$4$--- we get that ${x^2\over 4}\ge x$. Then ---taking the square root of
both sides--- we get $\sqrt{{x^2\over 4}}\ge\sqrt{x}$. Since
$\sqrt{{x^2\over 4}}={x\over 2}$, we have shown that 
${x\over 2}\ge \sqrt{x}$, which is ``the same as''
$\sqrt{x}\le{x\over 2}$. {\bf END OF PROOF}

\bigbreak\noindent {\bf Is this a truly convincing argument?}

\medbreak\noindent The answer is ``yes'' if all the steps are convincing. 
But {\bf if one step is not convincing, then the whole argument fails.}
(This is why you should {\bf not} expect to get partial credit for a wrong
proof if most steps are O.K. but just a few of them are wrong.  A proof is
like a chain. And {\bf a chain is no stronger than its weakest link.}  If a
16-Ton weight is hanging right above your head, held by a chain, and the
chain breaks because of one weak link, and the weight falls on your head
and kills you, do you think the chain manufacturer should be entitled to
``partial credit'' because ``most links were O.K.; only one broke''?  If
the driver of Lady Diana's car didn't drink every night, but only drank on
the particular night of the accident, should he be entitled to partial
credit because ``after all, most of the time he didn't drink''?)

\bigbreak\noindent So, {\bf are {\mymagtwobf ALL} the steps in the above 
proof truly convincing?}

\medbreak\noindent In the proof, we have given justifications for each
step. For example, at one point we went from 

$$x\ge 4$$
to 
$$x^2\ge 4x\,,$$
and we said that this was done 
``multiplying both sides by $x$.''   Now, why is this step permissible?

Suppose I tell you that ``I don't believe that you can do this''? 
How do you persuade me?   Well, let us give it a try. 
You can say: ``because there is a general rule that says 
if you take an inequality and multiply both sides by a number, then
you still get a valid inequality.''  (Or, in more symbolic notation:
$$
(\forall a)(\forall b)(\forall c)(a\ge b\Longrightarrow ac\ge bc)\,,
$$
using the language of a predicate calculus that we will discuss later.)  

This, however, will not do. This ``rule'' is NOT VALID.   
(For example, if $a=5$, $b=2$ and $c=-2$, then $a\ge b$, but 
it's not true that $ac\ge bc$, since $ac=-10$ and $bc=-4$, 
so $ac<bc$.)      So WE CANNOT USE IT.   

Now, here is a second try.  You can say: ``because there is a general rule
that says if you take an inequality and multiply both sides by a
nonnegative number, then you still get a valid inequality, that is,
using symbolic notation: 
$$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)\,.$$


It turns out that this time our rule IS VALID.  But notice that, in order
to apply this rule, we must be sure that $c\ge 0$. In our case, the number
that plays the role of $c$ is $x$, and it is indeed true that $x\ge 0$,
because $x\ge 4$.  (Since this is a crucial fact for us to be able to apply
the above ``rule,'' it would have been better to include in our proof a
little remark highlighting this. For example, instead of ``multiplying
both sides by $x$,'' it would have been better to say ``multiplying both
sides by $x$, which is possible because $x\ge 0$, since we are assuming
that $x\ge 4$.'')

But we haven't solved our problem.  Remember that your proof is really a
two-person game that you are playing against me, your opponent, in which
you have to convince me. What if I say ``How do we know 
that $(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)$?   I am not really convinced 
that this is true.''?

You could try to prove for me that 
$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b\wedge c\ge 0)\Longrightarrow 
ac\ge bc\Bigr)$  is indeed true.   But I am sure that in our proof you
will use other ``rules,''  and I, your opponent, could pick any of these 
and claim, once again, that I am not convinced of its validity.  

\medbreak
{\mymagtworm 

$\,$$\,$$\,$$\,$$\,$$\,$IF I WANTED TO, \ \ $\,\,$I COULD GO ON LIKE THIS 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$FOREVER. NO MATTER WHAT JUSTIFICATION 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$YOU $\,$$\,$GIVE $\,$$\,$FOR $\,$$\,$A $\,$$\,$STEP
$\,$$\,$IN $\,$$\,$YOUR $\,$$\,$PROOF, $\,$$\,$$\,$$\,$I 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$COULD ALWAYS ASK: $\,$$\,$$\,$$\,$``WHY IS THAT
VALID?''} 

\medbreak\noindent
In other words, if your opponent is truly stubborn and uncooperative, then
there is {\it nothing} you can do. Maybe you will come up with some reason,
but this reason will be based on some other facts that look obviously true
to you.  But your opponent can still refuse to accept these other facts,
and this process can go on for ever.

To take care of this ``problem of the opponent that doesn't play fair,'' 
mathematicians have come up with the concept of a {\bf formal proof}.
We give very precise rules defining what a ``formal proof'' is, 
and if you prove something correctly, and strictly following the rules, 
then you are {\it guaranteed} that your proof will be accepted as valid. 

If you do {\it not} obey the rules, then you are on uncertain ground: you
might be able to convince an instructor or grader, but you cannot be sure,
because any time you take a step one of us might ask ``how do you 
justify that step?''

You should think of giving formal proofs as similar to playing a game
such as chess: you are given very strict, precise rules, and you are only
allowed to make moves following the rules. As long as you obey the rules,
your opponent does not have the right to reject your moves. (If your
opponent unilaterally decides that a move of yours is not acceptable,
because ``you are not allowed to move a bishop if in the previous move you
moved a knight,'' then you would say that your opponent is not playing
fair, because this ``rule'' was not in the list of rules that had been
agreed upon before the game started.) 

{\bf In Mathematics we play fair.} We tell you what the rules are, and if
you prove something correctly following the rules, then you can be sure
that it will be accepted.  You may think that having to follow the rules is
a restriction to your freedom. But this is not so.  You should think of it
as {\bf your own protection against unfair play.}  If you correctly justify
every step according to the rules, then we promise you that we will not
change the rules in the middle of the game, kick the board, or come up with
other surprises.

\medbreak\noindent REMARK. I said that you should think of giving formal
proofs of theorems as a game similar to chess, in which you are given a
small number of very strict rules telling you which moves are permitted,
and each move has to be justifiable by explaining exactly which rule you
are using and how you are using it.  However, exactly as in the case of
chess, if all you know is the rules, and you always apply the rules
carefully and scrupulously, you may still lose all the time.  If you want
to {\bf win} (that is, in the case of a proof, if you want to get to the
desired conclusion) then you have to {\bf think}. You have to plan a
strategy, make a move now because it will be useful five or six steps
later, etc.

The one big difference between chess and theorem-proving is that {\bf chess
is a zero-sum game}, but {\bf theorem-proving is a cooperative game.} 
What this means is that in chess your opponent wants you to lose,
whereas in theorem-proving you and your opponent are working together,
trying to find out whether something is true, and to understand why.

That's why in chess you don't explain openly why you make a move.  You do
{\sl not} say to your opponent things like: ``I am making this move now so
that then in the next move I can do this other thing, so that three moves
later I can do that, and then I will checkmate you.'' {\sl You don't want your
opponent to be able to figure out your strategy.}

In theorem-proving you definitely {\bf do} want your opponent to understand
your strategy.  Your opponent is going to be ruthless in criticizing you if
you don't obey the rules, but as long as you obey them you will get full
coopertaion. In fact, usually you don't have to go through the whole
business of explaining every detail of every application of every rule.
{\sl As long as you persuade your opponent that something can be done using
the rules, there is no need to go through all the steps and actually do
it.}  And, since your opponent wants to be persuaded, the more you do to 
explain why you are doing something, the better. That's why {\sl in 
theorem-proving it's always nice to explain your strategy, so that your
opponent can understand where you are headed.}    

Because theorem-proving is a cooperative game, when writing proofs it is
always nice to use the first person plural ---the so-called ``editorial
we''--- as we did in our first example of a proof (``we pick,''\ \ ``we are
assuming,''\ \ ``we have shown'').  This is a good way to convey the idea
that the writer and the reader are {\sl working together}, that {\bf the writer
is leading the reader gently along a path of discovery.} \proofend

\vfill\eject

\centerline{{\mymagtwobf 2.\ VALID INFERENCES}} 

\bigbreak\noindent{\bf Proofs} involve {\bf steps} in which you move
from some statements (the {\it premisses}) to a new statement (the
conclusion).     We also say that you {\it infer}, or {\it derive}, or {\it
deduce} the conclusion from the premisses, and the steps where you 
do that are called {\it inferences}. 

So the first thing we need to discuss is:

\bigbreak
\centerline{{\mymagonebf WHAT IS A VALID INFERENCE?}}

\medbreak\noindent An {\bf inference} is a way of passing  from
certain statements (the {\bf premisses}) to another statement (the {\bf
conclusion}).   

\medbreak

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf VALID INFERENCES}}\\
(Informal definition)
\end{center}

A {\bf valid inference} is an inference for
which it is guaranteed  that, if the premisses are true, then the
conclusion is also true.
\end{minipage}

}

\end{center}

\medbreak\noindent For example:

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (1)}

\noindent\hbox to 7cm{\hfill Socrates is mortal}

\medbreak\noindent  is an example of a valid inference known as a {\it
syllogism}.  The premisses are 
``All men are mortal'' and ``Socrates is a man,''  and the conclusion is 
``Socrates is mortal.''

\medbreak\noindent  Roughly speaking, we would like to give the following 
definition of ``logically valid inference'': 

\bigbreak

\begin{center}
\fbox{
\begin{minipage}{4.6in}

\parindent=0pt

\null
\vspace{0.1mm}
\begin{center}
{\large{\bf LOGICALLY VALID INFERENCES}}\\
(Informal definition)
\end{center}

An inference is {\bf logically valid} if the conclusion follows
from the premisses {\bf by virtue of the ``form'' of the statements
involved.}
\end{minipage}
}
\end{center}

\medbreak\noindent  Eventually, we will say exactly that. But we are not
ready to say it yet, because we still do not know precisely
what we mean by the ``form'' of a statement. 
   
Before we talk about ``form,'' let us go on with examples to illustrate 
what we are trying to say. We would like to 
call the syllogism of (1) logically valid {\bf because in (1) it does not
matter what ``Socrates'' or ``man'' or ``mortal'' mean.}  It would be
equally valid to say

\medbreak\noindent\hbox to 7cm{\hfill All men are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (2)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All pigs are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a pig}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (3)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Donald Duck is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (4)}

\noindent\hbox to 7cm{\hfill Donald Duck  is mortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All borogoves are mimsy}

\noindent\hbox to 7cm{\hfill The Mad Hatter is a borogove}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (5)}

\noindent\hbox to 7cm{\hfill The Mad Hatter is mimsy}

\medbreak\noindent It does not matter whether the premisses are true (in
some of the above examples they are true, in others they are not) or
whether the conclusion is true.  It doesn't even matter whether you know
what a ``borogove'' is or what ``mimsy'' means.     The point is:

\bigbreak

\bi
\item{} {\it The four examples of syllogism given above all are of the
same {\bf logical form}, namely,}
\ei

\medbreak\noindent\hbox to 7cm{\hfill All $X$ are $Y$}

\noindent\hbox to 7cm{\hfill $Z$ is an $X$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (6)}

\noindent\hbox to 7cm{\hfill $Z$ is a $Y$}

\medbreak\noindent To be able to talk about an inference being ``logically 
valid'' we have to be able to talk about the {\bf logical form} (or, even
better, ``a'' logical form, rather than ``the'' logical form) of an
inference. 

For example, we could say that all five inferences (1), (2), (3), (4), (5)
are of the form (6), and any reasoning of this form is valid.  

Another example of a valid inference form would be:

\medbreak\noindent\hbox to 7cm{\hfill $X$ is  $Z$}

\noindent\hbox to 7cm{\hfill $Y$ is $Z$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (7)}

\noindent\hbox to 7cm{\hfill $X$ and $Y$ are $Z$}

\medbreak\noindent Examples of inferences of the form (7) are:

\medbreak\noindent\hbox to 7cm{\hfill John is  happy}

\noindent\hbox to 7cm{\hfill Mary is happy}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (8)}

\noindent\hbox to 7cm{\hfill John and Mary are happy}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is smart}

\noindent\hbox to 7cm{\hfill Hillary Clinton is smart}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (9)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are smart}

\medbreak\noindent  So it would seem that we have two examples of ``valid
forms'', namely, (6) and (7).  Consider, however, the following
examples: 

\medbreak\noindent\hbox to 7cm{\hfill All men are equal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (10)}

\noindent\hbox to 7cm{\hfill Socrates is equal}

\medbreak\noindent and 

\medbreak\noindent\hbox to 7cm{\hfill Al Gore is married}

\noindent\hbox to 7cm{\hfill Hillary Clinton is married}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (11)}

\noindent\hbox to 7cm{\hfill Al Gore and Hillary Clinton are married}

\medbreak\noindent Superficially, (10) is ``of the form (6),'' and
(11) is ``of the form (7).'' Something is clearly wrong here, because
in (110) the conclusion is clearly false, even though the premisses are
true, and in (10) the conclusion is meaningless. 

What is wrong here is that ``statement forms'' such as 
``All $X$ are $Y$'' or ``$X$ and $Y$ are $Z$'' are not good enough 
to capture logical distinctions that matter.   For example, the statements
``Al Gore and Hillary Clinton are married'' and
``Al Gore and Hillary Clinton are smart'' both appear to be of the same
``form,'' namely, ``$X$ and $Y$ are $Z$,'' but they really are not. 
``Smart'' is a {\bf one-variable predicate}. (If $x$ is a person, you can
say that ``$x$ is smart.'')  But ``married'' is {\it both} 
a {\it one-variable predicate} (in ``Al Gore is married'') and a
2-variable predicate (in ``Al Gore and Hillary Clinton are married''). 
In normal English discourse, when we say that ``$x$ and $y$ are married''
we do not mean that each one, separately, is married. We mean that 
``$x$ and $y$ are married to each other.''  (If you don't believe that,
ask yourself the following: when someone asks a couple ``are you two 
married?'', isn't it clear that this means ``to each other''?)

So, if we really want to talk about {\bf forms} of statements and
inferences, then we need a more sophisticated language.  We have to be able
to distinguish, for example, between one-variable predicates and two-variable
predicates. The languages that do that are the {\bf first-order predicate
calculi}. (``Calculi'' is the plural of ``calculus.'') Before we discuss
these rather complicated languages, we will say a few words about 
{\bf propositional calculi}, which are simpler. Both propositional calculi 
and first-order predicate calculi are examples of {\bf formal languages}, 
so we will start by talking about formal languages in general. 

\bigbreak
\centerline{\hbox to 5cm{\hrulefill}}



\vskip 0.7cm
\centerline{{\mymagtwobf 5.\ WHAT ARE PROPOSITIONAL CALCULI}}

\centerline{{\mymagtwobf GOOD FOR?}}

\bigbreak\noindent So far, we have just given some rules for writing
certain weird strings of symbols and calling them 
``propositional forms.''  What do we want to with all that?

The answer is: {\bf we use propositional forms to represent 
forms of propositions or statements.}  

For example, the statement 

\medbreak
\bi
\item{(S1)}  Abraham Lincoln was a great president 
and all the serious historians who have studied his career think very highly 
of him.
\ei

\medbreak\noindent could be represented by the propositional form $P\wedge 
Q$, where the 
propositional variable $P$ stands for ``Abraham Lincoln was a great
president'' and the propositional variable $Q$ stands for ``all the serious
historians who have studied Abraham Lincoln's career think very highly of
him.''

\medbreak Notice that (S1) could also be represented in the form 
$P$, where the propositional variable $P$ stands for ``Abraham Lincoln was
a great president and all the serious historians who have studied his
career think very highly of him.''

So {\bf the representation of a statement by means of a propositional form
is not unique.} There may be more than one way to represent the statement,
depending on how deeply you want to go into the details its structure.  In
our example, ``$P$'' is a {\bf coarser} propositional-form representation
of (S1) than ``$P\wedge Q$'' (and ``$P\wedge Q$'' is {\bf finer} than ``$P$'').

As explained in the book, a propositional form has a {\bf truth table}.    
The truth table tells you how to assing a ``truth value'' (True or False) 
to a propositional form if you assign truth values to the propositional
variables.        

\medbreak\noindent A {\bf tautology} is a propositional form that has the 
value True for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent A {\bf contradiction} is a propositional form that has the 
value False for all possible ways of assigning truth values to its
propositional variables. 

\medbreak\noindent Often, we also apply the words ``tautology'' and 
``contradiction'' to statements in ordinary English. In that case, what we
mean by a ``tautology'' is ``a statement that has a propositional form
which is a tautology.''  Also, what we mean by a ``contradiction'' is ``a
statement that has a propositional form which is a contradiction.''

For example: the statement 

\medbreak

\bi
\item{(S2)}  Hector Sussmann is a great teacher.
\ei


\medbreak\noindent is true (of course!), but it is of the form $P$, which is
not a tautology, and it has no other, finer propositional form.  So 
(S2) is not a tautology. 

\medbreak On the other hand,  the statement 

\medbreak
\bi
\item{(S3)}  Either Hector Sussmann is a great teacher
or he is not.
\ei


\medbreak\noindent has the propositional form $P$, which is
not a tautology, but is also has the form $(P\vee(\sim P))$, which is a
tautology.   So (S3) {\bf is} a tautology.  

\medbreak\noindent{\bf Remark.} The example of (S3) shows that, for
something to be a tautology, it suffices that {\bf one} of its
propositional forms be a tautology.  In the case of statement (S3), we can
find at least three different propositional form for it, namely, $P$,
$(P\vee Q)$, and $(P\vee(\sim P))$, which are progressively finer, and only
the third one is a tautology. This, however, suffices to make (S3) a 
tautology. \proofend

Let us look at another example of a formal representation. Consider the
statement (in which we assume that $f$ is a fixed function described
earlier, and $x_0$ is some given number)

\medbreak
\bi
\item{(S4)} If $f$ has a relative minimum at $x_0$ and
$f$ is differentiable at $x_0$ then $f'(x_0)=0$.  
\ei

\medbreak\noindent This is of the form $P$  (taking $P$ to stand for the
whole statement), or of the form $P\imply Q$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$ and $f$ is differentiable at $x_0$''
and $Q$ to stand for ``$f'(x_0)=0$''), 
or of the form $(P\wedge Q)\imply R$  (taking $P$ to stand for 
``$f$ has a relative minimum at $x_0$,'' $Q$ to stand for 
``$f$ is differentiable at $x_0$,'' and $R$ to stand for ``$f'(x_0)=0$''). 
Once again, notice that the logical form is not unique.


\bigbreak
%\centerline{\hbox to 5cm{\hrulefill}}


\vskip 0.7cm
\vbox{

\centerline{{\mymagtwobf 6.\ SOME TYPES OF VALID INFERENCES}}

\bigbreak\noindent Now that we know how to represent statements, we can
start giving examples of valid inferences of the propositional calculus.

\medbreak\noindent Assume that $f$ is a given function, and $n$ is a given
natural number. Then the inferences
}

\medbreak\noindent\hbox to 7cm{\hfill If $f$ is differentiable then 
$f$ is continuous.}

\noindent\hbox to 7cm{\hfill $f$ is differentiable.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (12)}

\noindent\hbox to 7cm{\hfill $f$ is continuous.}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill If $n$ is not prime
then $n$ is divisible by 3.}

\noindent\hbox to 7cm{\hfill $n$ is not a prime number.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (13)}

\noindent\hbox to 7cm{\hfill $n$ is divisible by 3.}

\medbreak\noindent are both of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (14)}

\noindent\hbox to 7cm{\hfill $Q$}

\medbreak\noindent This is a {\bf valid inference form.} Why is it valid?  
Because {\it if the premisses are true then the conclusion must be true}. 
Indeed: if $P\imply Q$ is true and $P$ is true then $Q$ is true, because
if $Q$ was false then $P\imply Q$ would be false (since $P$ is true) 
because the truth table for $\imply$ gives an $F$ to $P\imply Q$ if $P$ is
$T$ and $Q$ is $F$. 

\medbreak\noindent{\bf Remark.} Notice that in (13) the first premiss is 
usually not true.  (For example, it is not true if $n=25$, because 
$25$ is not a prime number and it's not divisible by $3$. So, if 
$n=25$, the first premiss says ``If $25$ is not a prime number
then $25$ is divisible by 3.'' This is of the form $P\imply Q$, where 
$P$ stands for ``$25$ is not a prime number'' and $Q$ for 
``$25$ is divisible by 3.'' Then $P$ is true and $Q$ is false, so $P\imply
Q$ is false.)

The inference form (14) is one of the most basic ones of all logic. It is
called {\bf modus ponens.}

What do we mean, in general, by a ``valid inference form''?

First of all, an {\bf inference form} is a list of propositional forms
(the premisses), followed by a horizontal line and then a propositional
form (the conclusion).  

A {\bf valid inference form} is an inference form such that the conclusion 
is true for all assignments of truth values to the propositional variables
that render the premisses true. 

Here is another example: the form 

\medbreak\noindent\hbox to 7cm{\hfill $P\imply Q$}

\noindent\hbox to 7cm{\hfill $\sim Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim P$}

\medbreak\noindent is valid. Examples of inferences of this form are:


\medbreak\noindent\hbox to 7cm{\hfill If I was rich I would own a Mercedes.}

\noindent\hbox to 7cm{\hfill I don't own a Mercedes.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill I am not rich.}

\medbreak\noindent and


\medbreak\noindent\hbox to 7cm{\hfill If $x>3$ then $x>2\,$.}

\noindent\hbox to 7cm{\hfill $\sim x>2\,$.}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $\sim x>3\,$.}


\bigbreak
\centerline{\hbox to 5cm{\hrulefill}}


\vskip 0.7cm
\centerline{{\mymagtwobf 7.\ FIRST-ORDER PREDICATE CALCULI}}

\bigbreak\noindent We now turn to the formal languages that are really
important for this course, namely, {\mymagtworm first-order predicate
calculi.} Inferences such as (1) above cannot be justified 
by just looking at their propositional form. If we look at propositional 
form, then (1) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $P$}

\noindent\hbox to 7cm{\hfill $Q$}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (15)}

\noindent\hbox to 7cm{\hfill $R$}

\medbreak\noindent which is not a valid inference form of the propositional
calculus. In a first-order predicate calculus, (1) will have associated
to it a much {\bf finer} form, namely

\medbreak\noindent\hbox to 7cm{\hfill $(\forall x)(p(x)\imply q(x))$}

\noindent\hbox to 7cm{\hfill $p(a)$}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (16)}

\noindent\hbox to 7cm{\hfill $q(a)$}

\medbreak\noindent (In (1), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is mortal,'' and ``$a$'' for ``Socrates.'' )

On the other hand, {\it (10) is not if the form (16)}.  In fact, to
represent (10) correctly we need a binary predicate $E(x,y)$ (meaning ``$x$
and $y$ are equal to each other''), and then {\it (10) cannot even be
represented in the language.}  Using a 1-variable predicate $M(x)$ for
``$x$ is a man'', and the symbol $a$ for ``Socrates'', we can represent the
premisses as $(\forall x)(\forall y)((M(x)\wedge M(y))\imply E(x,y))$ and
$M(a)$.  But {\it there is no way to represent the conclusion}, which is
fine because the conclusion is meaningless. In other words: {\bf by trying
to represent an inference such as (10) in a first-order predicate calculus,
we discover that it cannot be done, and this explains why it is mistaken 
to think that (10) is valid.}


Similarly, (8) and (9) are of the form

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (17)}

\noindent\hbox to 7cm{\hfill $p(x)\wedge p(y)$}

\medbreak\noindent which is valid, but (11) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (18)}

\noindent\hbox to 7cm{\hfill $q(x,y)$}

\medbreak\noindent where $q$ is a {\it different predicate} from $p$.
(In fact, $q$ is a two-variable predicate, and $p$ is a one-variable
predicate. Naturally, $p$ and $q$ are
related, because $p(x)$ ---i.e. ``$x$ is married''--- is a way of saying
$(\exists y)q(x,y)$, i.e. ``$x$ is married to somebody,'' or ``there exists
a person that $x$ is married to,'' or ``there exists a $y$ such that $x$ is
married to $y$.'')

Once again, {\bf we looked at an inference that appeared to be valid but
really should not have been.  We tried to represent it in a first-order
predicate calculus, and we discovered {\mymagtwobf} why it is not valid.
the reason is that a word that appeared in it was being used with two
different meanings, namely, the one-variable predicate ``married'' and the
two-variable predicate ``married.''}  The attempt to formalize our
inference in a first-order predicate calculus  forced us to be precise
about what kind of predicate ``married'' is, and we discovered that it 
is really two different predicates.  This solved our problem. 

\bigbreak We now turn to a more precise description of 
{\bf PREDICATE CALCULI}.

\bigbreak
\centerline{{\sl (to be continued)}}

\newpage

\def\rr{{\rm I}\kern-.2em{\rm R}}
\def\nn{{\rm I}\kern-.2em{\rm N}}
\def\eps{\varepsilon}
\def\imply{\Longrightarrow}
\font\mymagtworm=cmr10 scaled\magstep 2
\font\mymagonebf=cmbx10 scaled\magstep 1
\font\mymagtwobf=cmbx10 scaled\magstep 2
\font\mymagtwoit=cmti10 scaled\magstep 2
\centerline{{\mymagtwobf MATHEMATICS 300 --- FALL 1997}}
\bigbreak
\centerline{{\mymagtwoit Introduction to Mathematical Reasoning}}
\medbreak
\centerline{Instructor: H. J. Sussmann} 

\medbreak
\centerline{{\bf HANDOUT NO. 2.}  September 15, 1997}

\medbreak
\centerline{Revised version, October 6, 1997}

\medbreak

\raggedbottom
\def\rrbf{{\bf R}}
\def\rr{{\bf R}}
\def\mitem{\medbreak\item}
\def\ste{\enspace\enspace\enspace\enspace}
\newcount{\statco}
\newcount{\quesco}
\newcount{\defco}
\newcount{\notco}
\newcount{\remco}
\newcount{\rulco}
\newcount{\drulco}
%\def\co#1{#1}
\def\rulo{\global\advance\rulco by 1
\medbreak\noindent{\bf  RULE
\the\rulco.}\ }
\def\drulo{\global\advance\drulco by 1
\medbreak\noindent {\bf DERIVED RULE
\the\drulco.}\ }
\def\justo{
\medbreak\noindent {\bf JUSTIFICATION$\,$:}\ \ }
\def\co#1{}
\def\ep{\varepsilon}
\def\lto{\Longleftrightarrow}
\def\clo{{\rm Clos}}
\defco=0
\statco=0
\remco=0
\rulco=0
\drulco=0
\notco=0
\quesco=1
\def\ino{{\rm Int}}

\bigbreak

\centerline{{\mymagonebf PROOFS}} 

\medbreak\noindent A {\bf proof} of a statement is a {\bf 
convincing argument} that establishes {\bf beyond the shadow of a 
doubt} that the statement is true.  

You should think of proving statements as a {\bf game} that you are playing
against an opponent who does not believe that the statement you are proving
is true.  You make statements, your opponent may find them convincing or
not, and it is up to you to clear up all your opponent's doubts by 
arguing your case persuasively every time your opponent isn't convinced. 
(Maybe you should think that I, the instructor, am the ``opponent.''
{\bf You} have to convince {\bf me}. If you convince me, then your proof
has worked. Otherwise it has not.)

The trouble with all this is that it is not very clear what will count as a
``convincing argument.''  Suppose your opponent says, after you have given
what looks to you like a very convincing proof: ``I am not convinced; I
don't see why what you just said proves anything.''  How would you deal
with something like that?

Moreover, any argument you give will have to {\it start} somewhere.  (You
cannot prove anything starting from nothing.)  What can you do if your
opponent refuses to accept your starting point?  

\bigbreak\noindent EXAMPLE: suppose you want to prove that, 

\medbreak
\bi
\item{} {\sl if a real number is greater than or equal to $4$, then 
the square root of the number is less than or equal to half the number.}
\ei

\medbreak\noindent Equivalently ---and more concisely--- you want to show that 

\medbreak
\bi
\item{} {\sl if $x$ is any real number such
that $x\ge 4$, then $\sqrt{x}\le{x\over 2}$.}
\medbreak
\ei

\noindent Or ---even more concisely--- you want to prove that 

$$
(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)\,.
$$

\noindent To prove this, you can proceed as follows: 

\noindent 
To prove $(\forall x)\Bigl(x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}\Bigr)$,
we pick an arbitrary $x$ and try to prove that 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$.

\noindent 
So, let $x$ be arbitrary.  We want to prove 
$x\ge 4\Longrightarrow\sqrt{x}\le{x\over 2}$. To prove this, we assume 
$x\ge 4$ and try to prove that $\sqrt{x}\le{x\over 2}$. 

\noindent 
Now we are assuming that $x\ge 4$.  From this it follows ---multiplying
both sides by $x$--- that $x^2\ge 4x$. Then ---dividing both sides by
$4$--- we get that ${x^2\over 4}\ge x$. Then ---taking the square root of
both sides--- we get $\sqrt{{x^2\over 4}}\ge\sqrt{x}$. Since
$\sqrt{{x^2\over 4}}={x\over 2}$, we have shown that 
${x\over 2}\ge \sqrt{x}$, which is ``the same as''
$\sqrt{x}\le{x\over 2}$. {\bf END OF PROOF}

\bigbreak\noindent {\bf Is this a truly convincing argument?}

\medbreak\noindent The answer is ``yes'' if all the steps are convincing. 
But {\bf if one step is not convincing, then the whole argument fails.}
(This is why you should {\bf not} expect to get partial credit for a wrong
proof if most steps are O.K. but just a few of them are wrong.  A proof is
like a chain. And {\bf a chain is no stronger than its weakest link.}  If a
16-Ton weight is hanging right above your head, held by a chain, and the
chain breaks because of one weak link, and the weight falls on your head
and kills you, do you think the chain manufacturer should be entitled to
``partial credit'' because ``most links were O.K.; only one broke''?  If
the driver of Lady Diana's car didn't drink every night, but only drank on
the particular night of the accident, should he be entitled to partial
credit because ``after all, most of the time he didn't drink''?)

\bigbreak\noindent So, {\bf are {\mymagtwobf ALL} the steps in the above 
proof truly convincing?}

\medbreak\noindent In the proof, we have given justifications for each
step. For example, at one point we went from ``$x\ge 4$'' to 
``$x^2\ge 4x\,$,''  and we said that this was done 
``multiplying both sides by $x$.''   Now, why can we do that?

Suppose I tell you that ``I don't believe that you can do this''? 
How do you persuade me?   Well, let us give it a try. 
You can say: ``because there is a general rule that says 
if you take an inequality and multiply both sides by a number, then
you still get a valid inequality.''  (Or, in more symbolic notation:
$(\forall a)(\forall b)(\forall c)(a\ge b\Longrightarrow ac\ge bc)$.)  

This, however, will not do. This ``rule'' is NOT VALID.   
(For example, if $a=5$, $b=2$ and $c=-2$, then $a\ge b$, but 
it's not true that $ac\ge bc$, since $ac=-10$ and $bc=-4$, 
so $ac<bc$.)      So WE CANNOT USE IT.   

Now, here is a second try.  You can say: ``because there is a general rule
that says if you take an inequality and multiply both sides by a
nonnegative number, then you still get a valid inequality.''  (Or, in more
symbolic notation: $(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)$.)

It turns out that this time our rule IS VALID.  But notice that, in order
to apply this rule, we must be sure that $c\ge 0$. In our case, the number
that plays the role of $c$ is $x$, and it is indeed true that $x\ge 0$,
because $x\ge 4$.  (Since this is a crucial fact for us to be able to apply
the above ``rule,'' it would have been better to include in our proof a
little remark highlighting this. For example, instead of ``multiplying
both sides by $x$,'' it would have been better to say ``multiplying both
sides by $x$, which is possible because $x\ge 0$, since we are assuming
that $x\ge 4$.'')

But we haven't solved our problem.  Remember that your proof is really a
two-person game that you are playing against me, your opponent, in which
you have to convince me. What if I say ``How do we know 
that $(\forall a)(\forall b)(\forall c)\Bigl((a\ge b
\wedge c\ge 0)\Longrightarrow ac\ge bc\Bigr)$?   I am not really convinced 
that this is true.''?

You could try to prove for me that 
$(\forall a)(\forall b)(\forall c)\Bigl((a\ge b\wedge c\ge 0)\Longrightarrow 
ac\ge bc\Bigr)$  is indeed true.   But I am sure that in our proof you
will use other ``rules,''  and I, your opponent, could pick any of these 
and claim, once again, that I am not convinced of its validity.  

\medbreak
{\mymagtworm 

$\,$$\,$$\,$$\,$$\,$$\,$IF I WANTED TO, \ \ $\,\,$I COULD GO ON LIKE THIS 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$FOREVER. NO MATTER WHAT JUSTIFICATION 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$YOU $\,$$\,$GIVE $\,$$\,$FOR $\,$$\,$A $\,$$\,$STEP
$\,$$\,$IN $\,$$\,$YOUR $\,$$\,$PROOF, $\,$$\,$$\,$$\,$I 

\vskip 0.7mm
$\,$$\,$$\,$$\,$$\,$$\,$COULD ALWAYS ASK: $\,$$\,$$\,$$\,$``WHY IS THAT
VALID?''} 

\medbreak\noindent
In other words, if your opponent is truly stubborn and uncooperative, then
there is {\it nothing} you can do. Maybe you will come up with some reason,
but this reason will be based on some other facts that look obviously true
to you.  But your opponent can still refuse to accept these other facts,
and this process can go on for ever.

To take care of this ``problem of the opponent that doesn't play fair,'' 
mathematicians have come up with the concept of a {\bf formal proof}.
We give very precise rules defining what a ``formal proof'' is, 
and if you prove something correctly, and strictly following the rules, 
then you are {\it guaranteed} that your proof will be accepted as valid. 

If you do {\it not} obey the rules, then you are on uncertain ground: you
might be able to convince an instructor or grader, but you cannot be sure,
because any time you take a step one of us might ask ``how do you 
justify that step?''

You should think of giving formal proofs as similar to playing a game
such as chess: you are given very strict, precise rules, and you are only
allowed to make moves following the rules. As long as you obey the rules,
your opponent does not have the right to reject your moves. (If your
opponent unilaterally decides that a move of yours is not acceptable,
because ``you are not allowed to move a bishop if in the previous move you
moved a knight,'' then you would say that your opponent is not playing
fair, because this ``rule'' was not in the list of rules that had been
agreed upon before the game started.) 

{\bf In Mathematics we play fair.} We tell you what the rules are, and if
you prove something correctly following the rules, then you can be sure
that it will be accepted.  You may think that having to follow the rules is
a restriction to your freedom. But this is not so.  You should think of it
as {\bf your own protection against unfair play.}  If you correctly justify
every step according to the rules, then we promise you that we will not
change the rules in the middle of the game, kick the board, or come up with
other surprises.

\medbreak\noindent REMARK. I said that you should think of giving formal
proofs of theorems as a game similar to chess, in which you are given a
small number of very strict rules telling you which moves are permitted,
and each move has to be justifiable by explaining exactly which rule you
are using and how you are using it.  However, exactly as in the case of
chess, if all you know is the rules, and you always apply the rules
carefully and scrupulously, you may still lose all the time.  If you want
to {\bf win} (that is, in the case of a proof, if you want to get to the
desired conclusion) then you have to {\bf think}. You have to plan a
strategy, make a move now because it will be useful five or six steps
later, etc.

The one big difference between chess and theorem-proving is that {\bf chess
is a zero-sum game}, but {\bf theorem-proving is a cooperative game.} 
What this means is that in chess your opponent wants you to lose,
whereas in theorem-proving you and your opponent are working together,
trying to find out whether something is true, and to understand why.

That's why in chess you don't explain openly why you make a move.  You do
{\sl not} say to your opponent things like: ``I am making this move now so
that then in the next move I can do this other thing, so that three moves
later I can do that, and then I will checkmate you.'' {\sl You don't want your
opponent to be able to figure out your strategy.}

In theorem-proving you definitely {\bf do} want your opponent to understand
your strategy.  Your opponent is going to be ruthless in criticizing you if
you don't obey the rules, but as long as you obey them you will get full
coopertaion. In fact, usually you don't have to go through the whole
business of explaining every detail of every application of every rule.
{\sl As long as you persuade your opponent that something can be done using
the rules, there is no need to go through all the steps and actually do
it.}  And, since your opponent wants to be persuaded, the more you do to 
explain why you are doing something, the better. That's why {\sl in 
theorem-proving it's always nice to explain your strategy, so that your
opponent can understand where you are headed.}    

Because theorem-proving is a cooperative game, when writing proofs it is
always nice to use the first person plural ---the so-called ``editorial
we''--- as we did in our first example of a proof (``we pick,''\ \ ``we are
assuming,''\ \ ``we have shown'').  This is a good way to convey the idea
that the writer and the reader are {\sl working together}, that {\bf the writer
is leading the reader gently along a path of discovery.}


\bigbreak
\centerline{{\bf FORMAL PROOFS}} 

\medbreak\noindent A {\it theorem} has one or several {\it assumptions}
(or {\it hypotheses}, or {\it premisses}), and a {\it conclusion}.  A {\it
formal proof} of a theorem is a sequence of statements (the ``steps'' of
the proof), ending with the conclusion, such that each step is permitted by
the rules, given the previous steps.  

Later, we will give a complete list of the rules that tell you 
when a step is permitted. For the time being, let us start with 
three examples of such rules.

\medbreak
\bi
\item{(P.a)} it is always permitted to bring in an 
{\it axiom} (or ``postulate'') as a new step,  

\item{(P.b)} it is always permitted to bring in one of the hypotheses
as a new step, 

\item{(P.c)} it is always permitted to bring in as a new step
something that follows from previous steps by applying a valid {\it rule of
inference}.
\ei

\medbreak\noindent For all this to make sense, I have to tell you what are
the axioms and what are the rules of inference.  We will start with the
rules of inference.  There will be exactly {\bf eight} of them.   

\bigbreak
\centerline{{\bf WHAT IS A VALID INFERENCE?}}

\medbreak\noindent An {\it inference} is a way of passing  from
certain statements (the {\it premisses}) to another statement (the {\it
conclusion}).   A {\it valid inference} is one for which it is guaranteed
that, if the premisses are true, then the conclusion is also true.

For example:

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (1)}

\noindent\hbox to 7cm{\hfill Socrates is mortal}

\medbreak\noindent  is an example of a valid inference known as a {\it syllogism}. 

\medbreak\noindent  Roughly speaking, we would like to say that an
inference is {\it logically valid} if the conclusion follows from the
premisses {\it by virtue of the ``form'' of the statements involved}.   

For example, the syllogism of (1) is logically valid, because it does not matter
what ``Socrates'' or ``man'' or ``mortal'' mean.   It would be equally
valid to say 

\medbreak\noindent\hbox to 7cm{\hfill All men are immortal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (2)}

\noindent\hbox to 7cm{\hfill Socrates is immortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All men are mortal}

\noindent\hbox to 7cm{\hfill Donald Duck is a man}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (3)}

\noindent\hbox to 7cm{\hfill Donald Duck  is mortal}

\medbreak\noindent or

\medbreak\noindent\hbox to 7cm{\hfill All borogoves are mimsy}

\noindent\hbox to 7cm{\hfill The Mad Hatter is a borogove}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (4)}

\noindent\hbox to 7cm{\hfill The Mad Hatter is mimsy}

\medbreak\noindent It does not matter whether the premisses are true (in
some of the above examples they are true, in others they are not) or
whether the conclusion is true.  It doesn't even matter whether you know
what a ``borogove'' is or what ``mimsy'' means.     The point is:

\bi
\item{} {\it The four examples of syllogism given above all are of the
same {\bf logical form}, namely,}
\ei

\medbreak\noindent\hbox to 7cm{\hfill All $X$ are $Y$}

\noindent\hbox to 7cm{\hfill $Z$ is an $X$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (5)}

\noindent\hbox to 7cm{\hfill $Z$ is a $Y$}

\medbreak\noindent To be able to talk about an inference being ``logically 
valid'' we have to be have a way to say what the {\bf form} (or maybe ``a'' form,
rather than ``the'' form) of an inference is.    For example, we could say
that all four inferences (1), (2), (3), (4) are of the form (5), and 
any reasoning of this form is valid.  Another example of a valid 
inference form would be:

\medbreak\noindent\hbox to 7cm{\hfill $X$ is  $Z$}

\noindent\hbox to 7cm{\hfill $Y$ is $Z$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (6)}

\noindent\hbox to 7cm{\hfill $X$ and $Y$ are $Z$}

\medbreak\noindent Examples of inferences of the form (6) are:

\medbreak\noindent\hbox to 7cm{\hfill John is  happy}

\noindent\hbox to 7cm{\hfill Mary is happy}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (7)}

\noindent\hbox to 7cm{\hfill John and Mary are happy}

\medbreak\noindent and

\medbreak\noindent\hbox to 7cm{\hfill My mother is  smart}

\noindent\hbox to 7cm{\hfill My father is smart}

\noindent\hbox to 7cm{\hbox to 3cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (8)}

\noindent\hbox to 7cm{\hfill My mother and my father are smart}

\medbreak\noindent  So it would seem that we have two examples of ``valid
forms'', namely, (5) and (6).  Consider, however, the following
examples: 

\medbreak\noindent\hbox to 7cm{\hfill All men are equal}

\noindent\hbox to 7cm{\hfill Socrates is a man}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (9)}

\noindent\hbox to 7cm{\hfill Socrates is equal}

\medbreak\noindent and 

\medbreak\noindent\hbox to 7cm{\hfill Bob Dole is married}

\noindent\hbox to 7cm{\hfill Hillary Clinton is married}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (10)}

\noindent\hbox to 7cm{\hfill Bob Dole and Hillary Clinton are married}

\medbreak\noindent Superficially, (9) is ``of the form (5),'' and
(10) is ``of the form (6).'' Something is clearly wrong here, because
in (10) the conclusion is clearly false, even though the premisses are
true, and in (9) the conclusion is meaningless. 

What is wrong here is that ``statement forms'' such as 
``All $X$ are $Y$'' or ``$X$ and $Y$ are $Z$'' are not good enough 
to capture logical distinctions that matter.   For example, the statements
``Bob Dole and Hillary Clinton are married'' and
``Bob Dole and Hillary Clinton are smart'' both appear to be of the same
``form,'' namely, ``$X$ and $Y$ are $Z$,'' but they really are not. 
``Smart'' is a {\bf one-variable predicate}. (If $x$ is a person, you can
say that ``$x$ is smart.'')  But ``married'' is {\it both} 
a {\it one-variable predicate} (in ``Bob Dole is married'') and a
2-variable predicate (in ``Bob Dole and Hillary Clinton are married''). 
In normal English discourse, when we say that ``$x$ and $y$ are married''
we do not mean that each one, separately, is married. We mean that 
``$x$ and $y$ are married to each other.''  (If you don't believe that,
ask yourself the following: when someone asks a couple ``are you two 
married'', isn't it clear that this means ``to each other''?)

So, if we really want to talk about {\bf forms} of statements and
inferences, then we need a more sophisticated {\it formal language}.  

One formal language that does the required job rather well
is the {\it predicate calculus}. 
Using this language, we can see that (1), (2), (3) and (4) are of the form

\medbreak\noindent\hbox to 7cm{\hfill $(\forall x)(p(x)\imply q(x))$}

\noindent\hbox to 7cm{\hfill $p(a)$}

\noindent\hbox to 7cm{\hbox to 2cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (11)}

\noindent\hbox to 7cm{\hfill $q(a)$}

\medbreak\noindent (In (1), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is mortal,'' and ``$a$'' for ``Socrates.''  
In (2), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is immortal,'' and ``$a$'' for ``Socrates.''  
In (3), ``$p(x)$'' stands for ``$x$ is a man,'' 
``$q(x)$'' for ``$x$ is mortal,'' and ``$a$'' for ``Donald Duck.''  
In (4), ``$p(x)$'' stands for ``$x$ is a borogove,'' 
``$q(x)$'' for ``$x$ is mimsy,'' and ``$a$'' for ``the Mad Hatter.'')  

On the other hand, {\it (9) is not if the form (11)}.  In fact, to
represent (9) correctly we need a binary predicate $E(x,y)$ (meaning ``$x$
and $y$ are equal to each other''), and then (9) cannot even be represented
in the language.  (Using a 1-variable predicate $M(x)$ for ``$x$ is a
man'', and the symbol $a$ for ``Socrates'', we can represent the premisses
as $(\forall x)(\forall y)((M(x)\wedge M(y))\imply E(x,y))$ and $M(a)$.  But
{\it there is no way to represent the conclusion}, which is fine because
the conclusion is meaningless.)


Similarly, (7) and (8) are of the form

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (12)}

\noindent\hbox to 7cm{\hfill $p(x)\wedge p(y)$}

\medbreak\noindent which is valid, but (10) is of the form 

\medbreak\noindent\hbox to 7cm{\hfill $p(x)$}

\noindent\hbox to 7cm{\hfill $p(y)$}

\noindent\hbox to 7cm{\hbox to 5cm{\hfill}\hrulefill}\hbox to 4cm{\hfill (12)}

\noindent\hbox to 7cm{\hfill $q(x,y)$}

\medbreak\noindent where $q$ is a {\it different predicate} from $p$.
(In fact, $q$ is binary and $p$ is unary.  Naturally, $p$ and $q$ are
related, because $p(x)$ ---i.e. ``$x$ is married''--- is a way of saying
$(\exists y)q(x,y)$, i.e. ``$x$ is married to somebody,'' or ``there exists
a person that $x$ is married to,'' or ``there exists a $y$ such that $x$ is
married to $y$.'')

So from on we will use the formalism of the {\bf predicate calculus}
to formulate our rules. 

\bigbreak
\centerline{{\bf PREDICATE CALCULI}}

\medbreak\noindent A predicate calculus is a formal language whose
expressions (or ``words'') are finite  sequences made of 
some {\sl basic symbols}.   (Remember that the plural of ``calculus''
is ``calculi.'' That's why this section is called ``predicate calculi,''
because there are lots of different such calculi, depending on which
symbols you use.)

For the predicate calculi to be considered in this course, 
there will be {\sl basic symbols}, which are combined to form {\sl terms}
and {\sl formulas}.  

The basic symbols will be:

\medbreak
\bi
\item{1.} constants;

\item{2.} variables;

\item{3.} function symbols;

\item{4.} predicate symbols;

\item{5.} the logical connectives $\forall$, $\exists$, $\wedge$, $\vee$,
$\Rightarrow$, $\Longleftrightarrow$, $\sim$;

\item{6.} the left and right parentheses, ``$($'' and ``$)$,'' and the
comma ``,''.  
\ei


\medbreak{\sl VARIABLES} are letters such as $x$, $y$, $z$, $m$, $n$,
$k$, $\eps$,  $X$, $Y$, etc. In fact, any letter (or even other 
symbols, such as $\diamond$, or $\aleph$, or even a sequence of
letters such as ``grrrr'' can be used as a variable, provided 
it is {\sl declared} to be a variable.  Often, there is no need to declare
a symbol to be a variable, because the context makes it clear that it is a
variable. For example, if I write $(\forall x)(x^2\ge 0)$, then 
$x$ {\sl has} to be a variable, because only variables can appear in
quantifiers such as $(\forall x)(x^2\ge 0)$. 


\medbreak{\sl CONSTANTS} are letters such as $a$, $b$, $c$, $\alpha$, 
$A$, $B$, etc. Actually, any letter (or even other symbols, such as
$\diamond$, or $\aleph$, or $\eps$, or $x$, or $y$, or $\pi$, or $e$, or
$0$, or $1$, or $2$, or $3$, or $\emptyset$, or even a sequence of letters
such as ``grrrr'' or ``$34612$) can be used as a constant, provided it is
{\sl declared} to be a constant.  Often, there is no need to declare a
symbol to be a constant, because the context makes it clear that it is a
constant. For example, if I write ``let $a$ be an arbitrary real number,''
or ``let $a=3$,'' or ``suppose $A$ and $B$ are sets,'' then I am declaring
that $a$, $b$, $A$, $B$ are going to be treated as constants.

\medbreak Notice that, if a symbol is a constant, then there are
certain things you cannot do with it.  For example, after 
you have declared ``$a$'' to be a constant by saying, for example,  
``let $a$ be an arbitrary real number,'' {\sl you cannot 
use quantifiers such as $(\forall a)$}, because $a$ is now fixed,
not variable.    

\medbreak Similarly, if a symbol is a variable, then there 
things you cannot do with it.  For example, after 
you have said in a proof $(\forall x)(\forall y)(x+y=y+x)$, 
you cannot write in your proof something like $x+y=y+x$, 
because {\sl this is not a proposition}.  (This means that, if $x$ and $y$
are variables, then ``$x+y=y+x$'' does not have  
a definite truth value. It doesn't make sense to say that 
$x+y=y+x$'' is true or false unless $x$ and $y$ are given specific 
values.    And, if you give $x$ and $y$ specific, definite values, then
they become constants.) 

\medbreak\noindent{\sl FUNCTION SYMBOLS} are letters such as $f$, $g$,
$F$, $G$, or other symbols such as $+$, $\cdot$, $\cap$, $\cup$,
$\sim$. Each function symbol admits a {\sl fixed} number of arguments.
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} symbol. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} symbol.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} symbol. And so on.)   If you write 
a function symbol, followed by a ``$($,'' then as many arguments
as the function symbol admits, separated by commas, and then 
``$)$,'' you get a {\sl term}.  The strings that can be used 
as arguments are variables, or constants, or more general terms.

\medbreak\noindent {\bf EXAMPLE.} Suppose you want to have a predicate 
calculus to talk about kinship and  comparing ages of people.
You could introduce one-variable function
symbols $f$ and $g$ to stand for ``father of'' and ``mother of'' and, say,
a binary predicate $y$  to stand for ``the younger of two people.''
Then, for example, the term $f(Bill)$ would stand for 
``Bill's father,'' the term $m(Bill)$ would stand for 
``Bill's mother,'' and the term $y(Bill, John)$ would stand for ``the younger
of Bill  and John'' (that is,  $y(Bill, John)=Bill$ if Bill is younger than
John, and  $y(Bill, John)=John$ if John is younger than Bill).   As
explained before,  function letters can themselves admit terms as argument,
so for example you could write $f(f(Bill))$, which would stand for 
``Bill's paternal grandfather,'' $f(m(Bill))$, which would stand for 
``Bill's maternal grandfather,'' or even more complicated expressions such
as 
$$
f(m(y(Bill,y(John,Patricia))))\,,
$$
which stands for ``the maternal grandfather of the youngest 
of Bill, John and Patricia.''  (QUESTION: could we have written 
$f(m(y(Bill,John,Patricia)))$. Yes? No? Why?) 

\medbreak\noindent {\bf ANOTHER EXAMPLE.} In mathematics, we use
a predicate calculus that has symbols for the basic arithmetic 
operations ``$+$'' and ``$\cdot$''  (``plus'' and ``times''),
which are {\sl binary} (that is, you add and multiply 
{\sl two} numbers).    In this case, it is customary 
to write things such as ``$a+b$,'' ``$a\cdot b$,'' 
rather than ``$+(a,b)$,'' ``$\cdot(a,b)$.''   There are, of course, many
other function symbols in common use.  For example, $|\ldots|$
(``absolute value of'') is a {\sl one-variable} function symbol,
so we use terms such as $|3|$.    As in our previous example, one can use
terms as arguments, so for example we caan form terms such as
$|(|3|+5).|7+8|+4|$.


\medbreak\noindent{\sl PREDICATE SYMBOLS} are letters such as $p$, $q$,
$P$, $Q$, or other symbols such as $=$, $>$, $<$, $\le$,
$\ge$, $\epsilon$, $\subseteq$. Each function symbol admits a {\sl fixed}
number of arguments. 
(If admits {\sl one} argument, then it is a {\sl unary}, 
or {\sl one-variable} predicate. 
If admits {\sl two} arguments, then it is a {\sl binary}, 
or {\sl two-variable} predicate.  
If admits {\sl three}  arguments, then it is a {\sl ternary}, 
or {\sl three-variable} predicate.  And so on.)   If you write 
a predicate symbol, followed by a ``$($,'' then as many arguments
as the predicate symbol admits, separated by commas, and then 
``$)$,'' you get an {\sl atomic formula}.  The strings that can be used 
as arguments are {\sl terms}.

\medbreak\noindent {\bf EXAMPLE.} Let us go on with 
the predicate calculus that enables us to talk about kinship and compare
ages of people.  We could introduce two-variable predicates 
$s$, $d$, $c$, so that 
$s(a,b)$ would stand for ``$a$ is $b$'s son'',
$d(a,b)$ would stand for ``$a$ is $b$'s daughter'',
$c(a,b)$ would stand for ``$a$ is $b$'s child.'' (Then 
$c(a,b)\lto(s(a,b)\vee d(a,b))$. Make sure you understand why!)
We could also have the binary predicate ``$=$'' with its usual meaning
(that is, ``$a=b$'' means that $a$ and $b$ are the same person). 
With these symbols, we could say lots of more complicated
things. For example, suppose I want to say ``$a$ is $b$'s uncle.''
I could say this by writing 
$$(\exists x)(\exists y)(\exists z)
\Bigl(s(a,x)\wedge s(a,y)\wedge(\sim x=y)\wedge(\sim a=z)
\wedge c(z,x)\wedge c(z,y)\wedge c(b,z)\Bigr)\,.$$

\medbreak\noindent {\bf ANOTHER EXAMPLE.} In mathematics, the 
predicate calculi we use have symbols for the basic predicates
``$=$,'' ``$>$,'' ``$<$,'' ``$\ge$,'' ``$\le$,'' 
which are {\sl binary}. In this case, it is customary 
to write things such as ``$a>b$,'' ``$a=b$,'' 
rather than ``$>(a,b)$,'' ``$=(a,b)$.''   There are, of course, many
other predicates in common use.  For example, for integers
we could introduce a binary predicate symbol ``$|$'' that stands for
``divides.''   (That is, $m|n$ stands for $(\exists k)(k.m=n)$.)
These symbols admit terms as arguments, so for example we can write
things such as $(3+5).7=|50+6|$.  Here   
$(3+5).7$ and $|50+6|$ are terms, and $(3+5).7=|50+6|$ is an atomic
formula.) 

\medbreak\noindent Atomic formulas can be combined with the logical
connectives to yield {\sl formulas}.  For example, 
$$(\exists x)(\exists y)(\exists z)
\Bigl(s(a,x)\wedge s(a,y)\wedge(\sim x=y)\wedge(\sim a=z)
\wedge c(z,x)\wedge c(z,y)\wedge c(b,z)\Bigr)$$
is {\sl not} an atomic formula, but is obtained from the atomic formulas
$s(a,x)$, $s(a,y)$, $x=y$, $a=z$, $c(z,x)$, $c(z,y)$,
$c(b,z)$ by combining them via the connectives $\wedge$, 
$\sim$ and $\exists$.  

\bigbreak
\centerline{{\bf TERMS VS. FORMULAS}}

\medbreak\noindent{\bf Terms} are supposed to be 
names of ``things.''  {\bf Formulas} are
expressions that ---when we plug in values for the
open variables--- become sentences or statements.

For example, the following are terms: 

\medbreak
\noindent\hbox to 3.5cm{\hfill $3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to 3.5cm{\hfill $x$}\hbox to 3.5cm{\hfill $5xyz$}

\noindent\hbox to 3.5cm{\hfill $x+3$}\hbox to 3.5cm{\hfill $3+5$}\hbox to
3.5cm{\hfill $x^2+3y+7$}\hbox to 3.5cm{\hfill $\{x:x^2<3y\}$}\ ,

\noindent and the following are formulas:

\medbreak
\noindent\hbox to 3.5cm{\hfill $3<5$}\hbox to 3.5cm{\hfill $3+5=8$}\hbox to 3.5cm{\hfill $x+3=7y$}\hbox to 3.5cm{\hfill $5xyz<x+z$}

\noindent\hbox to 3.5cm{\hfill $(\forall x)(x+3>x)$}\hbox to 3.5cm{\hfill $(\forall x)(x+y>x)$}\hbox to 3.5cm{\hfill $(\exists x)(x+y>xz)$}\hbox to 3.5cm{\hfill $(\forall x)(x\!>\!y\imply y\!=\!z)$}\ .

\medbreak\noindent If you want to see a completely precise definition of
``term''  and ``formula'' then you have to consult a mathematical logic
textbook. In this course, we will take these concepts for granted.

How can you tell whether something is a term or a formula? 

\medbreak\noindent TERMS.  A {\bf term}
stands for a {\bf definite thing} (it may be a number, or a set, or a line,
or a function, or some other entity such a person or an animal), or for a
``variable thing'' that becomes definite once you plug in values for the
variables.  (For example, ``$3+5$'' denotes a definite number, whereas
``$3+x$'' denotes a ``variable number'' that becomes definite once you plug
in values for $x$: when you plug in $x=5$, $3+x$ has the value $8$; when
you plug in $x=-3$, $3+x$ has the value $0$; when you plug in $x=-20$,
$3+x$ has the value $-17$; and so on.)  For a more complicated example 
of a term, consider the expression ``$f(x)$''.   This is a term 
that contains two letter variables, namely, $f$ and $x$, standing,
respectively, for a function and a number. So, when you plug in a specific
function for $f$, and a specific number for $x$, you get a number. 
(For example, if $f$ is the ``squaring'' function, i.e. the function such
that $f(y)=y^2$ for all $y$, and $x=3$, then $f(x)$ has the value $9$. 
If $f$ is the ``exponential'' function, i.e. the function such
that $f(y)=e^y$ for all $y$, and $x=6$, then $f(x)$ has the value $e^6$. 
If $f$ is the function such
that $f(y)=y^2+3y+7$ for all $y$, and $x=6$, then $f(x)$ has the value 
$61$.) 


\medbreak\noindent FORMULAS.  A {\bf formula} makes an {\bf assertion},
which may be 
true or false.  So a formula has a {\bf definite truth value} (``true'' or
``false,'' often abbreviated $T$ or $F$) or a ``variable truth value'' that
becomes definite once you plug in values for the variables.  (For example,
``$3+5=8$'' has a definite truth value, namely, T, and ``$3+5=6$'' has a
definite truth value, namely, F.  On the other hand, ``$3+x=y$'' has a
variable truth value, which becomes definite once you plug in values for
the variables $x$ and $y$: when you plug in $x=5$ and $y=8$, $3+x=y$ is T;
when you plug in $x=2$ and $y=8$, $3+x=y$ is F; when you plug in $x=2$ and
$y=5$, $3+x=y$ is T; and so on.)

\bigbreak
\centerline{{\bf VARIABLES AND CONSTANTS}}

\medbreak\noindent As explained before, 
the basic constituents of terms are {\sl variables} and {\sl constants}.

{\sl Constants} are symbols that stand ---permanently or temporarily--- for one
{\sl fixed}, {\sl definite} thing.  
{\sl Variables} are letters such as $x$, $y$ that are meant to stand 
for things that only become definite when you assign a specific {\sl value}
to the variable.  

In principle, any symbol could be used for a constant, and also for a
variable. What matters is that in each case it should be clear which use is
intended, e.g. by {\sl declaring} a symbol to be a constant or a
variable. Usually, letters such as $x$, $y$, $z$, are used as variables for
real numbers, letters such as $k$, $m$, $n$, $p$ are used as variables for
natural numbers, and letters such as $X$, $Y$, $Z$ are used as variables
for sets.  So, for example, we would write formulas such as 
$$
(\forall x)(\exists y)(y>x)\,,
$$
$$
(\forall m)(\forall n)(\forall k)((k|m\wedge k|n)\Rightarrow k|m+n)\,,
$$
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,.
$$
But there is nothing sacred about this particular use.   We could, for
example, write 
$$
(\forall a)(\exists b)(b>a)\,,
$$
$$
(\forall x)(\forall y)(\forall z)((z|x\wedge z|y)\Rightarrow z|x+y)\,,
$$
$$
(\forall \alpha)(\forall \beta)(\forall \gamma)((\alpha\subseteq\beta\wedge
\beta\subseteq\gamma)\Rightarrow\alpha\subseteq\gamma)\,,
$$
provided that we specify that ``$a$, $b$ are real variables, 
$x$, $y$, $z$ are natural-number variables, $\alpha,\beta,\gamma$ are set
variables.'' 

One way to make it clear that something is a variable and to specify its
range is to put the range in the quantifier.  For example, if we write
$$
(\forall a\in\rr)(\exists b\in\rr)(b>a)\,,
$$
$$
(\forall x\in\nn)(\forall y\in\nn)(\forall z\in\nn)((z|x\wedge
z|y)\Rightarrow z|x+y)\,, 
$$
then it is clear that $a$, $b$ are real variables, 
and $x$, $y$, $z$ are natural-number variables.   

For our third sentence, involving sets, the situation is a little bit more
complicated. The obvious thing to do would be to write something like
$$
(\forall X\in{\cal S})(\forall Y\in{\cal S})(\forall Z\in{\cal S})
((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,
$$
where ${\cal S}$ is ``the set of all sets.''  This, however, is {\sl
forbidden}, because the ``set of all sets'' has to be avoided, due to the
Russell Paradox (see below). 

So we can get around this by writing, instead, 
$$
(\forall X)(\forall Y)(\forall Z)((X\subseteq Y\wedge Y\subseteq Z)
\Rightarrow  X\subseteq Z)\,,\,\,\,\,{\rm 
where\,\,\,}X,Y,Z\,\,\,{\rm are\,\,\,set\,\,\,variables},
$$
or
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,}
(X\subseteq Y\wedge Y\subseteq Z)\Rightarrow  X\subseteq Z\,.
$$

Naturally, we could even use more informal expressions such as 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,then\,\,\,if}
\,\,\,X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z\,\,\,{\rm it\,\,\,follows\,\,\,that\,\,\,}
\Rightarrow  X\subseteq Z\,,
$$
or 
$$
{\rm If}\,\,\,X,Y,Z\,\,\,{\rm are\,\,\, sets,\,\,\,and\,\,\,}
X\subseteq Y\,\,\,{\rm and}\,\,\,
Y\subseteq Z,\,\,\,{\rm then\,\,\,}
\Rightarrow  X\subseteq Z\,.
$$

More complicated terms can be constructed from the basic constants 
and variables by means of (i)~function symbols, and (ii)~the 
``curly bracket'' set notation.  For example, the expression 
$x+3$ is a term, obtained by combining the variable $x$ and the constant 
$3$ by means of the binary (i.e. 2-argument) function symbol $+$.
The expression $\{x\in\rr:x^2>2\}$ is a term, that stands 
for a certain set of numbers.  {\bf A more detailed description of 
how function symbols work, and how to build more complicated terms 
and work with them, will be given later. For the time being, 
let us forget about function symbols until further notice, 
and concentrate on predicates, constants, variables,
and the logical connectives.}




\bigbreak

\vbox{
\centerline{{\bf DERIVED RULES (SHORTCUTS)}}

\medbreak\noindent In a formal proof you are supposed to use the 
eight basic rules and nothing else.  However, if you really had to do
this all the time, then your proofs would become too long.  This is why one
often ``skips steps,'' or ``carries out several steps together.''  In fact,
what you do is observe that in many cases there are certain sequences of
steps that appear over and over again.  So you know in advance that
whenever you have such-and-such formulas you can always go to some other
formula.  We collect this information into a list of {\bf derived rules}.
}


\medbreak\noindent EXAMPLE: Suppose you want to go from $(\forall
x)(p(x)\imply q(x))$ 
and $p(a)$ to $q(a)$.   (For example, you want to deduce the conclusion
that ``Socrates is mortal'' from  the premisses ``All men are mortal''
and ``Socrates is a man.''   In formal language, if $p(x)$ stands 
for ``$x$ is a man,'' $q(x)$ stands for ``$x$ is mortal,''  and
$a$ stands for ``Socrates,'' then the premisses are precisely 
$(\forall x)(p(x)\imply q(x))$ and $p(a)$, and what you want to get is
$q(a)$.)  Then you could do it as follows:

\medbreak\noindent (1)\ \ \ $(\forall x)(p(x)\imply q(x))$\ \ \ \ \ \ \ \ (hypothesis);

\noindent (2)\ \ \ $p(a)\,$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (hypothesis);

\noindent (3)\ \ \ $p(a)\imply q(a)\,\,$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Rule 4,
specialization of (1) to $x=a$);

\noindent (4)\ \ \ $q(a)$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Rule 2, from (2) and (3)).

\medbreak\noindent However, this situation is going to occur over and over
again, and you do not want to repeat the above four steps every time, so
you can state a 

\drulo    If you have $(\forall x)(p(x)\imply q(x))$ and $p(a)$ then
you can go to $q(a)$.

\justo Steps (1), (2), (3), (4)  above.

\medbreak\noindent {\sl Now that you have Derived Rule 1, every time 
you need to go from (1) and (2) as above you can skip (3), and go
directly to (4), invoking the derived rule.}

\medbreak\noindent Here are a few more derived rules. There are lots of
them.  Hundreds of them.  But you do not need to memorize them, because
they are all rather trivial and reflect plain common sense,
and you can derive them whenever you want to.  For example, Derived Rule
10 is the basis of ``proofs by contradiction,'' but what it says is rather
obvious: 

\medbreak

{\sl 

\centerline{(E)~ If you put yourself in a scenario where $q$ is not true, and}

\centerline{$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$end  up showing
that something impossible must $\,$happen,}

\centerline{$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$$\,$then this scenario
cannot be possible, so $q$ must be true.}    

}

\medbreak\noindent {\bf Students sometimes find 
proofs by contradiction hard to justify.  Read the previous explanation 
(E) a couple of times. You should have no trouble understanding that 
the justification of proofs by contradiction is quite trivial.}  

Another example: Derived Rule 15 says another completely obvious 
and trivial thing:  suppose that $q$ necessarily must be true 
when $p$ is false, and also when $p$ is true. Then $q$ must be true
(because $p$ is either true or false, and in either case $q$ is true.)


\drulo    If you have $p\wedge q$ then you can go to $p$. 

\justo $p\wedge q\imply p$ is a tautology, so you can bring it in any time
you want.  If in addition you have $p\wedge q$, then from this and 
$p\wedge q\imply p$ you get $p$ by Modus Ponens (Rule 2).

\drulo    If you have $p\wedge q$ then you can go to $q$. 

\justo $(p\wedge q)\imply q$ is a tautology, so you can bring it in any time
you want.  If in addition you have $p\wedge q$, then from this and 
$(p\wedge q)\imply q$ you get $q$ by Modus Ponens (Rule 2).


\drulo    If you have $p$ and $q$ then you can go to $p\wedge q$. 

\justo $p\imply(q\imply (p\wedge q))$ is a tautology, so you can bring it
in any time you want.  If in addition you have $p$ and $q$, then from 
$p$ and $p\imply(q\imply (p\wedge q))$ you get $q\imply (p\wedge q)$
by Modus Ponens (Rule 2), and then from 
$q$ and $q\imply (p\wedge q)$ you get $p\wedge q$.

\vbox{

\drulo    If you have $p$ then you can go to $p\vee q$. 

\justo $p\imply(p\vee q)$ is a tautology, so you can bring it
in any time you want.  If in addition you have $p$, then from 
$p$ and $p\imply(p\vee q)$ you get $p\vee q$
by Modus Ponens (Rule 2).
}

\vbox{
\drulo    If you have $q$ then you can go to $p\vee q$. 

\justo $q\imply(p\vee q)$ is a tautology, so you can bring it
in any time you want.  If in addition you have $q$, then from 
$q$ and $q\imply(p\vee q)$ you get $p\vee q$
by Modus Ponens (Rule 2).
}

\drulo    If you have $p\imply q$ and $q\imply p$ 
then you can go to $p\Longleftrightarrow q$. 

\justo $(p\imply q)\imply((q\imply p)\imply(p\Longleftrightarrow q))$ is a tautology,
so you can bring it in any time you want.  If in addition you have $p\imply q$,
then from $p\imply q$ and $(p\imply q)\imply((q\imply p)\imply(p\Longleftrightarrow q))$ 
you get $(q\imply p)\imply(p\Longleftrightarrow q)$ by Modus Ponens (Rule 2).
If you also have $q\imply p$,
then from $q\imply p$ and $(q\imply p)\imply(p\Longleftrightarrow q)$ 
you get $p\Longleftrightarrow q$ by Modus Ponens (Rule 2).

%\vfill\eject
\drulo    If you have $p\Longleftrightarrow q$ 
then you can go to $p\imply q$.

\justo $(p\Longleftrightarrow q)\imply(p\imply q)$ is a tautology,
so you can bring it in any time you want.  If in addition you have 
$p\Longleftrightarrow q$, 
then from $p\Longleftrightarrow q$ and 
$(p\Longleftrightarrow q)\imply(p\imply q)$ 
you get $p\imply q$ by Modus Ponens (Rule 2).

\vbox{
\drulo    If you have $p\Longleftrightarrow q$ then you can go to $q\imply p$.

\justo $(p\Longleftrightarrow q)\imply(q\imply p)$ is a tautology,
so you can bring it in any time you want.  If in addition you have 
$p\Longleftrightarrow q$, 
then from $p\Longleftrightarrow q$ and 
$(p\Longleftrightarrow q)\imply(q\imply p)$ 
you get $q\imply p$ by Modus Ponens (Rule 2).

}


\vbox{
\drulo    If assuming $\sim q$ you get a contradiction, then 
you can go to $q$.

\justo Suppose that assuming $\sim q$ you get $c$, which is
a contradiction. Then you can go to $\sim q\imply c$ by Rule 3.   
Now, if $c$ is a contradiction, then $(\sim q\imply c)\imply q$
is a tautology (why?), so you can bring it in. Since you have 
$\sim q\imply c$, then from $\sim q\imply c$ and $(\sim q\imply c)\imply q$
you get $q$ by Modus Ponens (Rule 2).
}

\vbox{
\drulo    From $\sim p$ and $p\vee q$ you can go to $q$. 

\justo $(\sim p)\imply((p\vee q)\imply q)$ 
is a tautology, so you can bring it in any time you want.  If in addition
you have $\sim p$, then you get $(p\vee q)\imply q$ 
by Modus Ponens (Rule 2).  If you also 
have $p\vee q$, then you get $q$ by Modus Ponens (Rule 2).
}

\vbox{
\drulo    From $\sim q$ and $p\vee q$ you can go to $p$.

\justo $(\sim q)\imply((p\vee q)\imply p)$ 
is a tautology, so you can bring it in any time you want.  If in addition
you have $\sim q$, then you get $(p\vee q)\imply p$ 
by Modus Ponens (Rule 2).  If you also 
have $p\vee q$, then you get $p$ by Modus Ponens (Rule 2).
}

\vbox{
\drulo    If assuming $p$ and $\sim q$ you get $\sim p$, then 
you can go to $p\imply q$.

\justo Assume $p$. Then assuming $\sim q$ you can 
get $\sim p$, and you can also get $p$, so you can get $p\wedge \sim p$ by
Derived Rule 4.  But 
$p\wedge \sim p$ is a contradiction.    So you can get $q$ by 
Derived Rule 12, still assuming $p$. In other words, assuming $p$ we get
$q$. By Rule $3$, we get $p\imply q$. 
}

\vbox{
\drulo    If assuming $\sim p$ you get $q$, then 
you can go to $p\vee q$.

\justo $((\sim p)\imply q)\imply(p\vee q)$ 
is a tautology, so you can bring it in any time you want.  If you can prove
$q$ assuming $\sim p$, then you get $(\sim p)\imply q$ 
by Rule 3.   But then from $(\sim p)\imply q$  and 
$((\sim p)\imply q)\imply(p\vee q)$ you get  $p\vee q$  
by Modus Ponens (Rule 2).  
}


\vbox{
\drulo    If assuming $\sim p$ you get $q$, and assuming $p$ you also get $q$, 
then you can go to $q$ assuming nothing.

\justo $((\sim p)\imply q)\imply ((p\imply q)\imply q)$ 
is a tautology, so you can bring it in any time you want.  If you can prove
$q$ assuming $\sim p$, then you get $(\sim p)\imply q$ 
by Rule 3.   But then from $(\sim p)\imply q$  and 
$((\sim p)\imply q)\imply ((p\imply q)\imply q)$ you get  
$(p\imply q)\imply q$ by Modus Ponens (Rule 2).  
If you can also prove
$q$ assuming $p$, then you get $p\imply q$ 
by Rule 3.   But then from $p\imply q$  and 
$(p\imply q)\imply q$ you get $q$ by Modus Ponens (Rule 2).  
}


\vbox{
\drulo    If assuming $p$ you get $q$, and assuming $q$ you get $p$, 
then you can go to $p\Longleftrightarrow q$.

\justo $(p\imply q)\imply ((q\imply p)\imply(p\Longleftrightarrow q))$ 
is a tautology, so you can bring it in any time you want.  If you can prove
$q$ assuming $p$, then you get $p\imply q$ by Rule 3.  But then from $p\imply q$
and $(p\imply q)\imply ((q\imply p)\imply(p\Longleftrightarrow q))$ you get $(q\imply
p)\imply(p\Longleftrightarrow q)$ by Modus Ponens (Rule 2).  If you can also prove
$p$ assuming $q$, then you get $q\imply p$ by Rule 3.  But then from $q\imply p$
and $(q\imply p)\imply(p\Longleftrightarrow q)$ you get $p\Longleftrightarrow q$ by Modus
Ponens (Rule 2).
}

\vbox{
\drulo    If you have $(\forall x\in S)p(x)$ and you have
$s\in S$, you can go to $p(s)$. 

\justo $(\forall x\in S)p(x)$ is an abbreviation of
$(\forall x)(x\in S\imply p(x))$. If you have 
$(\forall x)(x\in S\imply p(x))$, then by specialization (Rule 4)
you can get $s\in S\imply p(s)$.  So, if you have in addition 
$s\in S$, then by Modus Ponens (Rule 2) you get $p(s)$. 
}

\vbox{
\drulo    To prove a statement of the form $(\forall x\in S)p(x)$, you
start by saying ``Let $a$ be an arbitrary element of $S$'' (that is,
``let $a$ be arbitrary; assume $a\in S$'') and then try to
prove $p(a)$. (Here, again, instead of ``$a$'' you could use ``$b$,'' or
``$z$,'' or ``$\alpha$,'' or ``$\beta$,'' or ``$\aleph$,'' or
``$\diamond$,'' or any symbol you want.  What is important that what you do
should apply to a {\it completely arbitrary} element of $S$.)

\justo $(\forall x\in S)p(x)$ is an abbreviation of $(\forall
x)(x\in S\imply p(x))$.  If, starting from ``let $a$ be arbitrary'' and
``assume $a\in S$'' you get $p(a)$, it means that starting with ``let $a$
be arbitrary'' you can get $a\in S\imply p(a)$ by Rule 3.  So by Rule 5 you
have proved that $(\forall x)(x\in S\imply p(x))$, i.e. that $(\forall x\in
S)p(x)$.
}

\vbox{
\drulo    If you have $(\exists x\in S)p(x)$ then you can introduce a
new object and give it a letter name such as $a$ or $\alpha$ or anything you
want, provided that this name has not been used before, and stipulate that
$a\in S$ and $p(a)$.  

\justo $(\exists x\in S)p(x)$ is an abbreviation of $(\exists
x)(x\in S\wedge p(x))$.  So, if you have $(\exists x\in S)p(x)$, then you
have $(\exists x)(x\in S\wedge p(x))$.  So by Rule 6 
you can introduce a new
object and give it the letter name $a$, and stipulate that
$a\in S\wedge p(a)$.  But then, by Derived Rules 2 and 3,
once you got $a\in S\wedge p(a)$, you can go to
$a\in S$ and to $p(a)$. 
}

\vbox{
\drulo     If you have $p(a)$ and $a\in S$ then you can go to $(\exists
x\in S)p(x)$. 

\justo $(\exists x\in S)p(x)$ is an abbreviation of $(\exists
x)(x\in S\wedge p(x))$.  So, if you have $a\in S$ and $p(a)$ then 
you can go to $a\in S\wedge p(a)$ by Derived Rule 4. Then 
you get $(\exists x)(x\in S\wedge p(x))$ by Rule 7.
}

\vbox{
\drulo     If you have $(\forall x)p(x)$ and $y$ is some other variable
that does not appear in $p(x)$, then you can go to 
$(\forall y)p(y)$.

\justo Let $a$ be a new letter.    Then  you can introduce a new
constant by saying ``let $a$ be arbitrary,'' and then by specialization
(Rule 4) you can go from $(\forall x)p(x)$ to $p(a)$.   Since $a$ was
arbitrary, you can go to $(\forall y)p(y)$.
}


\vbox{
\drulo     If you have $(\exists x)p(x)$ and $y$ is some other variable
that does not appear in $p(x)$, then you can go to 
$(\exists y)p(y)$.

\justo Let $a$ be a new letter.    Then,
since you have $(\exists x)p(x)$,  you can introduce a new
constant via Rule 6 by saying ``let $a$ be such that 
$p(a)$.''    Then by Rule 7 you can go to $(\exists y)p(y)$.
}


Why do we insist, in Derived Rules 21 and 22, that $y$ cannot occur in
$p(x)$?   Suppose, for example, that $p(x)$ is the formula
$(\exists y)(y\neq x)$.  Then  ``$(\forall x)p(x)$'' 
says that ``$(\forall x)(\exists y)(y\neq x)$.''
We can of course introduce an arbitrary constant $a$, and then use
Rule 4 to get $p(a)$, i.e. $(\exists y)(y\neq a)$. 
Now, can we apply Rule 5 to go to 
$(\forall y)(\exists y)(y\neq a)$? No! Why? Because Rule 5 tells
us that if we have proved  $q(a)$, then we can go to 
$(\forall y)q(y)$, {\sl if $q(y)$ is a formula having 
no $y$-quantifiers and $q(a)$ is what you get 
from $q(y)$ by substituting $a$ for $y$ wherever $y$ occurs}. 



\bigbreak
\centerline{AN EXAMPLE OF SOMETHING YOU {\bf CANNOT} DO}

\medbreak\noindent Many students seem to think that the specialization 
rule  allows you to go from

\medbreak
\pline{\noindent\ \ \ \ \ $\sim(\forall x)p(x)$\ \ \ \ \ to 
\ \ \ \ \ $\sim p(s)$\hfill\hfill (A)}

\medbreak\noindent if $s$ is a term that satisfies restriction (R). 

Can one do this?  The answer is {\bf NO!!!!!}.  Why not? Is it because
the teacher says so?  NO!   It does not matter what I say!   You cannot do it
because plain common sense tells you that it cannot be done.  All you need 
is to think a little bit.  Here are some special instances of Inference  
(A).   (Note: the (A) stands for ``awful''.)    


\medbreak
\bi
\item{1.} From ``not everybody is a Democrat'' we can go to 
``I am not a Democrat.''
\medbreak\item{2.} From ``not everybody is an actor" you can go to
``Tom Hanks is not an actor.''
\ei

\medbreak\noindent Silly,  isn't it?   That's why you cannot do it! Because it
suffices to {\bf think} about it for ten seconds to see that it's silly!  
Never mind what I tell you or what the book tells you!  If I told you that
(A) is permitted, then you should reject what I have told you.   

Incidentally: can you {\bf prove} that (A) is {\bf not}, cannot be, a valid
derived rule?   Yes, sure you can.   Every single one of our basic rules 
takes you from true statements to true statements. Every derived rule
does the same, because a derived rule never gives you anything that you cannot
get by applying the basic rules.    So every valid rule, basic or derived, 
will yield true statements when applied to true statements. ``Rule (A)'' 
does not do this, so it is not a valid rule.
 
\bigbreak
\centerline{HOW DO YOU KNOW WHETHER SOMETHING THAT LOOKS}

\centerline{LIKE A VALID RULE REALLY IS VALID?}

\medbreak\noindent Let us look at two  concrete examples:

\medbreak
\bi
\item{(1)}  Can you go from   $(\forall x)p(x)\wedge(\forall x)q(x)$ to
$(\forall x)(p(x)\wedge q(x))$? 

\medbreak\item{(2)}  Can you go from   $(\exists x)p(x)\wedge(\exists
x)q(x)$ to $(\exists x)(p(x)\wedge q(x))$? 
\ei


\medbreak\noindent Let us settle (2) first. The answer is {\bf NO!!!}.
Why?  {\bf Think!}     If you know that ``some people are rich''
and ``some people are happy,''   does it follow that there is a person
that is {\bf both} rich and happy?  It doesn't, right?   So, you see,
you {\bf cannot} go from   $(\exists x)p(x)\wedge(\exists x)q(x)$ to
$(\exists x)(p(x)\wedge q(x))$.

Now, how about (1)? Here, if you think for a minute, you see right away
that {\it every time $(\forall x)p(x)\wedge(\forall x)q(x)$ is true,
then $(\forall x)(p(x)\wedge q(x))$ has to be true too.} (Example:
if you know that ``everybody is rich and everybody is happy'' then you
can conclude that ``everybody is both rich and happy.'')    

In order to be completely sure, let us check that you can do this using our
rules:


\medbreak
\bi
\item{} Starting from $(\forall x)p(x)\wedge(\forall x)q(x)$,

\item{} you can go to $(\forall x)p(x)$, by Derived Rule 2,

\item{} and to $(\forall x)q(x)$ by Derived Rule 3.

\item{} \ \ \ \ \ \ \ \ Now, let $a$ be arbitrary.   

\item{} \ \ \ \ \ \ \ \ Since we have $(\forall x)p(x)$, by specialization
we get $p(a)$. 

\item{} \ \ \ \ \ \ \ \ Since we have $(\forall x)q(x)$, by specialization
we get $q(a)$. 

\item{} \ \ \ \ \ \ \ \ Since we have $p(a)$ and $q(a)$, by Derived Rule 4
we get $p(a) \wedge q(a)$.

\item{} So we have proved $p(a)\wedge q(a)$ for an arbitrary $a$. Therefore
we can go to $(\forall x)(p(x)\wedge q(x))$ by Rule 5.
\ei


\medbreak\noindent As you can see, {\bf you can always 
go from   $(\forall x)p(x)\wedge(\forall x)q(x)$ to
$(\forall x)(p(x)\wedge q(x))$ by applying the rules.}   In order words, 
you can make a new derived rule:

\drulo From   $(\forall x)p(x)\wedge(\forall x)q(x)$ you can go to
$(\forall x)(p(x)\wedge q(x))$.


\medbreak\noindent The justification of this derived rule is the sequence of steps given above.


\medbreak Do you need to memorize millions and millions of derived rules? NO!!!!!
Do you need to memorize millions and millions of examples of things that 
are {\it not} valid?  NO!!!! As the above examples show, it's always quite easy for you 
to figure out, in each case, whether you are dealing with a valid rule or not.




\bigbreak
\centerline{\hbox to5cm{\hrulefill}}

\bigbreak
\centerline{{\bf TWO EXAMPLES OF PROOFS FOLLOWING THE RULES}}

\medbreak\noindent
Here are two simple examples of proofs. 
We will show that, if $q(x)$ is a predicate over a domain of discourse $U$ and 
$r$ is a proposition, then 
$$(\exists x)\Bigl(r\wedge q(x)\Bigr)\Longleftrightarrow r\wedge 
(\exists x)(q(x))\eqno(I)$$ 
and 
$$(\forall x)\Bigl(r\vee q(x)\Bigr)\Longleftrightarrow r\vee 
(\forall x)(q(x))\eqno(II)$$ 
are theorems of the predicate calculus. 
Let us prove these two things using our rules.

\def\po{\phantom{1}}


{\parindent=0pt

\vbox{

\bigbreak\noindent PROOF OF (I), WITH SHORTCUTS:   


\pline{\po(1)\ste\ste\ Assume $(\exists x)\Bigl(r\wedge q(x)\Bigr)$.
\hfill(Preparation to apply Rule 3)} 

\pline{\po(2)\ste\ste\ste\ Let $a$ be such that $r\wedge q(a)$.\hfill
(Rule 6, from (1))} 

\pline{\po(3)\ste\ste\ste\  $r$. \hfill (Derived Rule 2, from (2))}

\pline{\po(4)\ste\ste\ste\ $q(a)$. \hfill (Derived Rule 3, from (2))}

\pline{\po(5)\ste\ste\ste \ $(\exists x)q(x)$. \hfill (Rule 7, from (4)) }

\pline{\po(6)\ste\ste\ste \ $r\wedge (\exists x)q(x)$. \hfill (Derived Rule 4, from
(3) and (5))}

\pline{\po(7)\ste\ste \ $r\wedge (\exists x)q(x)$. \hfill (Rule 6, from (6))}

\medbreak
\pline{\po(8)\ste\ste \ Assume $r\wedge (\exists x)q(x)$. 
\hfill(Preparation to apply Rule 3)} 

\pline{\po(9)\ste\ste  \ $r$. \hfill (Derived Rule 2, from (8))}

\pline{(10)\ste\ste \ $(\exists x)q(x)$.\hfill (Derived Rule 3, from (8))}

\pline{(11)\ste\ste\ste \ Let $a$ be such that $q(a)$. \hfill (Rule 6, from
(10)) } 

\pline{(12)\ste\ste\ste \ $r\wedge q(a)$. \hfill (Derived Rule 4, from (9)
and (11))} 

\pline{(13)\ste\ste\ste $(\exists x)(r\wedge q(x))$.\hfill (Rule 7, from (12))}

\pline{(14)\ste\ste \ $(\exists x)(r\wedge q(x))$.\hfill (Rule 6, from (6))}

\pline{(13)\ste \ $(\exists x)\Bigl(r\wedge q(x)\Bigr)\Longleftrightarrow 
r\wedge (\exists x)q(x)$. \hfill (Derived Rule 16, because we got}

\vskip -1mm
\pline{\ \ \ste\ste \hfill to (6) from (1) and to (12) from (7))}

\medbreak 
Q.E.D.
}


\bigbreak

\vbox{

\noindent PROOF OF (II), WITH SHORTCUTS:        


\pline{\po(1)\ste \ Assume $(\forall x)\Bigl(r\vee q(x)\Bigr)$. 
\hfill(Preparation to apply Rule 3)} 

\pline{\po(2)\ste\ste \ Assume $\sim r$.
\hfill(Preparation to apply Rule 3)} 

\pline{\po(3)\ste\ste\ste \ Let $a$ be arbitrary.
\hfill(Preparation to apply Rule 5)} 


\pline{\po(4)\ste\ste\ste \ $r\vee q(a)$.  \hfill (Rule 4, specializing to $x=a$ in  (1))}

\pline{\po(5) \ste\ste\ste $q(a)$. \hfill (Derived Rule 11, from (2) and (4))}

\pline{\po(6) \ste\ste \ $(\forall x)q(x)$. \hfill (Rule 5, from (3) and (5))}

\pline{\po(7) \ste \ $r\vee (\forall x)q(x)$. \hfill (Derived Rule 14,
bec. we proved  $(\forall x)q(x)$ assuming $\sim r$)}

\pline{\po(8)\ste \ Assume $r\vee (\forall x)q(x)$.
\hfill(Preparation to apply Rule 3)} 

\pline{\po(9)\ste\ste \ Assume $\sim r$.
\hfill(Preparation to apply Rule 3)} 


\pline{(10) \ste\ste $(\forall x)q(x)$. \hfill (Derived Rule 11, from (8) and (9))}

\pline{(11)\ste\ste\ste \ Let $a$ be arbitrary.
\hfill(Preparation to apply Rule 5)} 


\pline{(12)\ste\ste\ste  \ $q(a)$. \hfill (Rule 4, specializing to $x=a$ in  (10))}

\pline{(13)\ste\ste\ste  \ $r\vee q(a)$. \hfill (Derived Rule 6, from (12))}

\pline{(14)\ste\ste  \ $(\forall x)(r\vee q(x))$. \hfill (Rule 5, because we proved $r\vee
q(a)$ for arbitrary $a$) }

\pline{(15)\ste\ste \ Assume $r$.
\hfill(Preparation to apply Rule 5)} 

\pline{(16)\ste\ste\ste \ Let $a$ be arbitrary.
\hfill(Preparation to apply Rule 5)} 

\pline{(17)\ste\ste\ste  $r\vee q(a)$. \hfill (Derived Rule 5, from (16))}

\pline{(18)\ste\ste  $(\forall x)(r\vee q(x))$. \hfill (Rule 5, because we
proved $r\vee q(a)$ for arbitrary $a$) }

\pline{(19)\ste  $(\forall x)(r\vee q(x))$. \hfill (Derived Rule 15, because we
proved the formula}

\vskip-1mm
\pline{\ \ \ste\hfill $(\forall x)(r\vee q(x))$ 
assuming $\sim r$ and also assuming $r$)}

Now let us write $A$ for $(\forall x)(r\vee q(x))$, and $B$ for $r\vee (\forall
x)q(x)$.

\pline{(20) $A\Longleftrightarrow B$. \hfill (Derived Rule 16, because we
proved  $A$ assuming $B$ and $B$ assuming $A$.)}

\medbreak 
Q.E.D.
}}


\bigbreak\noindent NOTE: You may think that these proofs are very
complicated. They are not! Read them very carefully, and you will see that
all we are doing is very simple and obvious things. The point of writing
these particular proofs is to show you the rules ---and the shortcuts---
in action.  

To understand what these two proofs really are about, let us rewrite
them in plain English, i.e. in ``informal language:''

\vbox{
\medbreak\noindent INFORMAL PROOF OF (I):        


\medbreak\noindent We want to prove an equivalence of the form
$A\Longleftrightarrow B$, so we prove that we can get $B$ from $A$ 
and then that we can get
$A$ from $B$.  To get from $A$ to $B$, we notice that
$A$ says that $(\exists x)(r\wedge q(x))$, and we want to prove $B$,
i.e. that $r\wedge(\exists x)q(x)$. Now, if $(\exists x)(r\wedge q(x))$ is
true then we can pick an object $a$ such that $r\wedge q(a)$ is true. So
both $r$ and $q(a)$ are true.  But since $q(a)$ is true then $(\exists
x)q(x)$ is true, and since $r$ is true we get $r\wedge(\exists x)q(x)$,
i.e. $B$. To get from $B$ to $A$ we use a similar strategy.  $B$ says that
$r\wedge(\exists x)q(x)$. So if $B$ is true then $r$ is true and $(\exists
x)q(x)$. But then we can pick an object $a$ such that $q(a)$ holds. But
then $r$ and $q(a)$ hold, so $r\wedge q(a)$ is true. Therefore we can
conclude that $(\exists x)(r\wedge q(x))$ holds, i.e. that $A$ holds.  

\noindent {\bf END OF PROOF}
}

\vskip 0.8cm


\vbox{
\noindent INFORMAL PROOF OF (II):        


\medbreak\noindent We want to prove an equivalence of the form
$A\Longleftrightarrow B$, so we prove that we can get $B$ from $A$ 
and then that we can get
$A$ from $B$.  To get from $A$ to $B$, we notice that
$A$ says that $(\forall x)(r\vee q(x))$, and we want to prove $B$,
i.e. that $r\vee(\forall x)q(x)$.  Assume $A$ is true. 
Consider first the case when $r$ is true.  
Then $r\vee(\forall x)q(x)$ holds, i.e. $B$ holds. Now suppose $r$ is false.
Let $a$ be arbitrary.  Then $r\vee q(a)$ is true.  
Since $r$ is false, $q(a)$ must be true. But $a$ is completely arbitrary,
so $(\forall x)q(x)$ holds. So $r\vee(\forall x)q(x)$ holds, i.e. $B$
holds.   So we have shown that $B$ follows from $A$ both when $r$ is
true and when it is false. So $B$ follows from $A$.  Now let show that
$A$ followsfrom $B$.  Suppose $B$ is true. Then 
$r\vee(\forall x)q(x)$.  Consider first the case when $r$ is false.  
Then $(\forall x)q(x)$ is true.  So, if $a$ is arbitrary, we have
$q(a)$, and then we get $r\vee q(a)$.  So $(\forall x)(r\vee q(x))$ holds.
Now suppose $r$ is true.  Then for arbitrary $a$ 
$r\vee q(a)$ is true, so once again $(\forall x)(r\vee q(x))$ holds.
In both cases ---that is, when $r$ is false and when $r$ is true---
we have shown that $A$ follows from $B$.
{\bf END OF PROOF}
}

\bigbreak\noindent What is the difference between a formal proof
and an informal one?  {\bf As long as you are able to translate an informal
argument into a formal proof, you are free to write informal proofs.}

The trouble is  (and believe me, {\bf I know this from experience}),
students usually write informal proofs with vague, imprecise
justifications,  and the reason is that they {\bf do not know}
what they are really trying to say.   For example, students will
often write justifications of steps such as the following:

\medbreak\noindent (1) By the definition of real number, 

\medbreak\noindent (2) by the definition of equality, 

\medbreak\noindent (3) by tautology,

\medbreak\noindent (4) by specialization,

\medbreak\noindent (5) because it's in the book,

\medbreak\noindent (6) because the teacher said something like that in class.

\bigbreak\noindent {\mymagtwobf In this course, these tricks do not work:}


\medbreak If you tell me that something is true ``by the definition of real
number,'' then I will ask you to state precisely your definition of real
number, and to tell me how it applies.  If you cannot answer, then I will
take it that you really did {\bf not} know how to justify you step.

If you tell me that something is true ``by the definition of equality,''
then I will ask you to state precisely your definition of equality, and to
tell me how it applies.  %(This actually did happen in class, remember? And
%the student who proposed that ``justification'' had no idea of it applied.)
If you cannot answer, then I will take it that you really did {\bf not}
know how to justify you step.

If you tell me that something is true ``by tautology,'' then I will ask you
to tell me exactly which tautology you have in mind, and if you cannot
answer then I will take it that you really did {\bf not} know how to
justify you step.

If you tell me that something is true ``by specialization,'' then I will
ask you to tell me exactly which variable you are specializing, into which
value, and in which formula.  If you cannot answer then I will take it that
you really did {\bf not} know how to justify you step.

If you tell me that something is true ``because it is in the book,'' then I
will ask you to show me exactly where it is and exactly how it applies
here.  If you cannot answer then I will take it that you really did {\bf
not} know how to justify you step.

If you tell me that something is true ``because the teacher said it in
class,'' remember that {\mymagonebf I am the teacher} and I usually remember
quite well what I said.  And if I said something wrong---which is
possible, because I can make mistakes too--- then it's wrong, and the fact
that I said it is not a justification.  

Keep the following general principle in mind:

\bigbreak 
\vbox{
{\mymagtwobf 

\centerline{WHAT \ \ \ LOOKS \ \ \ $\,$LIKE \ \ \ A \ \ \ $\,$CONFUSED}

\medbreak
\centerline{EXPRESSION \ OF \ $\,$ A \ $\,$CLEAR \ \ $\,$THOUGHT}

\medbreak
\centerline{\ $\,$IS \ $\,$OFTEN \ $\,$NO
\ $\,$MORE \ THAN \ $\,$ A \ CLEAR\ } 

\medbreak
\centerline{EXPRESSION OF A CONFUSED THOUGHT.}

}}

\bigbreak\noindent It is {\bf your job} to persuade me that your thoughts
are not confused. Good formal proofs will do the job.  An informal proof 
may do it too, provided you know how to make it formal.  If you do not
know how to do it, then I will almost certainly know that you don't know.


\bigbreak
\centerline{\hbox to5cm{\hrulefill}}
\medbreak

\noindent In the other direction, you should also remember that proofs with
shortcuts are just shortened versions of proofs with no shortcuts at all.
To make this point, here is a proof of (I) without any shortcuts.


\medbreak

{\parindent=0pt

\bigbreak\noindent PROOF OF (I), WITHOUT USING DERIVED RULES:   


\pline{\po(1)\ste \ Assume $(\exists x)\Bigl(r\wedge q(x)\Bigr)$. 
\hfill(Preparation to apply Rule 3)} 

\pline{\po(2)\ste\ste \ Let $a$ be such that $r\wedge q(a)$. 
\hfill(Rule 6, from (1))}

\pline{\po(3)\ste\ste\ $(r\wedge q(a))\imply r$ \hfill (Rule 1; instance
of the tautology  
$(\alpha\wedge\beta)\imply\alpha$)}

\pline{\po(4)\ste\ste \ $(r\wedge q(a))\imply q(a)$ \hfill (Rule 1; instance of the tautology 
$(\alpha\wedge\beta)\imply\beta$) }
 
\pline{\po(5)\ste\ste \ $r$. \hfill (Rule 2, from (2) and (3))}

\pline{\po(6)\ste\ste \ $q(a)$. \hfill (Rule 2, from (2) and (4))}

\pline{\po(7)\ste\ste \ $(\exists x)q(x)$. \hfill(Rule 7, from (6)) }

\pline{\po(8)\ste\ste \ $r\imply\biggl((\exists x)q(x)\imply\Bigl(r\wedge
(\exists x)q(x)\Bigr)\biggr)$. \hfill (Rule 1; tautology $\alpha\imply
(\beta\imply (\alpha\wedge\beta))$)}

\pline{\po(9)\ste\ste \ $(\exists x)q(x)\imply\Bigl(r\wedge (\exists
x)q(x)\Bigr)$.  \hfill (Rule 2, from (5) and (8))}

\pline{(10)\ste\ste \ $r\wedge (\exists x)q(x)$. \hfill (Rule 2, from (7)
and (9))} 

\pline{(11)\ste \ $r\wedge (\exists x)q(x)$. \hfill (Rule 6)} 

\pline{(12)\  $(\exists x)\Bigl(r\wedge q(x)\Bigr)\imply\Bigl( r\wedge (\exists
x)q(x)\Bigr)$. \hfill (Rule 3, from (1) and (11), because we have} 


\vskip-2mm
\pline{\hfill proved $r\wedge (\exists x)q(x)$ assuming $(\exists
x)\Bigl(r\wedge q(x)\Bigr)$)} 

\pline{(13)\ste \ Assume $r\wedge (\exists x)q(x)$. 
\hfill(Preparation to apply Rule 3)} 


\pline{(14)\ste  \ $\Bigl(r\wedge (\exists x)q(x)\Bigr)\imply r$. \hfill (Rule 1;
instance of the tautology $(\alpha\wedge\beta)\imply\alpha$)}

 
\pline{(15)\ste  \ $\Bigl(r\wedge (\exists x)q(x)\Bigr)\imply (\exists x)q(x)$. 
\hfill (Rule 1; instance of the tautology  $(\alpha\wedge\beta)\imply\beta$)} 
 
\pline{(16)\ste  \ $r$. \hfill (Rule 2, from (12) and (13))}

\pline{(17)\ste  \ $(\exists x)q(x)$. \hfill (Rule 2, from (13) and (15))}

\pline{(18)\ste\ste \ Let $a$ be such that $q(a)$.\hfill (Rule 6, from (17))} 

\pline{(19)\ste\ste \ $r\imply (q(a)\imply(r\wedge q(a)))$. \hfill (Rule 1;
instance of the tautology  $\alpha\imply (\beta\imply (\alpha\wedge\beta))$)}

\pline{(20)\ste\ste \ $q(a)\imply(r\wedge q(a))$. \hfill (Rule 2, from (16) and (19))}

\pline{(21)\ste\ste \ $r\wedge q(a)$. \hfill (Rule 2, from (18) and (20))}

\pline{(22)\ste\ste \  $(\exists x)(r\wedge q(x))$. \hfill 
(Rule 7, from (21))}

\pline{(23)\ste \  $(\exists x)(r\wedge q(x))$. \hfill 
(Rule 6)}

\pline{(24) $\Bigl( r\wedge (\exists x)q(x)\Bigr)\imply
(\exists x)\Bigl(r\wedge q(x)\Bigr)$.  \hfill 
(Rule 3, from (13) and (23), because we have}

\vskip-1.2mm
\pline{\hfill proved $(\exists x)\Bigl(r\wedge q(x)\Bigr)$ 
assuming $r\wedge (\exists x)q(x)$)} 

Now let us write $A$ for $(\exists x)\Bigl(r\wedge q(x)\Bigr)$,  
$B$ for $r\wedge (\exists x)q(x)$. 


\pline{(25) $(A\imply B)\imply((B\imply A)\imply (A\Longleftrightarrow B))$. \hfill (tautology)}

\pline{(26) $(B\imply A)\imply (A\Longleftrightarrow B)$. \hfill (Rule 2,
from (12) ---which says $A\imply B$--- and (26))}


\pline{(27) $A\Longleftrightarrow B$.\hfill (Rule 2, from (24) ---which says
$B\imply A$--- and (25))}

\medbreak 
Q.E.D.
}

 
%\vfill\eject

\bigbreak\noindent PROBLEM 1:  give a completely formal proof of
(II), without shortcuts.

\vbox{
\medbreak{\sl 
In the following problems, you must first decide whether the statement to
be proved is true for all possible choices for the predicates under
consideration. If your answer is ``no,'' then you must give an example of
a predicate or predicates for which the statement is false.  If your answer
is ``yes,'' then you must give a proof.  (Shortcuts are allowed, but you
should provide full justification of each step by explaining which rule or
derived rule you are using. You can create new derived rules if you wish, 
but then you have to give them a name and justify them.)} 
}


\vbox{
\medbreak\noindent PROBLEM 2:  prove or disprove that, if $p$ and $q$
are 1-variable predicates, then 
$$(\forall x)(p(x)\vee q(x))\Longleftrightarrow((\forall x)p(x)\vee 
(\forall x)q(x))\,.$$
}

\vbox{
\medbreak\noindent PROBLEM 3:  prove or disprove that, if $p$ is a 
1-variable predicate, then 
$$\Bigl(\sim(\forall x)p(x)\Bigr)\Longleftrightarrow(\exists x)(\sim
p(x))\,.$$.
}

\vbox{
\medbreak\noindent PROBLEM 4:  prove or disprove that, if $p$ and $q$
are 1-variable predicates, then 
$$(\forall x)(p(x)\imply q(x))\imply ((\forall x)p(x)\imply (\forall x)q(x))\,.$$
}

\vbox{
\medbreak\noindent PROBLEM 5:  prove or disprove that, if $p$ and $q$
are 1-variable predicates, then 
$$((\forall x)p(x)\imply (\forall x)q(x))\imply (\forall x)(p(x)\imply q(x))\,.$$
}

\bigbreak

\centerline{\hbox to 5cm{\hrulefill}}

\bigbreak
\centerline{{\bf A SIMPLE PROOF FROM ARITHMETIC}}

\medbreak\noindent We now consider a third example, namely, a simple proof
from arithmetic.   Here are some axioms about real numbers and
addition. (The universe of discourse is ${\bf R}$, the set of real
numbers.) {\sl You will get the full list of axioms for arithmetic
in the next handout.}

\bi
\mitem{Add1.} $(\forall x)(\forall y)(x+y=y+x)$.

\mitem{Add2.} $(\forall x)(\forall y)(\forall z)\Bigl(
x+(y+z)=(x+y)+z)\Bigr)$.

\mitem{Add3.} $(\forall x)(x+0=x)$.

\mitem{Add4.} $(\forall x)(\exists y)(x+y=0)$.
\ei

\medbreak\noindent NOTE: These axioms express, respectively, 
the {\it commutative property} of addition, 
the {\it associative property} of addition, 
the fact that {\it zero is an additive identity}, 
and the fact that {\it every number has an additive inverse}.

\medbreak\noindent In addition, we need three axioms about equality:

\bi
\mitem{Eq1.} $(\forall x)(x=x)$.

\mitem{Eq2.} $(\forall x)(\forall y)(x=y\imply y=x)$.

\mitem{Eq3.} $(\forall x)(\forall y)(\forall z)((x=y\wedge y=z)\imply x=z)$.
\ei

\bigbreak
Let us prove, using these axioms, the following

\medbreak\noindent THEOREM: {\sl The additive inverse of a real number
is unique.}

\medbreak\noindent PROOF (WITH LOTS OF SHORTCUTS): Suppose $x$ is a real
number, and $y$, $z$ are additive inverses of $x$.  We have to prove that
$y=z$.  Now, the fact that $y$ and $z$ are additive inverses of $x$ means
that $x+y=0$ and $x+z=0$.  Then

\begin{eqnarray*}
y&=&y+0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm
(by\,\,\,Axiom\,\,\,Add3)}\\ 
&=&0+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm (by\,\,\,Axiom\,\,\,Add1)}\\
&=&(x+z)+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm
by\,\,\,Rule\,\,\,8,\,because}\,\,\,0=x+z)\\ 
&=&(z+x)+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,Add1})\\
&=&z+(x+y)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,Add2})\\
&=&z+0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,({\rm by\,\,\,Rule\,\,\,8,\,because}\,\,\,x+y=0)\\
&=&z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,Add3})\,.
\end{eqnarray*}



\bigbreak
\vbox{
\noindent A PROOF WITHOUT SHORTCUTS: We have to prove the
following:

$$(\forall x)(\forall y)(\forall z)((x+y=0\wedge x+z=0)\imply y=z)\,.\eqno{(@)}$$

Here is a proof: 

%\vfill\eject

{\parindent=0pt

{\baselineskip 10.7pt 

(1) \ $(\forall x)(x+0=x)$.

(2) \ $(\forall x)(\forall y)(\forall z)\Bigl(x+(y+z)=(x+y)+z\Bigr)$.

(3) \ $(\forall x)(\forall y)\Bigl(x+y=y+x\Bigr)$.

(4) \ste \ Let $a$ be arbitrary.  

(5) \ste\ste \ Let $u$ be arbitrary.

(6) \ste\ste\ste  \ Let $v$ be arbitrary.

(7) \ste\ste\ste\ste  \ Assume $a+u=0\wedge a+v=0$. 

(8) \ste\ste\ste\ste \ $(a+u=0\wedge a+v=0)\imply a+u=0$. 

(9) \ste\ste\ste\ste \ $(a+u=0\wedge a+v=0)\imply a+v=0$. 

(10) \ste\ste\ste\ste $a+u=0$. 

(11) \ste\ste\ste\ste $a+v=0$. 

(12) \ste\ste\ste\ste  $v+0=v$ 

(13) \ste\ste\ste\ste $v+(a+u)=v$


(14) \ste\ste\ste\ste $(\forall y)(\forall z)\Bigl(v+(y+z)=(v+y)+z\Bigr)$.

(15) \ste\ste\ste\ste $(\forall z)\Bigl(v+(a+z)=(v+a)+z\Bigr)$.

(16) \ste\ste\ste\ste $v+(a+u)=(v+a)+u$.

(17) \ste\ste\ste\ste $(\forall y)\Bigl(v+y=y+v\Bigr)$.

(18) \ste\ste\ste\ste $v+a=a+v$.

(19) \ste\ste\ste\ste $v+(a+u)=(a+v)+u$.

(20) \ste\ste\ste\ste $v+(a+u)=0+u$.

(21) \ste\ste\ste\ste $(\forall y)\Bigl(0+y=y+0\Bigr)$.

(22) \ste\ste\ste\ste $0+u=u+0$.

(23) \ste\ste\ste\ste $v+(a+u)=u+0$.

(24) \ste\ste\ste\ste $u+0=u$.

(25) \ste\ste\ste\ste $v+(a+u)=u$.

(26) \ste\ste\ste\ste $u=v$. 

(27) \ste\ste\ste $(a+u=0\wedge a+v=0)\imply u=v$. 

(28) \ste\ste $(\forall z)(a+u=0\wedge a+z=0)\imply u=z$.

(29) \ste $(\forall y)(\forall z)(a+y=0\wedge a+z=0)\imply y=z$.

(30) \ste $(\forall x)(\forall y)(\forall z)(x+y=0\wedge x+z=0)\imply y=z$.

\medbreak
Q.E.D.
}}}


\bigbreak

\noindent PROBLEM 6:  provide the missing justifications for 
the steps in the preceding proof.

\bigbreak

\centerline{\hbox to 5cm{\hrulefill}}

\bigbreak

\vbox{
\centerline{{\bf SET AXIOMS}}

\bigbreak\noindent SET AXIOM 1: If $S$, $T$ are sets, then
$$S=T\Longleftrightarrow (\forall x)(x\in S\Longleftrightarrow x\in T)\,.$$

\bigbreak\noindent SET AXIOM 2: If $S$ is a set and 
$p(x)$ is any formula containing the
variable $x$, no quantifiers involving $x$, and no other free variable,
then $\{x:x\in S\wedge p(x)\}$ (that is, 
$\{x\in S:p(x)\}$) is a set. 

\bigbreak\noindent SET AXIOM 3: If $p(x)$ is any formula containing the
variable $x$, and no quantifiers involving $x$,  and $p(x)$ does 
not contain the variable $y$, then 
$$(\forall y)(y\in\{x:p(x)\}\Longleftrightarrow p(y))\,.$$
}

\bigbreak\noindent {\bf THE RUSSELL PARADOX}. Set Axiom 2 should look 
strange to you. It says that if you have a set $S$, then you can form
subsets of $S$ by considering, for example, the set of all $x\in S$ that
satisfy some property $p(x)$.  It would be much nicer if we could just
state a much simpler axiom as follows:

\bigbreak\noindent SET AXIOM 2': If $p(x)$ is any formula containing the
variable $x$, no quantifiers involving $x$, and no other free variable,
then $\{x:p(x)\}$ is a set. 

\bigbreak\noindent Unfortunately, this is bad. Betrand Russell discovered 
that if you accept an axiom such as Set Axiom 2', then you can prove a
contradiction.  And, as you know, if you can prove one contradiction then
you can prove everything, including things that are false, so proving
theorems becomes useless. 

\medbreak\noindent Here is Russell's argument:

\medbreak
\bi
\item{} Suppose Set Axiom 2' was valid. Form the set
$$S=\{x:x\notin x\}\,.$$ Does $S$ belong to $S$ or not? If $S\in S$ then
$S$ must be one of the $x$'s that satisfy the condition defining $S$,
namely, $x\notin x$.  So $S\notin S$. If $S\notin S$ then $S$ is one of the
$x$'s that satisfy the condition defining $S$, namely, $x\notin x$, so
$S\in S$.  So we have proved $S\in S\Longleftrightarrow S\notin S$, which is a
contradiction.  (Using our axioms, we can say this more neatly: let
$S=\{x:x\notin x\}$. Use Set Axiom 3, specializing to $y=S$, with $p(x)$
being $x\notin x$. Then $S\in\{x:x\notin x\}\Longleftrightarrow S\notin S$. By
Rule 8, since $\{x:x\notin x\}=S$, we get $S\in S\Longleftrightarrow S\notin
S$.)
\ei

\sepa

\centerline{{\mymagtwobf THE AXIOMS OF REAL NUMBER ARITHMETIC }}

\medbreak
\centerline{{\mymagtwobf AND OTHER NUMBER SYSTEMS }}
\bigbreak



\sepa

\noindent Now we would like to be able to prove various  properties of
these number systems.  We will do this by starting with some {\sl axioms} \
or {\sl postulates}, and then using them to prove lots of other things.

First, let us write down some basic properties that we know are true for
the {\sl real} numbers.  In the following list, you should think of the
universe of discourse as being ${\bf R}$, the set of real numbers.
However, notice that $\rr$ itself is never mentioned in these statements,
so we can ask whether they are valid over the other number systems that we
have listed above.

These properties will be used as {\sl axioms} for the real number system.
They involve 

\medbreak

\bi
\item{1.} two primitive constants, namely, ``$0$'' and ``$1$,'' 

\medbreak\item{2.} one primitive one-variable predicate, namely,
``$\in{\cal P}$''  (meaning ``is strictly positive''),

\medbreak\item{3.} two primitive 2-variable function symbols, namely, 
``$+$'' and  ``$.$'', 

\medbreak\item{4.} one primitive 2-variable predicate, namely, ``$=$''. 
\ei

\medbreak In addition, we use ``$S\neq T$''  ---if 
$S$, $T$ are terms\footnote{$\,^1$}{{\mysmall See Handout 2 for an
explanation of the meaning of the word ``term''}}--- as an abbreviation of
``$\sim S=T$'', and ``$S\notin{\cal P}$'' ---if $S$ is a term--- as an
abbreviation of ``$\sim S\in{\cal P}$''.


\bigbreak\noindent Because the universe of discourse is ${\bf R}$, 
an expression such as $(\forall x)$ is exacty equivalent 
to $(\forall x\in\rrbf)$.

\bigbreak\noindent There will be {\bf 16} axioms.   The first twelve of them 
do not involve ``$\in{\cal P}$'' at all, and are known as the {\bf field
axioms}.  The last four have to do with ``$\in{\cal P}$,'' i.e. with
positivity, which is what will enable us later to define the {\sl order
relation} $>$ on $\rr$, by letting $x>y$ if there is a $z$ such that
$x=y+z$ and $z\in{\cal P}$.  So Axioms 13 through 16 are the {\sl order
axioms}. More precisely,

\medbreak\noindent {\bf DEFINITION A.} A system of objects on which there are 
binary operations $+$ and $.\,\,$, and objects $0$, $1$, such that Axioms
A1 to A12 hold is called a {\sl field}.

\medbreak\noindent {\bf DEFINITION B.} A system of objects on which there are 
binary operations $+$ and $.\,\,$, objects $0$, $1$, and a subset ${\cal P}$,
such that  Axioms A1 to A16 hold is called an {\sl ordered field}. 

\medbreak\noindent So Axioms A13,14,15,16 are the extra postulates that have 
to be added to those of a field in order to get an ordered field. So 
we will call them {\sl the order axioms of an ordered field.}
   

\bigbreak\noindent And here, finally,  are the 16 axioms: 

\bi
\mitem{A1.} $(\forall x)(\forall y)(\exists z)(x+y=z)$.

\mitem{A2.} $(\forall x)(\forall y)(\exists z)(x.y=z)$.

\mitem{A3.} $(\forall x)(\forall y)(x+y=y+x)$.

\mitem{A4.} $(\forall x)(\forall y)(x.y=y.x)$.

\mitem{A5.} $(\forall x)(\forall y)(\forall z)\Bigl(
(x+y)+z=x+(y+z)\Bigr)$.

\mitem{A6.} $(\forall x)(\forall y)(\forall z)\Bigl((x.y).z=x.(y.z)\Bigr)$.

\mitem{A7.} $(\forall x)(\forall y)(\forall z)\Bigl((x+y).z=x.z+y.z\Bigr)$.

\mitem{A8.} $(\forall x)(0+x=x)$.

\mitem{A9.} $(\forall x)(\exists y)(x+y=0)$.

\mitem{A10.} $(\forall x)(1.x=x)$.

\mitem{A11.} $(\forall x)\Bigl(x\neq 0\rightarrow(\exists y)(x.y=1)\Bigr)$. 

\mitem{A12.} $1\neq 0$.

\mitem{A13.} $(\forall x)(\forall y)\Bigl((x\in{\cal P}\wedge
y\in{\cal P})\imply x+y\in{\cal P}\Bigr)$.

\mitem{A14.} $(\forall x)(\forall y)\Bigl((x\in{\cal P}\wedge
y\in{\cal P})\imply x.y\in{\cal P}\Bigr)$. 

\mitem{A15.} $(\forall x)(x\in{\cal P}\vee x=0\vee(\forall y)(x+y=0\imply y\in{\cal P}))$. 

\mitem{A16.} $0\notin{\cal P}$.

\ei


\bigbreak\noindent In addition, we will always assume the validity of
the three axioms about equality, already presented in handout No. 2:

\bi
\mitem{Eq1.} $(\forall x)(x=x)$.

\mitem{Eq2.} $(\forall x)(\forall y)(x=y\imply y=x)$.

\mitem{Eq3.} $(\forall x)(\forall y)(\forall z)((x=y\wedge y=z)\imply x=z)$.
\ei


{\parindent=0pt

\bigbreak\noindent COMMENTS:\ Axioms A1 and A2 say that  addition and 
multiplication are binary operations {\sl defined on the whole universe of 
numbers}.  To understand thet importance of this,  we can compare with 
other operations that can be defined on the real numbers, such 
as division or exponentiation, that are only
``partially defined.''   For example, the quotient
${x\over y}$ is not going to be defined for all $x$ and $y$, but 
only when $y\neq 0$.  (So at some point we will prove:
$(\forall x)(\forall y)\Big(y\neq 0\imply(\exists z)(z={x\over y})\Big)$.)
Similarly,  the exponential $x^y$ is going to be defined 
for all real numbers $x$ and all {\sl positive} real numbrs $y$
(which means that there will be a theorem saying that 
$(\forall x)(\forall y)\Big(y\in{\cal P}\imply(\exists z)(z=x^y\Big)$.)


Thanks to these two axioms,  if you have a term such as $1+1$,  
you can be sure (by specializing (A1) to $x=1$, $y=1$) that 
$(\exists z)(1+1=z)$.  Once you know this, Rule 6 enables you to introduce
a constant symbol (for example, the symbol ``$2$'') for which 
$$1+1=2\,.$$  

Axioms A3 and A4 are the {\it commutative laws} of addition and
multiplication. 

Axioms A5 and A6 are the {\it associative laws} of addition and
multiplication. 

Axiom A7 is the {\it distributive law} of multiplication with respect to
addition.

Axiom A8 says that $0$ is an additive identity, that is, that 
adding $0$ to any number leaves the number unchanged. 

Axiom A9 says that every real number has an additive inverse.

Axiom A10 says that $1$ is a mutiplicative identity, that is, that 
mutiplying any number by $1$ leaves the number unchanged. 

Axiom A11 says that every nonzero real number has a multiplicative inverse.

Axiom A12 says that $0$ and $1$ are different. If you do not
believe that A12 is necessary, take a look at the other axioms. Notice that
they would all be satisfied if $\rrbf$ consisted of just $0$ and nothing
else, and $1$ was equal to $0$.  So unless we put in Axiom A12 we will
never be able to prove that $1\neq 0$.

Axiom A13 says that the sum of two positive numbers is positive. 

Axiom A14 says that the product of two positive numbers is positive. 

Axioms A15 says that every number $x$ is either positive, negative, or
zero.  We say it in a roundabout way because we have not yet defined what
we mean by a ``negative number.''  Instead of saying ``$x$ is negative,''
we say ``$(\forall y)(x+y=0\imply y\in{\cal P})$.''

Axiom A16 says that $0$ is not positive.

}

\bigbreak\noindent
If you recall the definitions of ``field'' and ``ordered field''
given above (Definitions A and B), we can summarize them as follows:  

\medbreak
\bi
\item{A.}
{\sl A {\bf field} is a system 
of objects  ---or ``numbers''--- on which two binary operations  
``+'' and ``,'' (``addition'' 
and ``multiplication'') are specified,
both of which are commutative and associative, and such that
multiplication is distributive with respect to addition, as well as 
two different\footnote{$\,\,^2$}{{\mysmall The fact that $0$ and $1$ are
different  
is Axiom A12.}}  numbers $0$ and $1$ that are, respectively, identity 
elements for the addition and multiplication operations, and are such 
that every number has an additive inverse and every nonzero number 
has a multiplicative inverse.}

\medbreak\item{B.}
{\sl An {\bf ordered field} is a field in which, in addition, 
a set ${\cal P}$ of numbers ---called the set of {\sl positive}
numbers--- is specified, in such a way that the sum and the product
of two positive numbers is positive, $0$ is not positive, and
every nonzero number is either positive or the additive inverse
of a positive number.} 
\ei

\bigbreak

\centerline{{\mymagonebf SOME EXAMPLES OF PROOFS}} 

\medbreak\noindent Here are some examples of proofs, using the axioms 
of real arithmetic plus the axioms on equality.

\medbreak\noindent First, let us define ``additive inverse'':  

\medbreak\noindent{\bf DEFINITION}:  An {\sl additive inverse} of a number
$x$ is a number $y$ such that $x+y=0$. 


\medbreak\noindent{\bf DEFINITION}:  A {\sl multiplicative inverse} of a number
$x$ is a number $y$ such that $x.y=1$. 

\medbreak\noindent We can then prove: 


\medbreak\noindent THEOREM 1.  {\sl The additive inverse of every 
number is unique.} In formal language:
$$(\forall x)(\forall y)(\forall z)\Bigl((x+y=0\wedge
x+z=0)\imply y=z\Bigr)\,.$$
(This just says that any two additive inverses of $x$ must be the same.)


\medbreak

\noindent PROOF (WITH SHORTCUTS): Suppose $x$ is  an arbitrary 
number,
and $y$, $z$ are  additive inverses of $x$.  We have to prove that $y=z$.
Now, the fact that $y$ and $z$ are  additive inverses of $x$ means that
$x+y=0$ and $x+z=0$.  Then 
\begin{eqnarray*}
y&=&0+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm (by\,\,\,Axiom\,\,\,A8)}\\
&=&(x+z)+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm
by\,\,\,Rule\,\,\,8,\,because}\,\,\,0=x+z)\\
&=&(z+x)+y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,A3})\\
&=&z+(x+y)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,A5})\\
&=&z+0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,({\rm by\,\,\,Rule\,\,\,8,\,because}\,\,\,x+y=0)\\
&=&0+z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,({\rm by\,\,\,Axiom\,\,\,A3})\\
&=&z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\rm by\,\,\,Axiom\,\,\,A8})\,.\\
&\quad\quad{\rm Q.E.D.}
\end{eqnarray*}

\medbreak\noindent{\bf REMARK.} \ The letters ``Q.E.D.'' are often used to
indicate the end of a proof. They stand for  the Latin words
``quod erat demonstrandum,'' which mean ``which is what was to be proved.''

\bigbreak
\medbreak\noindent THEOREM 2.  {\sl The multiplicative inverse of every 
number is unique.} In formal language:
$$(\forall x)(\forall y)(\forall z)\Bigl((x.y=1\wedge
x.z=1)\imply y=z\Bigr)\,.$$

\medbreak
\noindent PROOF (WITH SHORTCUTS): Suppose $x$ is  an arbitrary 
number,
and $y$, $z$ are  multiplicative inverses of $x$.  We have to prove that $y=z$.
Now, the fact that $y$ and $z$ are  multiplicative inverses of $x$ means that
$x.y=1$ and $x.z=1$.  Then 
\begin{eqnarray*}
y&=&1.y\\
&=&(x.z).y\\
&=&(z.x).y\\
&=&z.(x.y)\\
&=&z.1\\
&=&1.z\\
&=&z\,.\\
&\quad&\quad{\rm Q.E.D.}
\end{eqnarray*}

\medbreak\noindent A PROOF OF THEOREM 1 WITHOUT SHORTCUTS: We have to prove the
following:
$$(\forall x)(\forall y)(\forall z)((x+y=0\wedge x+z=0)\imply y=z)\,.\eqno{(@)}$$

\medbreak\noindent Here is a proof: 

{\parindent=0pt

{\baselineskip 10.7pt 

\ (1) \ $(\forall x)(0+x=x)$.

\ (2) \ $(\forall x)(\forall y)(\forall z)\Bigl(x+(y+z)=(x+y)+z\Bigr)$.

\ (3) \ $(\forall x)(\forall y)\Bigl(x+y=y+x\Bigr)$.

\ (4) \ste \ Let $a$ be arbitrary.  

\ (5) \ste\ste \ Let $u$ be arbitrary.

\ (6) \ste\ste\ste  \ Let $v$ be arbitrary.

\ (7) \ste\ste\ste\ste  \ Assume $a+u=0\wedge a+v=0$. 

\ (8) \ste\ste\ste\ste \ $(a+u=0\wedge a+v=0)\imply a+u=0$. 

\ (9) \ste\ste\ste\ste \ $(a+u=0\wedge a+v=0)\imply a+v=0$. 

(10) \ste\ste\ste\ste $a+u=0$. 

(11) \ste\ste\ste\ste $a+v=0$. 

(12) \ste\ste\ste\ste  $0+v=v$ 

(13) \ste\ste\ste\ste $(\forall y)\Bigl(v+y=y+v\Bigr)$.

(14) \ste\ste\ste\ste $v+0=0+v$.

(15) \ste\ste\ste\ste $v+0=v$.

(16) \ste\ste\ste\ste $v+(a+u)=v$

(17) \ste\ste\ste\ste $(\forall y)(\forall z)\Bigl(v+(y+z)=(v+y)+z\Bigr)$.

(18) \ste\ste\ste\ste $(\forall z)\Bigl(v+(a+z)=(v+a)+z\Bigr)$.

(19) \ste\ste\ste\ste $v+(a+u)=(v+a)+u$.

(20) \ste\ste\ste\ste $v+a=a+v$.

(21) \ste\ste\ste\ste $v+(a+u)=(a+v)+u$.

(22) \ste\ste\ste\ste $v+(a+u)=0+u$.

(23) \ste\ste\ste\ste $0+u=u$.

(24) \ste\ste\ste\ste $v+(a+u)=u$.

(25) \ste\ste\ste\ste $u+0=u$.

(26) \ste\ste\ste\ste $v+(a+u)=u$.

(27) \ste\ste\ste\ste $u=v$. 

(28) \ste\ste\ste $(a+u=0\wedge a+v=0)\imply u=v$. 

(29) \ste\ste $(\forall z)(a+u=0\wedge a+z=0)\imply u=z$.

(30) \ste $(\forall y)(\forall z)(a+y=0\wedge a+z=0)\imply y=z$.

(31) \ste $(\forall x)(\forall y)(\forall z)(x+y=0\wedge x+z=0)\imply y=z$.

\medbreak
Q.E.D.
}}


\bigbreak\noindent PROBLEM 1:  provide the missing justifications for 
the steps in the  proof with shortcuts of Theorem 2.


\bigbreak\noindent PROBLEM 2:  provide the missing justifications for 
the steps in the proof of Theorem 1 without shortcuts.


\bigbreak\noindent PROBLEM 3:  give a proof without shortcuts of Theorem 2.

\bigbreak\noindent Next, recall that the notation
$$
(\exists !x)p(x)
$$
means ``there is one and only one $x$ such that $p(x)$.''  (For example:
if our universe of discourse is the set of all people ---alive or dead---
and $p(x,y)$ means ``$y$ is $x$'s mother,'' $q(x,y)$ means 
``$y$ is $x$'s brother,''  and $g(x,y)$ means ``$y$ is 
$x$'s grandfather,'' then  $(\forall x)(\exists ! y)p(x,y)$
says that ``everybody has one and only one mother,'' which is true.
On the other hand, $(\forall x)(\exists ! y)q(x,y)$ says
that ``everybody has one and only one brother,'' which is not true,
because some people have no brothers at all, and other people have more
that one brother.  Finally, $(\forall x)(\exists ! y)g(x,y)$ says
that ``everybody has one and only one grandfather,'' which is not true,
because people\footnote{$\,\,^3$}{except in some rare cases, such as some 
ancient Egyptian kings.} have {\sl two} grandfathers.)

\bigbreak
\vbox{
\noindent THEOREM 3.  {\sl Every number has a 
unique  additive inverse.}  In formal language:
$$(\forall x)(\exists! y)(x+y=0)\,.$$
 

\medbreak\noindent PROOF: Suppose $x$ is  an arbitrary number. By Axiom A9, $x$ has
an additive inverse.  By Theorem 1, this inverse is unique. Q.E.D.
}



\bigbreak
\vbox{
\noindent THEOREM 4.  {\sl Every nonzero number has a 
unique  multiplicative inverse.}  In formal language:
$$(\forall x)\Big(x\neq 0\imply(\exists! y)(x.y=1)\Big)\,.$$
 

\medbreak\noindent PROOF: Suppose $x$ is  an arbitrary nonzero number.
By Axiom A11, $x$ has
a multiplicative inverse.  By Theorem 2, this inverse is unique. Q.E.D.
}


\bigbreak
\vbox{
\noindent THEOREM 5.  {\sl For every number $x$, we have 
$x.0=0$.}  In formal language:
$$(\forall x)(x.0=0)\,.$$
 

\medbreak\noindent PROOF: Suppose $x$ is  an arbitrary number.
Since $0+0=0$ by Axiom A8, we have $x.(0+0)=x.0$ (using 
Axiom Eq1, specialized to $x.0$, to get $x.0=x.0$, and 
then Rule 8 to plug in $0+0$ instead of the first $0$; the specialization
of $(\forall x)(x=x)$ to $x.0$ is justified because
Axiom A2 implies $(\exists y)(x.0=y)$, so we can introduce 
a constant $a$ for which $x.0=a$, and then we can specialize
$(\forall x)(x=x)$ to get $a=a$, and using Rule 8 we then get $x.0=x.0$).

Now Axiom A7 implies $x.(0+0)=x.0+x.0$. So Rule 8 gives us
$$
x.0+x.0=x.0\,.
$$ 
Now we can add te additive inverse of $x.0$ to both sides. More precisely,
we can specialize Axiom A9 to get $(\exists y)(x.0+y=0)$. (Why?)
Then we can use Rule 6 to introduce a constant $b$ such that 
$x.0+b=0$.   We then have 
$0=x.0+b=(x.0+x.0)+b=x.0+(x.0+b)=x.0+0=0+x.0=x.0$, using
Rule 8, Axiom A5, Rule 8,  Axiom A3 and Axiom A8, in that order.\quad Q.E.D.
} 

\medbreak\noindent PROBLEM 4:  Answer the question  that appears 
in the proof of Theorem 5. 

\bigbreak\noindent We know from Theorem 3 that every number has a unique 
additive inverse.   Let us give a name to the additive inverse of $1$. Let us 
call it $-1$.  


\bigbreak
\vbox{
\noindent THEOREM 6.    $(-1).(-1)=1$.

\medbreak\noindent PROOF.  We know that $1+(-1)=0$.  Axiom A3 then 
implies $(-1)+1=0$, so $1$ is the additive inverse of $-1$. 
Suppose we prove that 
$$
(-1)+(-1).(-1)=0\,.\eqno{(1)}$$
This  would tell us that $(-1).(-1)$ is also
an additive inverse of $-1$. Since the additive inverse is unique, 
we would get $(-1).(-1)=1$, as desired.

So all we need to do is prove (1). But this is easy:
$$(-1)+(-1).(-1)=1.(-1)+(-1).(-1)=(-1).1+(-1).(-1)=(-1).(1+(-1))=(-1).0=0\,$$
using Axioms A3, Rule 8, Axiom A7, and Theorem 5.  \quad Q.E.D.
}


\bigbreak
\vbox{
\noindent THEOREM 7.    $1\in{\cal P}$.

\medbreak\noindent PROOF.   Axiom A15 says that either $1\in{\cal P}$, or 
$1=0$, or $-1\in{\cal P}$.   The possibility that 
$1=0$ is excluded by Axiom A12.  If $1\in{\cal P}$ then we got what we wanted. 

So the only case left for us to consider is when 
$-1\in{\cal P}$.   But in this case $(-1).(-1)\in{\cal P}$ by Axiom A14.
Theorem 6 tells us that $(-1).(-1)=1$. So $1\in{\cal P}$.  So we have 
shown  that $1\in{\cal P}$ in all possible cases.\quad Q.E.D.
}

\bigbreak
\vbox{
\noindent THEOREM 8.    $-1\notin{\cal P}$. 

\medbreak\noindent PROOF.  Suppose $-1\in{\cal P}$.  We already know that
$1\in{\cal P}$.  Then Axiom A13 implies that $1+(-1)\in{\cal P}$. 
So $0\in{\cal P}$. But this is impossible, because Axiom A16 says that
$0\notin{\cal P}$.   
}

\medbreak\noindent We now introduce the symbols $2$, $3$ and $4$ as abbreviations 
for $1+1$, $2+1$ and $3+1$, respectively.   

\medbreak\noindent PROBLEM 5:  Prove that $2\neq 0$.

\medbreak\noindent PROBLEM 6:  Prove that $2+2=4$.

\medbreak\noindent PROBLEM 7:  Prove that $2.2=4$.

\medbreak\noindent PROBLEM 8:  Prove that 
$$
(\forall x)(x\in{\cal P}\imply(\exists y)(\exists z)(y\in{\cal P}\wedge
z\in{\cal P}\wedge y+z=x)\,.
$$
(Hint, you may wish to take $y=z={x\over 2}$.   But we don't yet know
what ``${x\over 2}$'' means, so you have to get around that. The result
of Problem 5 may be useful here.)  

\sepa

\centerline{{\bf A PROOF OF THEOREM 7 WITHOUT ANY SHORTCUTS}} 



\medbreak\noindent This is a pretty silly section, but you should 
read it anyhow.  

The point of this section is to remind you that {\bf a proof with shortcuts
is basically an abbreviated form of a proof without shortcuts.}
Ultimately, every proof with shortcuts can be translated into 
a proof without shortcuts.  The problem with this is that the proofs without
shortcuts  are usually enormously long, and once you know how to do such proofs
it would be a waste of time and energy to write them out in detail. 

You should think of theorem-proving without shortcuts as a {\bf ladder}
that you can climb until you get to the higher level of doing proofs
without shortcuts, in a more informal language.  Once you are confidently
settled at the top level, you can throw the ladder away, because you no
longer need it.  But, if you throw away the ladder before you get to the
top, you will fall.  (Or, if you prefer: {\bf when you make things up
without knowing how to justify them in terms of the basic rules and axioms,
it shows, and I can tell.})


So, to bring this point home, let us give a proof 
of Theorem 7 {\sl directly from first principles, not using 
any shortcuts, and not even using previous theorems.} We will 
just use the axioms and the rules of inference, and nothing else. 

Naturally, the strategy for the proof is basically the same as what we have 
already done. You can regard the proof of Theorem 7 given above as an 
{\sl outline} of the ``true'' proof that will be presented now.  

\medbreak\noindent WARNING: this is going to be painful, because it's very
long.  But it is quite easy.  Every step is in fact completely trivial.

\medbreak\noindent And here is the proof.

\medbreak  
\pline{\noindent \ (1) \ste$(\forall x)\Bigl(x\in{\cal P}\vee
x=0\vee(\forall y)(x+y=0\imply y\in{\cal P})\Bigr)$.  \hfill (Axiom A15.)}

\pline{\noindent \ (2) \ste$(\forall x)(\forall y)(\exists z)(x.y=z)$.
\hfill (Axiom A2.)}

\pline{\noindent \ (3) \ste$(\forall x)(\forall y)(x+y=y+x)$.
\hfill (Axiom A3.)}

\pline{\noindent \ (4) \ste$(\forall x)(\exists y)(x+y=0)$.  \hfill (Axiom A9.)}

\pline{\noindent \ (5) \ste$(\forall x)(\forall y)(\forall
z)\Bigl((x+y).z=x.z+y.z\Bigr)$.  \hfill (Axiom A7.)}

\pline{\noindent \ (6) \ste$(\forall x)(\forall y)\Bigl((x\in{\cal P}\wedge
y\in{\cal P})\imply x.y\in{\cal P}\Bigr)$. \hfill (Axiom A13).} 

\pline{\noindent \ (7) \ste$(\forall x)(1.x=x)$. \hfill (Axiom A10.)}

\pline{\noindent \ (8) \ste$(\forall x)(\forall y)(\forall z)\Bigl(
(x+y)+z=x+(y+z)\Bigr)$. \hfill (Axiom A5.)}

\pline{\noindent \ (9) \ste$\sim 1=0$. \hfill (Axiom A12.)}

\pline{\noindent (10) \ste$(\forall x)(0+x=x)$. \hfill (Axiom A8.)}

\pline{\noindent (11) \ste$1\in{\cal P}\vee
1=0\vee(\forall y)(1+y=0\imply y\in{\cal P})$.  \hfill (Rule 4, from (1),
specialized to $x=1$.)}

\pline{\noindent (12) \ste$(\forall y)(0+y=y+0)$.
\hfill (Rule 4, from (3), specialized to $x=0$.)}

\pline{\noindent (13) \ste$(\exists y)(1+y=0)$.  \hfill 
(Rule 4, from (4), specialized to $x=1$.)}

\pline{\noindent (14.1) \ste\ste Assume $(\forall y)(1+y=0\imply
y\in{\cal P})$. \hfill (Preparation to apply Rule 3.)} 

\pline{\noindent (14.2.1) \ste\ste\ste Let $a$ be such that $1+a=0$. \hfill
(Rule 6, from (13).)}

\pline{\noindent (14.2.2) \ste\ste\ste $1+a=0\imply a\in{\cal P}$. 
\hfill (Rule 4, from (14.1), specialized to $y=a$.)}

\pline{\noindent (14.2.3) \ste\ste\ste $a\in{\cal P}$. \hfill (Rule 2,
from (14.2.1) and (14.2.2).)}

\pline{\noindent (14.2.4) \ste\ste\ste $(\forall y)(\forall z)((0+y).z=
0.z+y.z)$ \hfill (Rule 4, from (5), specialized to $x=0$.)}

\pline{\noindent (14.2.5) \ste\ste\ste $(\forall z)((0+0).z=
0.z+0.z)$ \hfill (Rule 4, from (14.2.4), specialized to $y=0$.)}

\pline{\noindent (14.2.6) \ste\ste\ste $(0+0).a=
0.a+0.a$ \hfill (Rule 4, from (14.2.5), specialized to $z=a$.)}

\pline{\noindent (14.2.7) \ste\ste\ste $0+0=0$. \hfill 
(Rule 4, from (10), specialized to $x=0$.)}

\pline{\noindent (14.2.8) \ste\ste\ste $0.a=0.a+0.a$.
\hfill (Rule 8, from (14.2.6) and (14.2.7).)}

\pline{\noindent (14.2.9) \ste\ste\ste $(\forall y)(\exists z)(0.y=z)$.
\hfill (Rule 4, from (2), specialized to $x=0$.)}

\pline{\noindent (14.2.10) \ste\ste\ste$\!\!(\exists z)(0.a=z)$.
\hfill (Rule 4, from (14.2.9), specialized to $y=a$.)}


\pline{\noindent (14.2.11.1) \ste\ste\ste\ste Let $b$ be such that $0.a=b$.
\hfill (Rule 6, from (14.2.10).)}

\pline{\noindent (14.2.11.2) \ste\ste\ste\ste $b=b+b$.
\hfill (Rule 8, from (14.2.8) and (14.2.11.1).} 

\pline{\noindent (14.2.11.3) \ste\ste\ste\ste $(\exists y)(b+y=0)$.
\hfill (Rule 4, from (4), specialized to $x=b$.)}

\pline{\noindent (14.2.11.4.1) \ste\ste\ste\ste\ste Let
$c$ be such that $b+c=0$.
\hfill (Rule 6, from (14.2.11.3).)}

\pline{\noindent (14.2.11.4.2) \ste\ste\ste\ste\ste $(b+b)+c=0$.
\hfill (Rule 8, from (14.2.11.2) and (14.2.11.4.1).} 

\pline{\noindent (14.2.11.4.3) \ste\ste\ste\ste\ste $(\forall y)(\forall
z)\Bigl((b+y)+z=b+(y+z)\Bigr)$. \hfill (Rule 4, from (8), specialized to $x=b$.)}

\pline{\noindent (14.2.11.4.4) \ste\ste\ste\ste\ste $(\forall z)\Bigl(
(b+b)+z=b+(b+z)\Bigr)$. \hfill (Rule 4, from (14.2.11.4.3), spec. to
$y=b$.)} 

\pline{\noindent (14.2.11.4.5) \ste\ste\ste\ste\ste 
$(b+b)+c=b+(b+c)$. \hfill  (Rule 4, from (14.2.11.4.4), specialized to $z=c$.)}

\pline{\noindent (14.2.11.4.6) \ste\ste\ste\ste\ste $(b+b)+c=b+0$. \hfill 
(Rule 8, from (14.2.11.4.1) and (14.2.11.4.5).)}

\pline{\noindent (14.2.11.4.7) \ste\ste\ste\ste\ste $0=b+0$. \hfill (Rule
8, from (14.2.11.4.2) and (14.2.11.4.6).)} 

\pline{\noindent (14.2.11.4.8) \ste\ste\ste\ste\ste $0+b=b$. \hfill 
(Rule 4, from (10), specialized to $x=b$.)}

\pline{\noindent (14.2.11.4.9) \ste\ste\ste\ste\ste $0+b=b+0$. \hfill 
(Rule 4, from (3), specialized to $x=b$.)}

\pline{\noindent (14.2.11.4.10) \ste\ste\ste\ste\ste$\!\!b+0=b$. \hfill 
(Rule 8, from (14.2.11.4.8) and (14.2.11.4.9).)}

\pline{\noindent (14.2.11.4.11) \ste\ste\ste\ste\ste$\!\!0=b$. \hfill 
(Rule 8, from (14.2.11.4.7) and (14.2.11.4.10).)}

\pline{\noindent (14.2.11.4.12) \ste\ste\ste\ste\ste$\!\!0=0.a$. \hfill 
(Rule 8, from (14.2.11.1) and (14.2.11.4.11).)}

\pline{\noindent (14.2.11.5) \ste\ste\ste\ste$\!\!0=0.a$. \hfill 
(Rule 6, from (14.2.11.4.1) and (14.2.11.4.12.)}

\pline{\noindent (14.2.12) \ste\ste\ste$\!\!0=0.a$. \hfill 
(Rule 6, from (14.2.11.1) and (14.2.11.5)}

\pline{\noindent (14.2.13) \ste\ste\ste$\!\!0=(1+a).a$. \hfill 
(Rule 8, from (14.2.1) and (14.2.12).)}

\pline{\noindent (14.2.14) \ste\ste\ste$\!\!(\forall y)(\forall
z)\Bigl((1+y).z=1.z+y.z\Bigr)$.  \hfill  (Rule 4, from (5), specialized to
$x=1$.)} 

\pline{\noindent (14.2.15) \ste\ste\ste$\!\!(\forall z)\Bigl((1+a).z=
1.z+a.z\Bigr)$.  \hfill (Rule 4, from (14.2.14), specialized to $y=a$.)}

\pline{\noindent (14.2.16) \ste\ste\ste$\!\!(1+a).a=1.a+a.a$.  \hfill 
(Rule 4, from (14.2.15), specialized to $z=a$.)}

\pline{\noindent (14.2.17) \ste\ste\ste$\!\!0=1.a+a.a$. \hfill 
(Rule 8, from (14.2.13) and (14.2.16).)}

\pline{\noindent (14.2.18) \ste\ste\ste$\!\!1.a=a$. \hfill 
(Rule 4, from (7), specializing to $x=a$.)}

\pline{\noindent (14.2.19) \ste\ste\ste$\!\!0=a+a.a$. \hfill 
(Rule 8, from (14.2.17) and (14.2.18).)}

\pline{\noindent (14.2.20) \ste\ste\ste$\!\!0+1=1$. \hfill 
\hfill (Rule 4, from (10) specialized to $x=1$.)}

\pline{\noindent (14.2.21) \ste\ste\ste$\!\!0+1=1+0$. \hfill 
\hfill (Rule 4, from (12) specialized to $y=1$.)}

\pline{\noindent (14.2.22) \ste\ste\ste$\!\!1+0=1$. \hfill 
(Rule 8, from (14.2.20) and (14.2.21).)}

\pline{\noindent (14.2.23) \ste\ste\ste$\!\!1+(a+a.a)=1$. \hfill 
(Rule 8, from (14.2.19) and (14.2.22).)}


\pline{\noindent (14.2.24) \ste\ste\ste$\!\!(\forall y)(\forall z)\Bigl(
(1+y)+z=1+(y+z)\Bigr)$. \hfill 
(Rule 4, from (8), specialized to $x=1$.)}

\pline{\noindent (14.2.25) \ste\ste\ste$\!\!(\forall z)\Bigl(
(1+a)+z=1+(a+z)\Bigr)$. \hfill 
(Rule 4, from (14.2.24), specialized to $y=a$.)}

\pline{\noindent (14.2.26) \ste\ste\ste$\!\!(\forall y)(\exists z)(a.y=z)$.
\hfill (Rule 4, from (2), specialized to $x=a$.)}

\pline{\noindent (14.2.27) \ste\ste\ste$\!\!(\exists z)(a.a=z)$.
\hfill (Rule 4, from (14.2.26), specialized to $y=a$.)}

\pline{\noindent (14.2.28.1) \ste\ste\ste\ste Let $d$ be such that
$a.a=d$. \hfill (Rule 6, from (14.2.27).)}

\pline{\noindent (14.2.28.2) \ste\ste\ste\ste 
$(1+a)+d=1+(a+d)$. \hfill  (Rule 4, from (14.2.25), specialized to $z=d$.)}

\pline{\noindent (14.2.28.3) \ste\ste\ste\ste 
$1+(a+d)=1$. \hfill  (Rule 8, from (14.2.23) and (14.2.28.1).)}

\pline{\noindent (14.2.28.4) \ste\ste\ste\ste 
$(1+a)+d=1$. \hfill  (Rule 8, from (14.2.28.2) and (14.2.28.3).)}

\pline{\noindent (14.2.28.5) \ste\ste\ste\ste 
$0+d=1$. \hfill  (Rule 8, from (14.2.1) and (14.2.28.4).)}

\pline{\noindent (14.2.28.6) \ste\ste\ste\ste 
$0+d=d$. \hfill (Rule 4, from (10), specialized to $x=d$.)}

\pline{\noindent (14.2.28.7) \ste\ste\ste\ste 
$d=1$. \hfill  (Rule 8, from (14.2.28.5) and (14.2.28.6).)}

\pline{\noindent (14.2.28.8) \ste\ste\ste\ste $a.a=1$. \hfill 
(Rule 8, from (14.2.28.1) and (14.2.28.7).)}


\pline{\noindent (14.2.29) \ste\ste\ste$\!\!a.a=1$.
\hfill (Rule 6, from (14.2.28.1) and (14.2.28.8).)}

\pline{\noindent (14.2.30) \ste\ste\ste$\!\!(\forall y)\Bigl((a\in{\cal P}\wedge
y\in{\cal P})\imply a.y\in{\cal P}\Bigr)$. \hfill
\hfill (Rule 4, from (1), specialized to $x=a$.)}

\pline{\noindent (14.2.31) \ste\ste\ste$\!\!(a\in{\cal P}\wedge
a\in{\cal P})\imply a.a\in{\cal P}$. \hfill
\hfill (Rule 4, from (14.2.30), specialized to $x=a$.)}

\pline{\noindent (14.2.32) \ste\ste\ste$\!\!a\in{\cal P}\imply 
(a\in{\cal P}\wedge a\in{\cal P})$.
\hfill(Rule 1, tautology $\alpha\imply(\alpha\wedge\alpha)$.)}

\pline{\noindent (14.2.33) \ste\ste\ste$\!\!a\in{\cal P}\wedge a\in{\cal P}$.
\hfill(Rule 2, from (14.2.3) and (14.2.32).)}

\pline{\noindent (14.2.34) \ste\ste\ste$\!\!a.a\in{\cal P}$.  \hfill(Rule 2,
from (14.2.31) and (14.2.33).)}

\pline{\noindent (14.2.35) \ste\ste\ste$\!\!1\in{\cal P}$.  \hfill(Rule 8,
from (14.2.29) and (14.2.34).)}

\pline{\noindent (14.3) \ste\ste $1\in{\cal P}$.  \hfill(Rule 6,
from (14.2.1) and (14.2.35).)}

\pline{\noindent (15) \ste $(\forall y)(y+1=0\imply y\in{\cal P})\imply 1\in{\cal
P}$.  \hfill(Rule 3, from (14.1) and (14.3).)}

\pline{\noindent (16) \ste $\Bigl(1\in{\cal P}\vee
1=0\vee(\forall y)(y+1=0\imply y\in{\cal P})\Bigr)\imply$\hfill}

\pline{\hfill $\Biggl(\sim 1=0\imply\biggl(\Bigl(
(\forall y)(y+1=0\imply y\in{\cal P})\imply 1\in{\cal
P}\Bigr)\imply 1\in{\cal P}\biggr)\Biggr)$.  \hfill}

\pline{\hfill
(Instance of the tautology  $(p\vee q\vee r)\imply
\Bigl((\sim q)\imply((r\imply p)\imply p)\Bigr)$.)}

\pline{\noindent (17) \ste $\sim 1=0\imply\biggl(\Bigl(
(\forall y)(y+1=0\imply y\in{\cal P})\imply 1\in{\cal
P}\Bigr)\imply 1\in{\cal P}\biggr)$.  \hfill (Rule 2, from (11) and (16).)}

\pline{\noindent (18) \ \ $\Bigl(
(\forall y)(y+1=0\imply y\in{\cal P})\imply 1\in{\cal
P}\Bigr)\imply 1\in{\cal P}$.  \hfill (Rule 2, from (9) and (17).)}

\pline{\noindent (19) \ \ $1\in{\cal P}$.  
\hfill (Rule 2, from (15) and (18).)}

\medbreak\noindent{\bf WE GOT THERE!  WE DID IT! IT'S PROVED!  


\medbreak\noindent AND IT TOOK US {\mymagtwobf ONLY 78 STEPS.}}

\bigbreak\noindent{\mymagfourbf\ \ \ Q.E.D.}

\sepa

Of the sixteen axioms A1 to A16, which ones are true for our other number
systems?

Axioms A1,2,3,4,5,6,7, and A10 are true in {\bf all} of them. Axioms A8 and
A12 are true in all them also but, technically, $0\notin\nn$, so we should
not count A8 and A12 as holding there, because, when the universe is $\nn$,
$0$ does not exist.  Axiom A9 is true in $\zz$, $\qq$, $\rr$ and $\cc$, but
not in $\nn$ or $\nn_{\ge 0}$.  Axiom A11 is true in $\qq$, $\rr$ and
$\cc$, but not in $\nn$, $\nn_{\ge 0}$ or $\zz$. 

As for Axioms A13,1,4,15,16, they are true in $\rr$ {\sl and also in
$\qq$}, if we define ``positive'' in the usual way.  Are they true 
in other systems, e.g. in $\cc$?    Well, that would depend on what the set
${\cal P}$ is, that is, on how we might choose to define what it means for
a complex number to be ``positive.''  It turns out that {\sl there is no
way to do it.}

\medbreak\noindent PROBLEM 9:  Prove that there is no way to define a
subset ${\cal P}$ of $\cc$ so that Axioms A1 to A16 hold.  (In other words,
there is no way to make $\cc$ into an ordered field.)   Hint: 
suppose you could. Would $i$ be positive?   

\medbreak\noindent We have seen that all the axioms A1 to A16 hold 
on $\qq$ as well as on $\rr$.  This means that, as far as Axioms A1 to A16 
are concerned, one cannot distinguish between $\rr$ and $\qq$.   
So, {\bf as a system of axioms for $\rr$, A1-A16 is not strong enough 
to give us everything that is true in $\rr$, including those things that
aren't true in $\qq$.}

\medbreak For example: the statement
$$
(\forall x)(x\in{\cal P}\imply(\exists y)(y.y=x))
$$
(i.e. ``every positive number has a square root'') is true in
$\rr$ but not in $\qq$.  So {\bf this statement cannot be proved
from A1-A16 alone} (because, if it could be proved from A1-A16, then it
would be true in every number system where A1-A16 are true, and in
particular in $\qq$). 

\medbreak\noindent What happens with the integers modulo $m$? We proved in
class that

\medbreak

\bi
\item{1.} If $m$ is prime then $\zz_m$ is a field.

\medbreak\item{2.} If $m$ is not prime then $\zz_m$ is not a field.
\ei


\medbreak\noindent PROBLEM 10:  Let $m\in\nn$, $m>1$.
Prove that there is no way to define a subset ${\cal P}$ of $\zz_m$ so that
Axioms A1 to A16 hold.  (In other words, there is no way to make $\zz_m$
into an ordered field.)

\medbreak\noindent PROBLEM 11:  Find the multiplicative inverse
of 35 (i.e. the equivalence class of 35) in $Z_{89}$. 

\medbreak\noindent PROBLEM 12:  Find a nonzero member 
of $\zz_{124}$ that has no multiplicative inverse.

\end{document}





\bigbreak\noindent\fbox{
\begin{minipage}{5.6in}

\parindent=0pt

\null
\vspace{0.1mm}




\begin{center} 
{\bf AN EXAMPLE OF THE USE OF RULES F1 TO F9}
\end{center}

Let us look at the formula $F$ defined earlier. 
Recall that $F$ is
$$
(\exists y)c(y,y)\vee
(\exists x)(\forall y)(\exists z) (((s(a,x)\wedge s(a,y))\wedge((\sim
\!x\!=\!y)\wedge(\sim\! a\!=\!z))) \wedge ((c(z,x)\wedge c(z,y))\wedge c(b,z)))\,.
$$
Now, here is a ``derivation'' of $F$ from the rules: 

\medbreak

STEP $\,\,\,$1:$\,\,$ $c(y,y)$ is an a-formula by Rule F1.

STEP $\,\,\,$2:$\,\,$ $(\exists y)(c(y,y))$ is a q-formula by Rule F8.

STEP $\,\,\,$3:$\,\,$ $s(a,x)$ is an a-formula by Rule F1.

STEP $\,\,\,$4:$\,\,$ $s(a,y)$ is an a-formula by Rule F1.

STEP $\,\,\,$5:$\,\,$ $s(a,x)\wedge s(a,y)$ is a p-formula by Rule F4.

STEP $\,\,\,$6:$\,\,$ $x=y$ is an a-formula by Rule F1.

STEP $\,\,\,$7:$\,\,$ $\sim x=y$ is a p-formula by Rule F2.

STEP $\,\,\,$8:$\,\,$ $a=z$ is an a-formula by Rule F1.

STEP $\,\,\,$9:$\,\,$ $\sim a=z$ is a p-formula by Rule F2.

STEP 10:$\,\,$ $(\sim x=y)\wedge(\sim a=z)$ is a p-formula by Rule F6.

STEP 11:$\,\,$ $(s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z))$ 
is a p-formula by Rule F6.

STEP 12:$\,\,$ $c(z,x)$ is an a-formula by Rule F1.

STEP 13:$\,\,$ $c(z,y)$ is an a-formula by Rule F1.

STEP 14:$\,\,$ $c(z,x)\wedge c(z,y)$ is a p-formula by Rule F4.

STEP 15:$\,\,$ $c(b,x)$ is an a-formula by Rule F1.

STEP 16:$\,\,$ $(c(z,x)\wedge c(z,y))\wedge c(b,x)$ is a p-formula by Rule F4.

STEP 17:$\,\,$ $((s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x))$ is a\\
\phantom{STEP 17:$\,\,$}p-formula by Rule F4.

STEP 18:$\,\,$ $(\exists z)(((s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$ is\\
\phantom{STEP 17:$\,\,$}  a q-formula by Rule F8.

STEP 19:$\,\,$ $(\forall y)(\exists z)(((s(a,x)\wedge s(a,y))\wedge((\sim
x=y)\wedge(\sim a=z))) 
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$ 
\phantom{STEP 17:$\,\,$}  is a q-formula by Rule F7.

STEP 20:$\,\,$ $(\exists x)(\forall y)(\exists z)(((s(a,x)\!\wedge\!s(a,y))\!\wedge\!
((\sim\!x=y)\!\wedge\!(\sim\!a\!=\!z)))  
\!\wedge\!((c(z,x)\!\wedge\!c(z,y))\!\wedge\!c(b,x)))$ 
\phantom{STEP 17:$\,\,$} is a q-formula by Rule F7.

STEP 20:$\,\,$ $F$ is  a p-formula by Rule F4.

\vspace{0.1mm}

\end{minipage}
}


\bigbreak\noindent\fbox{
\begin{minipage}{5.6in}

\parindent=0pt

\null
\vspace{0.1mm}


\begin{center} 
{\bf AN EXAMPLE OF THE USE OF RULES F1 TO F9}
\end{center}

Let us look at the formula $F$ defined earlier. 
Recall that $F$ is
$$
(\exists y)c(y,y)\vee
(\exists x)(\forall y)(\exists z) (((s(a,x)\wedge s(a,y))\wedge((\sim
\!x\!=\!y)\wedge(\sim\! a\!=\!z))) \wedge ((c(z,x)\wedge c(z,y))\wedge c(b,z)))\,.
$$
Now, here is a ``derivation'' of $F$ from the rules: 

\medbreak

STEP $\,\,\,$1:$\,\,$ $c(y,y)$ is an a-formula by Rule F1.

STEP $\,\,\,$2:$\,\,$ $(\exists y)(c(y,y))$ is a q-formula by Rule F8.

STEP $\,\,\,$3:$\,\,$ $s(a,x)$ is an a-formula by Rule F1.

STEP $\,\,\,$4:$\,\,$ $s(a,y)$ is an a-formula by Rule F1.

STEP $\,\,\,$5:$\,\,$ $s(a,x)\wedge s(a,y)$ is a p-formula by Rule F4.

STEP $\,\,\,$6:$\,\,$ $x=y$ is an a-formula by Rule F1.

STEP $\,\,\,$7:$\,\,$ $\sim x=y$ is a p-formula by Rule F2.

STEP $\,\,\,$8:$\,\,$ $a=z$ is an a-formula by Rule F1.

STEP $\,\,\,$9:$\,\,$ $\sim a=z$ is a p-formula by Rule F2.

STEP 10:$\,\,$ $(\sim x=y)\wedge(\sim a=z)$ is a p-formula by Rule F6.

STEP 11:$\,\,$ $(s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z))$ 
is a p-formula by Rule F6.

STEP 12:$\,\,$ $c(z,x)$ is an a-formula by Rule F1.

STEP 13:$\,\,$ $c(z,y)$ is an a-formula by Rule F1.

STEP 14:$\,\,$ $c(z,x)\wedge c(z,y)$ is a p-formula by Rule F4.

STEP 15:$\,\,$ $c(b,x)$ is an a-formula by Rule F1.

STEP 16:$\,\,$ $(c(z,x)\wedge c(z,y))\wedge c(b,x)$ is a p-formula by Rule F4.

STEP 17:$\,\,$ $((s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x))$ is a\\
\phantom{STEP 17:$\,\,$}p-formula by Rule F4.

STEP 18:$\,\,$ $(\exists z)(((s(a,x)\wedge s(a,y))\wedge((\sim x=y)\wedge(\sim a=z)))
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$ is\\
\phantom{STEP 17:$\,\,$}  a q-formula by Rule F8.

STEP 19:$\,\,$ $(\forall y)(\exists z)(((s(a,x)\wedge s(a,y))\wedge((\sim
x=y)\wedge(\sim a=z))) 
\wedge((c(z,x)\wedge c(z,y))\wedge c(b,x)))$ 
\phantom{STEP 17:$\,\,$}  is a q-formula by Rule F7.

STEP 20:$\,\,$ $(\exists x)(\forall y)(\exists z)(((s(a,x)\!\wedge\!s(a,y))\!\wedge\!
((\sim\!x=y)\!\wedge\!(\sim\!a\!=\!z)))  
\!\wedge\!((c(z,x)\!\wedge\!c(z,y))\!\wedge\!c(b,x)))$ 
\phantom{STEP 17:$\,\,$} is a q-formula by Rule F7.

STEP 20:$\,\,$ $F$ is  a p-formula by Rule F4.

\vspace{0.1mm}

\end{minipage}
}





%\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\input mymacros
\def\lto{\rightarrow}
\def\rr{{\bf R}}
\def\nn{{\bf N}}
\def\rrr{{\mymagtwobf R}}
\def\nnn{{\mymagtwobf N}}
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 7.8cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\mymagtwobf MATHEMATICS 300}}

\bigbreak
\centerline{{\mymagtwobf INTRODUCTION TO MATHEMATICAL REASONING}}

\medbreak
\centerline{{\mymagtwobf SPRING 2000}}

\medbreak
\centerline{{\mymagtwoit INSTRUCTOR: H\'ector J.  Sussmann}}


\medbreak
\centerline{{\mymagtwoit HANDOUT NO. 3 --- March 6, 2000}}

\bigbreak
\centerline{{\mymagtworm I. ANSWERS TO THE QUESTIONS IN THE 
FIRST MIDTERM EXAM}}

\medbreak
\item{{\bf 1.}} {\sl In this problem, we use a formal
language which is a predicate calculus with
the usual logical connectives ---that is $\vee$, $\wedge$, $\sim$, $\to$,
$\leftrightarrow$, $\exists$ and $\forall$--- as well as 
the left and right parentheses ---``('' and ``)''--- 
and binary predicate symbols $s$, $d$ for the predicates ``son of,''
``daughter of,'' and ``is equal to'' (so $s(x,y)$ stands for ``$x$ is $y$'s
son,'' $d(x,y)$ stands for ``$x$ is $y$'s daughter,'' $x=y$ stands for
``$x$ and $y$ are the same person''), 
individual symbols (that is, the letters $x$, $y$, $z$, $u$, $v$, 
$n$, $m$, $p$, $q$, etc,), 
{\mymagtwobf and nothing else}. 
We take the universe of discourse $U$ to be the set of all human beings.}


\medbreak
\item{I.} Give formal definitions of the following four predicates:
``is male,'' ``is female,'' ``is the mother of,'' ``is a brother of.''
(Note: first you have to decide how many arguments ---i.e. how many
variables--- are involved in each case. EXAMPLE: if I ask you to give a
formal definition of ``has no children,'' then you would 
first remark that this is a one-variable
predicate, so the thing that must be translated into the formal language is
``$x$ has no children.'' Then a correct translation is 
$\sim\exists y(s(y,x)\vee d(y,x))$.$\,$)

\medbreak
\item{} ANSWERS: 

\item{} ``$x$ is male'' means ``$\exists y\,s(x,y)$'';

\item{} ``$x$ is female'' means ``$\exists y\,d(x,y)$'';

\item{} ``$x$ is the mother of $y$'' means 
``$\exists z\,(d(x,z)\wedge(s(y,x)\vee d(y,x)))$'';

\item{} ``$x$ is a brother of $y$'' means 

\itemitem{} ``$\exists z\exists w
(\sim z=w\wedge s(x,z)\wedge s(x,w)\wedge(s(y,z)\vee d(y,z))
\wedge(s(y,w)\vee d(y,w)))\wedge\sim x=y$.''

\bigbreak
\item{II.} Translate each of the following three English sentences into a
formula of the formal language:

\medbreak
\itemitem{i.} Some people have a son and a daughter.

\itemitem{} ANSWER:  $\exists x\exists y\exists z(s(y,x)\wedge d(z,x))$.

\itemitem{ii.} $x$ has exactly two children.

\itemitem{} ANSWER:  

\itemitem{} $\exists y\exists z\,((s(y,x)\vee d(y,x))
\wedge(s(z,x)\vee d(z,x))\wedge\sim y=z\wedge\forall w
((s(w,x)\vee d(w,x))\to(w=y\vee w=z)))$.

\itemitem{iii.} $x$ is $y$'s uncle.

\itemitem{} ANSWER:  $\exists z\exists w(s(x,z)\wedge(s(w,z)\vee d(w,z))
\wedge \sim x=w\wedge (s(y,w)\vee d(y,w)))$.


\medbreak
\item{{\bf 2.}} {\sl In this problem, we use the
same formal language as in Problem 1, with the same interpretation.} 
Translate each of the following five formulas into plain English,
using normal English sentences ({\it without} letter variables).

\medbreak
\noindent\ \ \ \ \ i.\ \ $\forall x\exists y\,s(x,y)$. ANSWER: everybody is somebody's son.

\medbreak
\noindent \ \ \ \ ii.\ \ $\forall x\exists y\,s(y,x)$. ANSWER: everybody has a son.

\medbreak
\noindent \ \ \ \ iii.\ \ $\forall x\exists y\Bigl(s(y,x)\vee d(y,x)\Bigr)$.
ANSWER: everybody has a son or a daughter. (Even better: ``everybody

\noindent \ \ \ \ \ \ \  \ \ has children.'')


\noindent \ \ \ \ iv.\ \ $\forall x\forall y\forall z\forall w
\Biggl(\Bigl(s(x,z)\wedge d(y,w)\Bigr)\to\sim x=y\Biggr)$.
ANSWER: a son and a daugther are not the same person.

\noindent \ \ \ \ v.\ \ $(\forall x)(\forall y)\Biggl(
\biggl((\exists z)
\Bigl(s(x,z)\wedge s(y,z)\Bigr)\biggr)\to 
\biggl((\exists z)(\exists w)
\Bigl(s(x,z)\wedge s(y,z)\wedge s(x,w)\wedge s(y,w)\wedge \sim z=w\Bigr)
\biggr)\Biggr)$. ANSWER: if two men have one parent in common then they have
two parents in common.

\bigbreak
\item{{\bf 3.}} {i.}  Define ``tautology.''  
ANSWER: A {\bf tautology} is a propositional form
(or ``sentence form'' or ``statement form'') whose corrresponding truth
function has the value T for all possible assignments of values to the
propositional variables. {\sl Note: the book says ``a {\bf tautology}
is a statement whose truth function has all $T$'s as outputs. We discussed
this in class in great detail and explained why this is not right.)}


\item{ii.}  Define ``contradiction.'' ANSWER: A {\bf contradiction} is a 
propositional form (or ``sentence form'' or ``statement form'') whose
corrresponding truth function has the value F for all possible
assignments of values to the propositional variables.


\item{iii.}  Explain the distinction between ``proposition'' and
``propositional form.'' ANSWER: A {\bf proposition} is a declarative
sentence that can be true or false. A {\bf propositional form} is an
expression in a formal language called ``propositional calculus'' (more
precisely, a well formed formula in a propositional calculus) meant to
represent the logical form of a proposition.  (Example: ``Patricia is 10 years
old'' is a proposition.  ``$P\wedge\sim P$'' is a propositional form.)

\bigbreak
\item{{\bf 4.}} In this problem, we use a formal
language which is a predicate calculus with the usual logical connectives,
that is $\vee$, $\wedge$, $\sim$, $\lto$, $\lr$, $\exists$
and $\forall$, as well as the comma, the left and right parentheses
---``('' and ``)''---the $=$ sign, acting as a binary predicate,
and the following additional predicate symbols: 
a binary (i.e. two-variable) predicate $p$, two 1-variable
predicates $q$, $r$, and 
individual symbols (that is, the letters $x$, $y$, $z$, $u$, $v$, 
$n$, $m$, $p$, $q$, etc,).  A formula is {\bf logically valid} (or a 
{\bf law of logic}) if it is true 
for every possible choice of the universe of discourse and every 
way of assigning a meaning to the predicate symbols.
For example: $(\forall x)q(x)\lto\Big(\sim(\exists x)(\sim q(x))\Big)$ 
is logically valid because, no matter what $q(x)$ might stand for, 
the statement is true: ``everybody eats lettuce'' implies 
``there is no person who does not eat lettuce;'' ``all people are mortal''
implies ``nobody is immortal,'' ``every natural number is positive''
implies ``there is no natural number which is not positive,'' and so on.
On the other hand, $(\exists x)q(x)\lto(\forall x)q(x)$ is {\bf not}
logically valid, because we could take $q(x)$ to stand for ``$x$ is an
elephant,'' in the universe of  animals, and then our sentence would say
that ``if there are elephants then every animal is an elephant,'' which is
plainly false.

\itemitem{} The following five formulas are
sentences of this predicate calculus:
$$\forall y\exists x\,p(x,y)\lto\exists x\forall y\,p(x,y)\,;
\eqno{(1)}$$
$$\exists x\forall y\,p(x,y)\lto\forall y\,\exists x\,p(x,y)
\,;\eqno{(2)}$$
$$\forall x\forall y\Big(x=y\lto (q(x)\lto q(y))\Big)
\,.\eqno{(3)}$$
$$\forall x\forall y\Big((q(x)\lto q(y))\lto x=y\Big)
\,.\eqno{(4)}$$
$$\Bigl(\sim (\exists x)q(x)\Bigr)\to\Bigl((\forall x)(q(x)\to r(x))\Bigr)
\,.\eqno{(5)}$$

\item{} For each of the five formulas, decide whether the formula is logically
valid, and give a convincing reason (that is, a proof, but not necessarily 
a formal proof) for your answer. 

\item{} ANSWERS: (1) is {\bf not} logically valid. PROOF: let $p(x,y)$ 
stand for ``$x$ is a friend of $y$.'' Then (1) says that
``if everybody has a friend then there somebody who is everybody's friend.''
This is clearly not true.

\item{} (2) {\bf is} logically valid. PROOF: Assume 
$\exists x\forall y\,p(x,y)$. We are going to show that
$\forall y\,\exists x\,p(x,y)$. Since $\exists x\forall y\,p(x,y)$ is true,
we can pick an $a$ such that $\forall y\,p(a,y)$. Let $b$ be arbitrary.
Since $\forall y\,p(a,y)$, we can conclude that $p(a,b)$. 
So $\exists x\,p(x,b)$. Since $b$ was arbitrary,we have shown that
$\forall y\,\exists x\,p(x,y)$.

\item{} (3) {\bf is} logically valid. PROOF: Let $x,y$ be arbitrary.
We prove that $x=y\lto (q(x)\lto q(y))$. Assume $x=y$. We prove that 
$q(x)\lto q(y)$. Assume $q(x)$. Since $x=y$, we have $q(y)$.

\item{} (4) is {\bf not} logically valid. PROOF: let $q(x)$ stand
for ``$x$ is very smart.'' We will prove that, in this particular
interpretation, (4) is false. For this purpose, we must exhibit
$x,y$ such that $(q(x)\lto q(y))\lto x=y$ is false.  Let
$x=$Bill~Bradley,  $y=$H\'ector~ Sussmann. Then 
$q(y)$ is (obviously) true, so $q(x)\to q(y)$ is true. (It does not matter
whether $q(x)$ is true or not.)  But $x=y$ is not true. So 
$(q(x)\lto q(y))\lto x=y$ is false. 

\item{} (5) {\bf is} logically valid. PROOF: Assume 
$\sim (\exists x)q(x)$, We are going to prove that
$(\forall x)(q(x)\to r(x))$. Let $x$ be arbitrary. We are going to prove
$q(x)\to r(x)$. Now, $q(x)$ is false, because we are assuming that there 
aren't any $u$'s for which $q(u)$. So 
$q(x)\to r(x)$ is true.

\bigbreak
\item{{\bf 5.}} In
this problem, the universe of discourse is $\rr$, the set of real numbers.
{\bf ``Number'' means ``real number,''} that is, member of $\rr$. 
We use LERNA, the {\bf language of elementary real-number arithmetic}.
This is a {\bf
formal language} which is a predicate calculus containing the usual logical
connectives ---$\wedge$, $\vee$, $\sim$, $\lto$, $\lr$, 
$\exists$ and $\forall$--- as well as the usual binary predicate symbols
$=$, $>$, $<$, $\ge$, $\le$, with their usual meanings, the constant
symbols $0$, $1$, the symbols $+$, $\cdot$ for addition and multiplication,
the left and right parentheses ---``('' and ``)'',
individual symbols (that is, the letters $x$, $y$, $z$, $u$, $v$, 
$n$, $m$, $p$, $q$, etc,), 
<{\mymagtwobf and nothing
else}.  (For example, we do not have the symbol $2$.  If you want to say
$2$, you have to say $1+1$. Also, we do not have the symbol $-$ for
subtraction, so if you want to say $x-y=z$ you have to manage without using
``$-$'', e.g. by saying $x=y+z$ instead.)

\item{} Translate each of the following English sentences into our formal
language, and decide if it is true or false. (Note: ``positive'' here means
``$>0$.'')

\medbreak
\itemitem{1.} Every number has a multiplicative inverse. TRANSLATION:
$\forall x\exists y(x.y=1)$. This is FALSE, because if $x=0$ then
$y$ does not exist.

\itemitem{2.} Every positive number has a fourth root. TRANSLATION:
$\forall x(x>0\to\exists y((y.y).(y.y)=x)$. This is TRUE.


\itemitem{3.} There exists a number whose square  added to the number 
is equal to $1$. 

\itemitem{} TRANSLATION: $\exists x(x.x+x=1)$. This is TRUE. (REASON:
the function $f$ given by $f(x)=x^2+x$ is continuous. Moreover, 
$f(0)=0$ and $f(1)=2$. So by the intermediate value theorem there exists
an $x$ such that $f(x)=1$. ANOTHER PROOF: you can compute $x$ directly, 
by solving the quadratic equation $x^2+x-1=0$. The solutions are
$x={-1\pm\sqrt{5}\over 2}$.)


\itemitem{4.} For any two numbers, there exists a unique number that added
to the second number gives the first number.  TRANSLATION:
$\forall x\forall y\exists z(y+z=x\wedge\forall u(y+u=x\to u=z))$. This is TRUE.

\itemitem{5.} There exists a smallest number. TRANSLATION: 
$\exists x\forall y(x\le y)$. This is FALSE.

\itemitem{6.} For any two distinct numbers, there is a number 
that is strictly between them (that is, between them and not equal 
to either of them). 

\itemitem{} TRANSLATION: $\forall x\forall y(\sim x=y\to
\exists z((x<z\wedge z<y)\vee(y<z\wedge z<x)))$. 

\bigbreak
\item{{\bf 6.}} In this problem, the universe of discourse is $\nn$, the
set of natural numbers, so {\bf ``number'' means ``natural number.''} (The
natural numbers are the integers that are strictly greater than zero, that
is, $1,2,3,\ldots$.) We use LENNA, the {\bf formal language of elementary
natural number aithmetic.} This is a {\bf formal language}
which is a predicate calculus
containing the usual logical connectives ---$\wedge$, $\vee$, $\sim$,
$\lto$, $\lr$, $\exists$ and $\forall$--- as well as the usual binary predicate
symbols $=$, $>$, $<$, $\ge$, $\le$, with their usual meanings, the
constant symbol $1$, the symbols $+$ and $\cdot$ for addition and
multiplication, the left and right parentheses---``('' and ``)'', 
individual symbols (that is, the letters $x$, $y$, $z$, $u$, $v$, 
$n$, $m$, $p$, $q$, etc,) \ {\mymagtwobf and nothing else}.

\item{6.a} Define ``prime number'' in terms of a sentence in the formal
language.  (That is, write a sentence in the formal language that says
``$n$ is a prime number.'')  

\item{} ANSWER: ``$n$ is prime'' means
``$n>1\wedge\forall p\forall q(n=p.q\to(p=1\vee q=1))$''.

\item{6.b} Translate each of the following English sentences into the formal
language, and indicate whether it is true or false.

\medbreak
\itemitem{1.} Every number has an additive inverse.

\itemitem{} ANSWER: $\forall x\exists y(x+y+1=1)$. (Comment: I do not 
say $\forall x\exists y(x+y=0)$ because $0$ is not in the list of permitted symbols.)
This statement is FALSE, because if $x=1$ then $y$ does not exist.

\itemitem{2.} There exist at least two prime numbers.  

\itemitem{} ANSWER: $\exists n\exists m\Big((
n>1\wedge\forall p\forall q(n=p.q\to(p=1\vee q=1)))$

\vskip-2mm
\itemitem{} \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\wedge
(m>1\wedge\forall p\forall q(m=p.q\to(p=1\vee q=1)))\wedge \sim n=m\Big)$.
This is TRUE.

\itemitem{3.} Every number which is greater than $1$
is divisible by some prime number.

\itemitem{} ANSWER: $\forall n(n>1\to\exists m
(m>1\wedge\forall p\forall q(m=p.q\to(p=1\vee q=1))))$.  This is TRUE.


\itemitem{4.} If two numbers are not equal then 
one of them is greater than or equal to $1$ plus the other one.

\itemitem{} ANSWER: $\forall n\forall m(\sim n=m\to(n\ge m+1\vee m\ge n+1))$. 
This is TRUE. 
	
\vbox{
\itemitem{5.} There is no smallest number.   

\itemitem{} ANSWER: $\sim\exists n\forall m(n\le m)$. This is
FALSE. REASON: there does exist a smallest number, namely, $1$.}

	
\itemitem{6.} There is no largest number. 

\itemitem{} ANSWER: $\sim\exists n\forall m(n\ge m)$. This is TRUE, because 
if $n$ was such that $\forall m(n\ge m)$,  then we could take $m=n+1$ and
get a contradiction. 

	
\itemitem{7.} The sum of four consecutive numbers is always divisible
by $4$. 

\itemitem{} ANSWER: $\forall n\exists m(n+(n+1)+(n+1+1)+(n+1+1+1)=4m)$.  This is
FALSE, because if $n=1$ then $n+(n+1)+(n+2)+(n+3)=10$ which is not divisible
by $6$. 


\item{{\bf 7.}} In each of the
following cases, check the appropriate box, and {\it justify your
answer}. (In most cases there is no need to write a complete truth table.
Recall that a {\bf contingency} is a formula of the propositional calculus 
that in neither a tautology nor a contradiction.)

\medbreak
\item{1.} \pltt{$P\lto(Q\lto (R\lto (S\lto (P\wedge Q\wedge R\wedge S))))$}

\item{} ANSWER: It's a {\bf tautology}. REASON: for
$P\lto(Q\lto (R\lto (S\lto (P\wedge Q\wedge R\wedge S))))$, $P$ has to be $T$
and $Q\lto (R\lto (S\lto (P\wedge Q\wedge R\wedge S)))$ has to be $F$. 
But then $Q$ has tgo be $T$ and $R\lto (S\lto (P\wedge Q\wedge R\wedge S))$ has to be $F$. 
Then $R$ is $T$ and $S\lto (P\wedge Q\wedge R\wedge S)$ has to be $F$. 
So  $S$ is $T$ and $P\wedge Q\wedge R\wedge S$ is  $F$. So we have 
seen that $P$, $Q$, $R$ and $S$ must be $T$ but $P\wedge Q\wedge R\wedge S$ is  $F$. 
This is impossible. 

\medbreak
\item{2.} \pltt{$(P\wedge(\sim P))\lto Q$}  

\item{} ANSWER: It's a {\bf tautology}. REASON: $P\wedge(\sim P)$ is always $F$.   

\medbreak
\item{3.} \pltt{$P\lto(\sim P)$}

\item{} ANSWER: It's a {\bf contigency}. REASON: $P\lto(\sim P)$ is $T$ if $P$ is $F$,
and  $F$ is $P$ is $T$.   

\medbreak
\item{4.} \pltt{$(P\wedge (\sim Q))\lto(P\lto Q)$}

\item{} ANSWER: It's a {\bf contigency}. REASON: 
$(P\wedge (\sim Q))\lto(P\lto Q)$ is $T$ if $P$ is $F$, and is $F$ is
$P$ is $T$ and $Q$ is $F$. 

\medbreak
\item{5.} \pltt{$((\sim P)\vee(\sim Q))\lto(P\lto(Q\lto R))$}

\item{} ANSWER: It's a {\bf tautology}. REASON: for 
$((\sim P)\vee(\sim Q))\lto(P\lto(Q\lto R))$ to be $F$, 
$(\sim P)\vee(\sim Q)$  has to be $T$ and $P\lto(Q\lto R)$ has to be $F$.
But if $(\sim P)\vee(\sim Q)$  is $T$, then $P$ or $Q$ must be $F$.
If $P$ is $F$ then $P\lto(Q\lto R)$ is $T$. 
If $Q$ is $F$ then $Q\lto R$ is $T$, so $P\lto(Q\lto R)$ is $T$. 


\medbreak
\item{6.} \pltt{$((\sim P)\wedge(\sim Q))\wedge(P\vee Q)$}

\item{} ANSWER: It's a {\bf contradiction}. REASON: for 
$((\sim P)\wedge(\sim Q))\wedge(P\vee Q)$ to be $T$, 
$\sim P$, $\sim Q$ and $P\vee Q$ must be $T$.  But then 
$P$ and $Q$ must be $F$, so $P\vee Q$ is $F$.  

\bigbreak
\item{{\bf 8.}} {\sl In this problem,
``number'' means ``natural number,'' as in problem 6 above.}

\itemitem{i.}  Prove or disprove: The sum of 6 consecutive numbers is
always divisible by $6$. 

\itemitem{}  ANSWER. This is FALSE. PROOF: $1+2+3+4+5+6=21$, and $21$ is
not divisible by $6$. 

\itemitem{ii.}  Prove or disprove: The sum of 7 consecutive numbers is
always divisible by $7$. 

\itemitem{}  ANSWER. This is TRUE. PROOF: let $n$ be an arbitrary  natural number.
Then 
$$
n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)=7n+1+2+3+4+5+6=7n+21=7(n+3)\,,
$$
so $n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)$  is divisible by $7$. 

\bigbreak
\item{{\bf 9.}} In this problem, the range of all the variables is the set $\rr$ of
real numbers. Translate each of the following English sentences into the
language LERNA described in Problem 5.

\medbreak
\itemitem{i.} $x$ is the absolute value of $y$. ANSWER: 
$x\ge 0\wedge(x=y\vee x+y=0)$. 

\medbreak\itemitem{ii.} $x$ is the largest of $y$ and $z$. ANSWER:
$x\ge y\wedge x\ge z\wedge(x=y\vee x=z)$. 

\medbreak\itemitem{iii.} The function $x^2-x$ attains a maximum value on
the closed interval $[0,1]$. ANSWER: 

\itemitem{} $\exists x\Big(0\le x\wedge x\le 1\wedge\forall y(
(0\le y\wedge y\le 1)\to x.x+y\ge y.y+x)\Big)$.

\bigbreak
\item{{\bf 10.}} Decide whether each of the following sentences is true, over the universe
of real numbers, and justify your statement by giving a reason. {\sl Note:
For Parts (v), (vi) and (vii) you may need to use some of the things you
learned in your Calculus courses.}

\medbreak
\itemitem{(i)} $\forall x\exists y(y>x)$. ANSWER: It is TRUE. PROOF:
let $x$ be arbitrary; pick $y=x+1$; then $y>x$. 

\itemitem{(ii)} $\forall x\exists y(y+1=x)$. ANSWER: It is TRUE. PROOF:
let $x$ be arbitrary; pick $y=x-1$; then $y+1=x$. 

\itemitem{(iii)} $\exists y\forall x(y+1=x)$. ANSWER: It is FALSE. PROOF:
Suppose $y$ exists. Pick one such $y$. Then $\forall x(y+1=x)$. Now apply this to 
$x=y$. We get $y+1=y$, which is false. 

\itemitem{(iv)} $\forall x\Big(x\ge 0\lto\exists y(y^2=x)\Big)$.
ANSWER: It is TRUE. REASON: this statement says that every nonnegative
number has a square root. 



\itemitem{(v)} $\exists x(x^4+32x+50=0)$. ANSWER: It is FALSE. REASON: the function
$f$ given by $f(x)=x^4+32x$ has a derivative $f'(x)=4x^3+32$. Then
$f'(x)<0$ of $x<-2$ and $f'(x)>0$ if $x>-2$. So $f$ is decreasing for $x<-2$ and
increasing for $x>-2$. So the minimum value of $f$ is attained at $x=-2$.
Now, $f(-2)=-48$.  Therefore $x^4+32x\ge -48$ for all $x$. So 
$x^4+32x+50\ge 2$ for all $x$. So there is no $x$ for which 
$x^4+32x+50=0$.


\itemitem{(vi)} $\exists x(x^4+32x+45=0)$. ANSWER: It is TRUE. REASON: let
$f$ be the same function as in the previous part. Then 
$f(-2)=-48$ and $f(0)=0$. It then follows from the intermediate value theorem
that $f(x)=-45$ for some $x$, that is, 
$x^4+32x+45=0$ for some $x$.



\itemitem{(vii)} $\exists x(x^7+93x^6+3x^4+32x+50=0)$ ANSWER: It is TRUE. REASON: let
$f(x)=x^7+93x^6+3x^4+32x+50$. Then $f(0)=50$ and $f(-100)<0$ (because,
if $x=100$, then $x^7+93x^6+3x^4+32x+50=
(x+93)x^6+3x^4+32x+50=-7000000000000+300000000-3200+50<0$).
It then follows from the intermediate value theorem
that $f(x)=0$ for some $x$, that is, 
$x^7+93x^6+3x^4+32x+50=0$ for some $x$.

\bigbreak
\centerline{\hbox to5cm{\hrulefill}}



\end
%\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\input mymacros
\def\rrr{{\bf R}}
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\mymagtwobf MATH 300}}

\medbreak
\centerline{{\mymagonebf INTRODUCTION TO MATHEMATICAL REASONING}}

\medbreak
\centerline{{\mymagonebf SPRING 1997}}

\bigbreak
\centerline{{\mymagonerm SECOND MIDTERM EXAM\ \ \ ---\ \ \ April 24, 1997 }}

\medbreak
\centerline{{\mymagtworm INSTRUCTOR: H\'ector J.  Sussmann}}

\bigbreak \noindent {\mymagtwosl Please make sure that your name 
and student I.D. number are clearly indicated  on every page, 
and that the pages are properly stapled.} 

\medbreak\noindent
{\mymagtwobf Please write
clearly, in complete sentences with correct grammar 
and punctuation.}

\medbreak \noindent {\mymagtwoit If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure to tell us that
you are continuing in the back.} 

\medbreak \noindent {\mymagtwobf This test consists of {\mymagfourbf 6}
problems. \ The \ test \ will \ be \ graded 

\noindent on \ a \ scale \ from 
{\mymagfourbf \ 0 \ to \ 100}. \ \ \ The \ point
\ values \ of \ all 

\noindent the \ problems \ add \ up \ to {\mymagfivebf \ 143}. \ That 
\ means \ that}

{\baselineskip 23pt

\bigbreak
\item{} {\mymagfourbf You 
have plenty of freedom to choose.  You do not need
to do everything right to get a perfect 100 score. It suffices to do 
70\%{} of the work right to get a perfect score.}

}

\bigbreak \noindent {\mymagtworm
In Pages 8 and 9 you will find the complete list of rules of inference,

\noindent derived rules and axioms that appear in Handout No. 10.}

\vskip 0.2cm
\noindent {\mymagtworm In this test, {\mymagtwobf R}, {\mymagtwobf N}, 
denote, respectively, the
sets of real numbers and 

\noindent natural numbers.}

\vfill

\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vfill
\startexpage


\item{{\bf 1.}} ({\sl 24 points total}.) Prove or disprove each of the following:

\medbreak\itemitem{(i)} ({\sl 8 points}.) For every $n\in{\bf N}$ such that $n\ge 3$
the inequality $n^3<3^n$ holds.

\medbreak\itemitem{(ii)} ({\sl 8 points}.)  For every $n\in{\bf N}$ such
that $n\ge 3$ the inequality $n^2+3n<4^{n-1}+4$ holds.

\medbreak\itemitem{(iii)} ({\sl 8 points}.) For every $n\in{\bf N}$, 
$$\sum_{k=1}^n{1\over k(k+1)}= {n\over n+1}\,.$$

\startexpage

\item{{\bf 2.}} ({\sl 20 points  total}.)  

\itemitem{(i)} ({\sl 5 points}.)  Define what an {\it interval} is.

\vfill

\itemitem{(ii)} Using the definition of ``interval'' given
in (i), prove or disprove each of the following:

\medbreak
{\parindent=30pt 
\itemitem{(a)} ({\sl 5 points}.)  If $a$ is a real number,
then the set $I=\{x\in{\bf R}: x\ge a\}$ is an interval.

\vfill

\itemitem{(b)} ({\sl 5 points}.)  The set ${\bf N}$ is an interval.

\vfill

\itemitem{(c)} ({\sl 5 points}.)  The union of two intervals is an interval.
}

\vfill

\startexpage

\item{{\bf 3.}} ({\sl 7 points for each of the three formulas, 
21 points total}.) In this problem, we use a formal
language which is a predicate calculus with the usual logical connectives
---that is $\vee$, $\wedge$, $\sim$, $\to$, $\leftrightarrow$, $\exists$
and $\forall$--- as well as the comma, the left and right parentheses
---``('' and ``)''--- and the following predicate symbols: 
a binary (i.e. two-variable) predicate $p$, and two 1-variable
predicates $q$, $r$.  The following three formulas are
sentences of this predicate calculus:
$$(\forall y)(\exists x)p(x,y)\to(\exists x)(\forall y)p(x,y)\,;
\eqno{(1)}$$
$$(\exists x)(\forall y)p(x,y)\to(\forall y)(\exists x)p(x,y)
\,;\eqno{(2)}$$
$$\Bigl(\sim (\exists x)q(x)\Bigr)\to\Bigl((\forall x)(q(x)\to r(x))\Bigr)
\,.\eqno{(3)}$$
For each of the three formulas, decide whether the formula is a theorem 
of the predicate calculus.  If it is a theorem prove it. (In the proofs,
you are allowed to use the 8 rules of inference {\bf and nothing else}.)
If it is not a theorem, then establish this fact by exhibiting an
interpretation of the predicates  for which the sentence is false. 


\vfill

\startexpage

\item{{\bf 4.}} ({\sl 5 points for each formula, 30 points total.}) In this
problem, we work with real 
number arithmetic.  The axioms are the ones of Handout No. 10. (You
will find the complete list of axioms in Page 9.)
``2'' means ``$1+1$.''  ``$x\neq y$'' means ``$\sim x=y$.''
The expression ``$x-y$'' is defined to mean ``$x+(-y)$'', where $-y$ is
characterized by the fact that $(-y)+y=0$.  Then ``$x<y$'' means
``$y-x\in{\cal P}$,''  and ``$x\le y$'' means ``$x<y\vee x=y$.''
The following six formulas
are sentences of real number arithmetic.  For each formula, there are four
possibilities:  (a)~the formula is a theorem (that is, it can be proved
from the axioms of Page 9); (b)~the negation of the formula is a
theorem; (c)~the formula is true as a statement about the real numbers but
cannot be proved from the axioms; (d)~the negation of the 
formula is true as a statement about the real numbers but
cannot be proved from the axioms.  For each of the formulas, (i)~translate 
the formula into plain English, without using letter variables or
mathematical symbols, (ii)~decide which of the above four possibilities 
holds, (iii)~if (a) holds, prove the formula (using the rules and the
axioms; shortcuts are allowed);  
(iv)~if (b) holds, prove the negation of the formula (using the rules and the
axioms;  shortcuts are allowed);  
(v)~if (c) or (d) holds, give a convincing reason.   
$$(\exists x)(\forall y)(x\le y)\,;\eqno{(1)}$$
$$(\forall y)(\exists x)(x<y)\,;\eqno{(2)}$$
$$(\forall x)(\forall y)(x<y\to x.x<y.y)\,;\eqno{(3)}$$
$$(\forall x)(0<x\to (\exists y)(y.y=x))\,;\eqno{(4)}$$
$$(\forall x)(\forall y)(\forall z)(x.y=x.z\to y=z)\,;\eqno{(5)}$$
$$(\forall x)(\forall y)(\forall z)\Bigl(x.y=x.z\to(x\neq 0\to
y=z)\Bigr)\,;\eqno{(6)}$$
\vfill

\startexpage
\item{{\bf 5.}} ({\sl 24 points total}.) (Informal proofs are O.K. here.) 

\medbreak
\itemitem{{\bf 5a.}} ({\sl 12 points}.) A set $I$ of real numbers is {\it
convex} if $(\forall x\in I)(\forall y\in I)(\forall t\in {\bf R})
(0<t<1\to t.x+(1-t)y\in I)$.  

\itemitem{} Prove or disprove: $I$ is convex $\leftrightarrow\,\,\,I$ is an
interval. 


\medbreak
\itemitem{{\bf 5b.}} ({\sl 12 points}.) A set $I$ of real numbers is {\it
bounded} if $(\exists x\in{\bf R})(\forall y\in I)(|y|\le x)$.  

\itemitem{} Prove or disprove: the union of two bounded sets is also
bounded.


\vfill

\startexpage
\item{{\bf 6.}} ({\sl 24 points}.) A set $S$ of points $(x,y)$ in
${\bf R}\times{\bf R}$ is {\it symmetric with respect to the origin}
if 
$$(\forall x)(\forall y)((x,y)\in S\to (-x,-y)\in S)\,.$$
(Here ``$(x,y)$'' means ``the ordered pair formed by $x$ and $y$.'' 
It does {\it not} mean ``the open interval from $x$ to $y$.'')

\item{} A real-valued function $f$ with domain ${\bf R}$ is {\it odd}
if $(\forall x)(f(-x)=-f(x))$.  The {\it graph} of a function $f$ is the
set $G(f)=\{(x,y):y=f(x)\}$. 

\item{} Prove or disprove each of the following 
statements (informal proofs are O.K.):

\itemitem{(i)} The function $f$ given by $f(x)=x^2$ is odd.

\itemitem{(ii)} The function $f$ given by $f(x)=x+x^3$ is odd.

\itemitem{(iii)} A function $f$ is odd if and only if $G(f)$ is symmetric 
with respect to the origin. 


\vfill\eject


\def\ste{\enspace\enspace\enspace\enspace}
\newcount{\statco}
\newcount{\quesco}
\newcount{\defco}
\newcount{\notco}
\newcount{\remco}
\newcount{\rulco}
\newcount{\drulco}
%\def\co#1{#1}
\def\rulo{\global\advance\rulco by 1
\medbreak\noindent{\bf  RULE
\the\rulco.}\ }
\def\drulo{\global\advance\drulco by 1
\medbreak\noindent {\bf DERIVED RULE
\the\drulco.}\ }
\def\justo{
\medbreak\noindent {\bf JUSTIFICATION$\,$:}\ \ }
\def\co#1{}
\def\ep{\varepsilon}
\def\lto{\leftrightarrow}
\def\clo{{\rm Clos}}
\defco=0
\statco=0
\remco=0
\rulco=0
\drulco=0
\notco=0
\quesco=1
\def\ino{{\rm Int}}

\centerline{{\bf INFERENCE RULES FOR PROOFS}}

\medbreak\noindent Here are the eight basic rules of inference 
that you are allowed to use in a formal proof.

\rulo You are allowed to bring in any statement which is
an instance of a tautology.

\rulo If you have $p$ and $p\to q$ then you can go to $q$. (This is called {\it
Modus Ponens}).

\rulo If you have proved $q$ under the assumption $p$ (that is, if you have
assumed $p$ and proved $q$), then you can go to $p\to q$.

\rulo If you have $(\forall x)p(x)$ and $s$ is a term that does not
contain any variable that occurs under the scope of a quatinfier in 
$p(x)$, then you can go 
to $p(s)$. (This is called {\it specialization}).

\rulo To prove a statement of the form $(\forall x)p(x)$, you start 
by saying ``Let $a$ be arbitrary,'' and then try to prove $p(a)$. 


\rulo If you have $(\exists x)p(x)$ then you can introduce a new
object and give it a letter name such as $a$ or $\alpha$ or anything you
want, provided that this name has not been used before, and stipulate that
$p(a)$.  (Here {\bf it's important that this new object should be given a
name that has not been used before.})

\rulo If you have an object $a$ such that $p(a)$,  then you can go to 
$(\exists x)p(x)$.  

\rulo You can substitute equals for equals.  That is, 
if you know that $a=b$, then you can go from a statement involving 
$a$ to the statement in which $b$ is substituted for $a$ everywhere,
and from a statement involving 
$b$ to the statement in which $a$ is substituted for $b$ everywhere.
(Example: After you have given the definition
$$[0,1]=\{x:0\le x\le 1\}\,$$
you can go from
$a\in [0,1]$ to $a\in\{x:0\le x\le 1\}$, and then you can use Set Axiom 2
below to get to $0\le a\le 1$.) 

\bigbreak

\centerline{{\bf DERIVED RULES (SHORTCUTS)}}

\drulo    If you have $(\forall x)(p(x)\to q(x))$ and $p(a)$ then
you can go to $q(a)$.

\drulo    If you have $p\wedge q$ then you can go to $p$. 

\drulo    If you have $p\wedge q$ then you can go to $q$. 

\drulo    If you have $p$ and $q$ then you can go to $p\wedge q$. 

\drulo    If you have $p$ then you can go to $p\vee q$. 

\drulo    If you have $q$ then you can go to $p\vee q$. 

\drulo    If you have $p\to q$ and $q\to p$ 
then you can go to $p\leftrightarrow q$. 

\drulo    If you have $p\leftrightarrow q$ 
then you can go to $p\to q$.

\drulo    If you have $p\leftrightarrow q$ then you can go to $q\to p$.

\drulo    If assuming $\sim q$ you get a contradiction, then 
you can go to $q$.

\drulo    From $\sim p$ and $p\vee q$ you can go to $q$. 

\drulo    From $\sim q$ and $p\vee q$ you can go to $p$.

\drulo    If assuming $p$ and $\sim q$ you get $\sim p$, then 
you can go to $p\to q$.

\drulo    If assuming $\sim p$ you get $q$, then 
you can go to $p\vee q$.

\drulo    If assuming $\sim p$ you get $q$, and assuming $p$ you also get $q$, 
then you can go to $q$ assuming nothing.

\drulo    If assuming $p$ you get $q$, and assuming $q$ you get $p$, 
then you can go to $p\leftrightarrow q$.

\drulo    If you have $(\forall x\in S)p(x)$ and you have
$s\in S$, you can go to $p(s)$. 

\drulo    To prove a statement of the form $(\forall x\in S)p(x)$, you
start by saying ``Let $a$ be an arbitrary element of $S$'' (that is,
``let $a$ be arbitrary; assume $a\in S$'') and then try to
prove $p(a)$. (Here, again, instead of ``$a$'' you could use ``$b$,'' or
``$z$,'' or ``$\alpha$,'' or ``$\beta$,'' or ``$\aleph$,'' or
``$\diamond$,'' or any symbol you want.  What is important that what you do
should apply to a {\it completely arbitrary} element of $S$.)

\drulo    If you have $(\exists x\in S)p(x)$ then you can introduce a
new object and give it a letter name such as $a$ or $\alpha$ or anything you
want, provided that this name has not been used before, and stipulate that
$a\in S$ and $p(a)$.  

\drulo     If you have $p(a)$ and $a\in S$ then you can go to $(\exists
x\in S)p(x)$. 

\bigbreak
\centerline{{\bf SET AXIOMS}}

\bigbreak\noindent SET AXIOM 1: If $S$, $T$ are sets, then
$$S=T\leftrightarrow (\forall x)(x\in S\leftrightarrow x\in T)\,.$$

\bigbreak\noindent SET AXIOM 2: If $S$ is a set and 
$p(x)$ is any formula containing the
variable $x$, no quantifiers involving $x$, and no other free variable,
then $\{x:x\in S\wedge p(x)\}$ (that is, 
$\{x\in S:p(x)\}$) is a set. 

\bigbreak\noindent SET AXIOM 3: If $p(x)$ is any formula containing the
variable $x$, and no quantifiers involving $x$,  and $p(x)$ does 
not contain the variable $y$, then 
$$(\forall y)(y\in\{x:p(x)\}\leftrightarrow p(y))\,.$$

\def\mitem{\item}
\def\rrbf{{\bf R}}
\bigbreak
\end


%\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\input mymacros
\def\rrr{{\bf R}}
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\bf MATH 300 (INTRODUCTION TO MATHEMATICAL REASONING), SPRING 1996}}

\medbreak
\centerline{FIRST MIDTERM EXAM.\ \  February 29, 1996 --- Waller 203, Cook Campus}

\medbreak
\centerline{INSTRUCTOR: H\'ector J.  Sussmann}

\bigbreak \noindent {\sl Please make sure that your name 
and student I.D. number are clarly indicated  on every page, 
and that the pages are properly stapled.} {\bf Please write
clearly, in complete sentences with correct grammar and punctuation.}

\medbreak \noindent {\bf If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure to tell us that
you are continuing in the back.} 

\medbreak \noindent {\bf This test consists of {\mymagtwobf 9}
problems. The test will be graded on a scale from 
{\mymagtwobf 0 to 100}. The point
values of all the problems add up to {\mymagthreebf 152}. That means that
{\mymagtwobf you have a lot of freedom to pick and choose.  You do not need
to do everything right to get a perfect 100 score. It suffices to do 
65\%{} of the questions right to get a perfect score.}}

\vskip 0.5cm
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vskip 0.2cm
\noindent In this test, ${\bf R}$, ${\bf N}$, denote, respectively, the
sets of real numbers and natural numbers.

\vskip 0.7cm

\item{{\bf 1.}} ({\sl 12 points}.) Explain what {\it Russell's paradox} is.

\vfill

\startexpage

\item{{\bf 2.}} ({\sl 26 points}. Note: this problem has {\it three} parts,
and occupies Pages 2 and 3.) In this problem, we use a formal
language which is a predicate calculus with
the usual logical connectives ---that is $\vee$, $\wedge$, $\sim$, $\to$,
$\leftrightarrow$, $\exists$ and $\forall$--- as well as 
the left and right parentheses ---``('' and ``)''--- 
and binary predicate symbols $s$, $d$ for the predicates ``son of,''
``daughter of,'' and ``is equal to'' (so $s(x,y)$ stands for ``$x$ is $y$'s
son,'' $d(x,y)$ stands for ``$x$ is $y$'s daughter,'' $x=y$ stands for
``$x$ and $y$ are the same person''), {\bf and nothing else}. 
We take the universe of discourse $U$ to be the set of all human beings.

{\parindent=25pt

\medbreak
\item{I.} Give formal definitions of the following four predicates:
``is male,'' ``is female,'' ``is the mother of,'' ``is a brother of.''
(Note: first you have to decide how many arguments
---i.e. how many variables--- are involved in each case.)


\vfill\vfill
\item{II.} Translate each of the following three English sentences into a
formula of the formal language:

\medbreak
\itemitem{i.} Some people have a son and a daughter.

\vfill

\itemitem{ii.} Everybody has a daughter. 

\vfill

\itemitem{iii.} Everybody has a father. 

\vfill

}

\startexpage

{\parindent=25pt
\noindent{\bf Problem 1, continued:}

\medbreak
\item{III.} Translate each of the following five formulas into plain English,
using normal English sentences ({\it without} letter variables).

\medbreak
\noindent\ \ \ \ \ i.\ \ $(\forall x)(\exists y)s(x,y)$.

\vfill

\noindent \ \ \ \ ii.\ \ $(\forall x)(\exists y)s(y,x)$.

\vfill

\noindent \ \ \ \ iii.\ \ $(\forall x)(\exists y)\Bigl(s(y,x)\vee d(y,x)\Bigr)$.

\vfill

\noindent \ \ \ \ iv.\ \ $(\forall x)(\forall y)(\forall z)(\forall w)
\Biggl(\Bigl(s(x,z)\wedge d(y,w)\Bigr)\to\sim x=y\Biggr)$.

\vfill

\noindent \ \ \ \ v.\ \ $(\forall x)(\forall y)\Biggl(
\biggl((\exists z)
\Bigl(s(x,z)\wedge s(y,z)\Bigr)\biggr)\to 
\biggl((\exists z)(\exists w)
\Bigl(s(x,z)\wedge s(y,z)\wedge s(x,w)\wedge s(y,w)\wedge \sim z=w\Bigr)
\biggr)\Biggr)$.

\vfill
}

\startexpage

\item{{\bf 3.}} ({\sl 12 points, but you get $-1$ for each wrong answer,
so if you don't know you are better off not guessing.}) In each of the
following cases, check the appropriate box, and {\it justify your
answer}. (In most cases there is no need to write a complete truth table.)

\medbreak
\item{1.} \pltt{$p\to(q\to (r\to (s\to (p\wedge q\wedge r\wedge s))))$}

\vfill
\item{2.} \pltt{$(p\wedge(\sim p))\to q$}

\vfill
\item{3.} \pltt{$p\to(\sim p)$}

\vfill
\item{4.} \pltt{$(p\wedge (\sim q))\to(p\to q)$}

\vfill
\item{5.} \pltt{$((\sim p)\vee(\sim q))\to(p\to(q\to r))$}

\vfill
\item{6.} \pltt{$((\sim p)\wedge(\sim q))\wedge(p\vee q)$}

\vfill

\startexpage
\item{{\bf 4.}} ({\sl 22 points}.) In this problem, the universal set 
$U$ is the set of all 
functions $f$ with domain $\rrr$ and real values.    We define sets 
$A,B,C,D,E$ as follows:

$A=\{f:f(1)>f(0)\}\,$;

$B=\{f:f\,\,\,{\rm is\,\,\,differentiable\,\,\,at\,\,\,every\,\,\,
point}\}\,$;

$C=\{f:f\in B\,\,\,{\rm and\,\,\,}f'(x)>0\,\,\,{\rm 
for\,\,\,every}\,\,\,x\in\rrr\,$;

$D=\{f:f\,\,\,{\rm is\,\,\,a\,\,\,linear\,\,\,function}\}\,$;

$E=\{f:f\in B\,\,\,{\rm and\,\,\,}f'(0)>0\}\,$.

\medbreak\itemitem{I.} List all possible inclusion relations among the 
five sets.  (There are 25 ways of picking two sets $X,Y$ 
from the above list. You have to tell us in which of the 25 cases
the inclusion $X\subseteq Y$ is true.)

\vfill\vfill\vfill\vfill\vfill

\medbreak\itemitem{II.} Describe, {\bf as precisely 
as possible}, the sets

\medbreak
\itemitem{i.} $E\cap D$,  

\vfill

\itemitem{ii.} $D-E$,  

\vfill

\itemitem{iii.} $A\cap E$,

\vfill

\itemitem{iv.} $B\cup C$.

\vfill

\medbreak\itemitem{III.} For each of the four sets of Part II, give an
example of a function  $f$ belonging to the set and an example of a function
$g$ not belonging to the set.

\vfill\vfill\vfill\vfill\vfill

\startexpage

\item{{\bf 5.}} ({\sl 18 points}.) Let $S$ be the set of all natural
numbers whose decimal expansion consists of {\bf exactly five} digits, at
least one of which appears at least twice.  (For example, $23042\in S$,
$55566\in S$, $11019\in S$, $90782\notin S$, $4553\notin S$, $658972\notin
S$.)  How many members does $S$ have?  Please {\it explain how you get
your answer}, and {\it make sure that you clearly indicate what your final
answer is}, and that {\it you carry out all arithmetic operations and
produce a number}. (An answer such as $41\times 38\times 32$ is {\it not}
acceptable.)

\vfill

\startexpage
\item{{\bf 6.}} ({\sl 20 points}.) Analyze each of the following two
arguments for logical validity.

\medbreak\item{i.} If $S\in S$ then $\sim S\in S$. If $\sim S\in S$ then
$S\in S$. Therefore the 1996 Republican presidential nominee will be Howard
Stern. 

\vfill

\medbreak\item{ii.} If Arthur says that he saw Judy kill Wendell, then either
Judy killed Wendell or Arthur is lying. If Arthur is lying, then either
Arthur killed Wendell or Arthur is trying to protect Lydia by 
falsely accusing Judy.  If  Arthur is trying to protect Lydia by 
falsely accusing Judy, then Arthur thinks that Lydia killed Wendell.
If  Arthur thinks that Lydia killed Wendell then either 
Lydia killed Wendell or Arthur saw Judy kill Wendell but mistook Judy for
Lydia. If  Arthur saw Judy kill Wendell but mistook Judy for Lydia, then
Judy killed Wendell.  For Judy to have killed Wendell, Judy would have had
to hate Wendell.  Judy did not hate Wendell. Lydia did not kill Wendell.
Therefore Arthur killed Wendell.

\vfill\vfill\vfill

\startexpage

\item{{\bf 7.}} ({\sl 18 points}.) In this problem, the universe of
discourse is $\rrr$, and we use a formal language which is a predicate
calculus containing the usual logical connectives ---$\wedge$, $\vee$,
$\sim$, $\to$, $\leftrightarrow$, $\exists$ and $\forall$--- as well as the
usual binary predicate symbols $=$, $>$, $<$, $\ge$, $\le$, with their
usual meanings, the constant symbols $0$, $1$, the symbols $+$, $-$,
$\cdot$ for addition, subtraction and multiplication, 
the left and right parentheses ---``(''  and ``)''--- 
{\bf and nothing else}.  (For example, we do not have the symbol $2$.  If
you want to say $2$, you have to say $1+1$.)

\item{} Translate each of the following English sentences into our formal
language, and decide if it is true or false.

\medbreak
\itemitem{1.} Every real number has a multiplicative inverse.

\vfill

\itemitem{2.} Every positive real number has a square root. (Note: ``positive''
means ``$\ge 0$.'')

\vfill

\itemitem{3.} Multiplication is associative. 

\vfill

\itemitem{4.} Subtraction is commutative.

\vfill
	
\itemitem{5.} The sum of two positive numbers is positive.

\vfill

\startexpage


\item{{\bf 8.}} ({\sl 14 points}.)

\itemitem{i.}  Define ``interval.''  

\vfill


\itemitem{ii.} Suppose a student defines ``interval''  as follows:

\medbreak\itemitem{} 
An {\it interval} is a set $I$ of real numbers such that whenever $a\in I$
and $b\in I$ it follows that there exists $c\in \rrr$ such that $a<c<b$ and
$c\in I$. 

\medbreak\itemitem{} 
Show that the student's definition is wrong, by giving an example of a set
$I$ of real numbers which is an interval according to the correct
definition, but isn't an interval according to the student's definition.

\vfill\vfill

\item{{\bf 9.}} ({\sl 10 points}.)

\itemitem{i.}  Define ``tautology.''  

\vfill

\itemitem{ii.} Suppose a student defines ``tautology''  as follows:

\medbreak\itemitem{} 
A {\it tautology} is a true statement.

\medbreak\itemitem{} 
Show that the student's definition is wrong, by giving an example of 
something that is {\it not} a tautology, but is 
a tautology  according to the student's definition.

\vfill

\vfill

\end


%\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\input mymacros
\def\rrr{{\bf R}}
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\bf MATH 300 (INTRODUCTION TO MATHEMATICAL REASONING), SPRING 1996}}

\medbreak
\centerline{FIRST MIDTERM EXAM.\ \  February 29, 1996 --- Waller 203, Cook Campus}

\medbreak
\centerline{INSTRUCTOR: H\'ector J.  Sussmann}

\bigbreak \noindent {\sl Please make sure that your name 
and student I.D. number are clarly indicated  on every page, 
and that the pages are properly stapled.} {\bf Please write
clearly, in complete sentences with correct grammar and punctuation.}

\medbreak \noindent {\bf If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure to tell us that
you are continuing in the back.} 

\medbreak \noindent {\bf This test consists of {\mymagtwobf 9}
problems. The test will be graded on a scale from 
{\mymagtwobf 0 to 100}. The point
values of all the problems add up to {\mymagthreebf 152}. That means that
{\mymagtwobf you have a lot of freedom to pick and choose.  You do not need
to do everything right to get a perfect 100 score. It suffices to do 
65\%{} of the questions right to get a perfect score.}}

\vskip 0.5cm
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vskip 0.2cm
\noindent In this test, ${\bf R}$, ${\bf N}$, denote, respectively, the
sets of real numbers and natural numbers.

\vskip 0.7cm

\item{{\bf 1.}} ({\sl 12 points}.) Explain what {\it Russell's paradox} is.

\vfill

\startexpage

\item{{\bf 2.}} ({\sl 26 points}. Note: this problem has {\it three} parts,
and occupies Pages 2 and 3.) In this problem, we use a formal
language which is a predicate calculus with
the usual logical connectives ---that is $\vee$, $\wedge$, $\sim$, $\to$,
$\leftrightarrow$, $\exists$ and $\forall$--- as well as 
the left and right parentheses ---``('' and ``)''--- 
and binary predicate symbols $s$, $d$ for the predicates ``son of,''
``daughter of,'' and ``is equal to'' (so $s(x,y)$ stands for ``$x$ is $y$'s
son,'' $d(x,y)$ stands for ``$x$ is $y$'s daughter,'' $x=y$ stands for
``$x$ and $y$ are the same person''), {\bf and nothing else}. 
We take the universe of discourse $U$ to be the set of all human beings.

{\parindent=25pt

\medbreak
\item{I.} Give formal definitions of the following four predicates:
``is male,'' ``is female,'' ``is the mother of,'' ``is a brother of.''
(Note: first you have to decide how many arguments
---i.e. how many variables--- are involved in each case.)


\vfill\vfill
\item{II.} Translate each of the following three English sentences into a
formula of the formal language:

\medbreak
\itemitem{i.} Some people have a son and a daughter.

\vfill

\itemitem{ii.} Everybody has a daughter. 

\vfill

\itemitem{iii.} Everybody has a father. 

\vfill

}

\startexpage

{\parindent=25pt
\noindent{\bf Problem 1, continued:}

\medbreak
\item{III.} Translate each of the following five formulas into plain English,
using normal English sentences ({\it without} letter variables).

\medbreak
\noindent\ \ \ \ \ i.\ \ $(\forall x)(\exists y)s(x,y)$.

\vfill

\noindent \ \ \ \ ii.\ \ $(\forall x)(\exists y)s(y,x)$.

\vfill

\noindent \ \ \ \ iii.\ \ $(\forall x)(\exists y)\Bigl(s(y,x)\vee d(y,x)\Bigr)$.

\vfill

\noindent \ \ \ \ iv.\ \ $(\forall x)(\forall y)(\forall z)(\forall w)
\Biggl(\Bigl(s(x,z)\wedge d(y,w)\Bigr)\to\sim x=y\Biggr)$.

\vfill

\noindent \ \ \ \ v.\ \ $(\forall x)(\forall y)\Biggl(
\biggl((\exists z)
\Bigl(s(x,z)\wedge s(y,z)\Bigr)\biggr)\to 
\biggl((\exists z)(\exists w)
\Bigl(s(x,z)\wedge s(y,z)\wedge s(x,w)\wedge s(y,w)\wedge \sim z=w\Bigr)
\biggr)\Biggr)$.

\vfill
}

\startexpage

\item{{\bf 3.}} ({\sl 12 points, but you get $-1$ for each wrong answer,
so if you don't know you are better off not guessing.}) In each of the
following cases, check the appropriate box, and {\it justify your
answer}. (In most cases there is no need to write a complete truth table.)

\medbreak
\item{1.} \pltt{$p\to(q\to (r\to (s\to (p\wedge q\wedge r\wedge s))))$}

\vfill
\item{2.} \pltt{$(p\wedge(\sim p))\to q$}

\vfill
\item{3.} \pltt{$p\to(\sim p)$}

\vfill
\item{4.} \pltt{$(p\wedge (\sim q))\to(p\to q)$}

\vfill
\item{5.} \pltt{$((\sim p)\vee(\sim q))\to(p\to(q\to r))$}

\vfill
\item{6.} \pltt{$((\sim p)\wedge(\sim q))\wedge(p\vee q)$}

\vfill

\startexpage
\item{{\bf 4.}} ({\sl 22 points}.) In this problem, the universal set 
$U$ is the set of all 
functions $f$ with domain $\rrr$ and real values.    We define sets 
$A,B,C,D,E$ as follows:

$A=\{f:f(1)>f(0)\}\,$;

$B=\{f:f\,\,\,{\rm is\,\,\,differentiable\,\,\,at\,\,\,every\,\,\,
point}\}\,$;

$C=\{f:f\in B\,\,\,{\rm and\,\,\,}f'(x)>0\,\,\,{\rm 
for\,\,\,every}\,\,\,x\in\rrr\,$;

$D=\{f:f\,\,\,{\rm is\,\,\,a\,\,\,linear\,\,\,function}\}\,$;

$E=\{f:f\in B\,\,\,{\rm and\,\,\,}f'(0)>0\}\,$.

\medbreak\itemitem{I.} List all possible inclusion relations among the 
five sets.  (There are 25 ways of picking two sets $X,Y$ 
from the above list. You have to tell us in which of the 25 cases
the inclusion $X\subseteq Y$ is true.)

\vfill\vfill\vfill\vfill\vfill

\medbreak\itemitem{II.} Describe, {\bf as precisely 
as possible}, the sets

\medbreak
\itemitem{i.} $E\cap D$,  

\vfill

\itemitem{ii.} $D-E$,  

\vfill

\itemitem{iii.} $A\cap E$,

\vfill

\itemitem{iv.} $B\cup C$.

\vfill

\medbreak\itemitem{III.} For each of the four sets of Part II, give an
example of a function  $f$ belonging to the set and an example of a function
$g$ not belonging to the set.

\vfill\vfill\vfill\vfill\vfill

\startexpage

\item{{\bf 5.}} ({\sl 18 points}.) Let $S$ be the set of all natural
numbers whose decimal expansion consists of {\bf exactly five} digits, at
least one of which appears at least twice.  (For example, $23042\in S$,
$55566\in S$, $11019\in S$, $90782\notin S$, $4553\notin S$, $658972\notin
S$.)  How many members does $S$ have?  Please {\it explain how you get
your answer}, and {\it make sure that you clearly indicate what your final
answer is}, and that {\it you carry out all arithmetic operations and
produce a number}. (An answer such as $41\times 38\times 32$ is {\it not}
acceptable.)

\vfill

\startexpage
\item{{\bf 6.}} ({\sl 20 points}.) Analyze each of the following two
arguments for logical validity.

\medbreak\item{i.} If $S\in S$ then $\sim S\in S$. If $\sim S\in S$ then
$S\in S$. Therefore the 1996 Republican presidential nominee will be Howard
Stern. 

\vfill

\medbreak\item{ii.} If Arthur says that he saw Judy kill Wendell, then either
Judy killed Wendell or Arthur is lying. If Arthur is lying, then either
Arthur killed Wendell or Arthur is trying to protect Lydia by 
falsely accusing Judy.  If  Arthur is trying to protect Lydia by 
falsely accusing Judy, then Arthur thinks that Lydia killed Wendell.
If  Arthur thinks that Lydia killed Wendell then either 
Lydia killed Wendell or Arthur saw Judy kill Wendell but mistook Judy for
Lydia. If  Arthur saw Judy kill Wendell but mistook Judy for Lydia, then
Judy killed Wendell.  For Judy to have killed Wendell, Judy would have had
to hate Wendell.  Judy did not hate Wendell. Lydia did not kill Wendell.
Therefore Arthur killed Wendell.

\vfill\vfill\vfill

\startexpage

\item{{\bf 7.}} ({\sl 18 points}.) In this problem, the universe of
discourse is $\rrr$, and we use a formal language which is a predicate
calculus containing the usual logical connectives ---$\wedge$, $\vee$,
$\sim$, $\to$, $\leftrightarrow$, $\exists$ and $\forall$--- as well as the
usual binary predicate symbols $=$, $>$, $<$, $\ge$, $\le$, with their
usual meanings, the constant symbols $0$, $1$, the symbols $+$, $-$,
$\cdot$ for addition, subtraction and multiplication, 
the left and right parentheses ---``(''  and ``)''--- 
{\bf and nothing else}.  (For example, we do not have the symbol $2$.  If
you want to say $2$, you have to say $1+1$.)

\item{} Translate each of the following English sentences into our formal
language, and decide if it is true or false.

\medbreak
\itemitem{1.} Every real number has a multiplicative inverse.

\vfill

\itemitem{2.} Every positive real number has a square root. (Note: ``positive''
means ``$\ge 0$.'')

\vfill

\itemitem{3.} Multiplication is associative. 

\vfill

\itemitem{4.} Subtraction is commutative.

\vfill
	
\itemitem{5.} The sum of two positive numbers is positive.

\vfill

\startexpage


\item{{\bf 8.}} ({\sl 14 points}.)

\itemitem{i.}  Define ``interval.''  

\vfill


\itemitem{ii.} Suppose a student defines ``interval''  as follows:

\medbreak\itemitem{} 
An {\it interval} is a set $I$ of real numbers such that whenever $a\in I$
and $b\in I$ it follows that there exists $c\in \rrr$ such that $a<c<b$ and
$c\in I$. 

\medbreak\itemitem{} 
Show that the student's definition is wrong, by giving an example of a set
$I$ of real numbers which is an interval according to the correct
definition, but isn't an interval according to the student's definition.

\vfill\vfill

\item{{\bf 9.}} ({\sl 10 points}.)

\itemitem{i.}  Define ``tautology.''  

\vfill

\itemitem{ii.} Suppose a student defines ``tautology''  as follows:

\medbreak\itemitem{} 
A {\it tautology} is a true statement.

\medbreak\itemitem{} 
Show that the student's definition is wrong, by giving an example of 
something that is {\it not} a tautology, but is 
a tautology  according to the student's definition.

\vfill

\vfill

\end



\magnification=\magstephalf
%\nopagenumbers
\font\mymagtwobf=cmbx10 scaled\magstep 2
\font\mymagthreebf=cmbx10 scaled\magstep 3
%\hsize 18true cm
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\bf MATH 300 (INTRODUCTION TO MATHEMATICAL REASONING), FALL 1995}}

\medbreak
\centerline{SECOND MIDTERM EXAM.\ \  December 4, 1995}

\bigbreak \noindent {\sl Please make sure that your name 
and student I.D. number are clarly indicated  on every page, 
and that the pages are properly stapled.} {\bf Please write
clearly, in complete sentences with correct grammar and punctuation.}

\medbreak \noindent {\bf If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure you tell us that
you are continuing in the back.} 

\medbreak \noindent {\bf This test will be graded on a scale from 0 to
100. The point values of all the problems add up to {\mymagthreebf
165}. That means that {\mymagtwobf you have a lot of freedom to pick and
choose.  You do not need to do everything right to get a perfect 100
score.}}


\vskip 0.5cm
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vskip 0.2cm
\noindent In this test, ${\bf R}$, ${\bf N}$, denote, respectively, the
sets of real numbers and natural numbers.

\vskip 0.7cm

\item{{\bf 1.}} ({\sl 8 points}.) Define {\it interval}.

\vfill

\item{{\bf 2.}} ({\sl 12 points}.) Define $\lim_{x\to a}f(x)=L$.

\vfill\vfill

\startexpage

\item{{\bf 3.}} ({\sl 15 points}.) Explain what {\bf Russell's paradox}
is. 

\vfill
\startexpage

\item{{\bf 4.}} ({\sl 15 points.}) Let $S=\{x\in{\bf R}:x>0\}$.
Define a binary operation $*$ on $S$ by letting
$$a*b={ab\over a+b}\,\,\,\,\,\,\,{\rm for}\,\,\,a,b\in S\,.$$
Prove or disprove each of the following two statements:

\medbreak
\itemitem{a.} $*$ satisfies the {\it associative law}, i.e. 
$(\forall x\in S)(\forall y\in S)(\forall z\in S)(x*(y*z)=(x*y)*z)$;

\medbreak
\itemitem{b.} there exists a member $m$ of $S$ which is an 
{\it identity} for $*$, i.e. is such that $x*m=x$ for all $x\in S$.

\startexpage
\item{{\bf 5.}} ({\sl 18 points.}) Prove that the limit
of a sum is the sum of the limits, for limits as $x\to+\infty$.  Precisely,
prove that if $f$ and $g$ are functions defined on ${\bf R}$ and $A$, $B$
are real numbers such that $\lim_{x\to+\infty}f(x)=A$ and
$\lim_{x\to+\infty}g(x)=B$  , then $\lim_{x\to+\infty}(f(x)+g(x))=A+B$.
(Note: by definition, $\lim_{x\to+\infty}h(x)=L$  means that
$(\forall\varepsilon>0)(\exists M)(\forall x)(x>M\to
|h(x)-L|<\varepsilon)$.) 

\startexpage
\item{{\bf 6.}} ({\sl 15 points}.) Using the definition of limit, prove that
$\lim_{x\to 1}x^2=1$. 

\vfill

\startexpage
\item{{\bf 7.}} ({\sl 15 points}.) Let $f$ be the function given by 
$f(x)=1$ for $x\neq 0$, $f(0)=0$. Prove or disprove: $\lim_{x\to 0}f(x)=1$. 

\vfill

\startexpage
\item{{\bf 8.}} ({\sl 21 points}.) Prove or disprove: if $S$ and $T$ are 
open subsets of ${\bf R}$, then $S\cap T$ is also an open subset of ${\bf
R}$.   The relevant definitions here are as follows:   

\medbreak
\itemitem{a.} An {\it open subset} of ${\bf R}$ is a subset $S$ of ${\bf
R}$ such that every $x\in S$ is an interior point of $S$. 

\medbreak
\itemitem{b.} If $x\in{\bf R}$ and $S$ is a subset of ${\bf R}$, then $x$
is an {\it interior point} of $S$ if $(\exists\delta>0)(N(x,\delta)\subseteq
S)$.    

\medbreak
\itemitem{c.} If $x\in{\bf R}$ and $\delta>0$, then $N(x,\delta)$ is
defined to be the open interval $(x-\delta,x+\delta)$, i.e. 
$N(x,\delta)=\{y\in{\bf R}:x-\delta<y<x+\delta\}$.


\vfill
\startexpage

\item{{\bf 9.}} ({\sl 16 points}.) Using the definition of interval,
prove or disprove each of the following:

\medbreak\itemitem{(i)} The set $A=\{x\in{\bf R}:1<x<3\}$ is an interval.

\medbreak\itemitem{(ii)} The set $B=\{1,3\}$ is an interval.

\vfill
\startexpage

\item{{\bf 10.}} ({\sl 9 points}.) Explain what is wrong with the following
proof (note: please avoid needless verbiage; a short, concise
answer ---not longer than one line--- is much better that a lot of talk): 

\medbreak\noindent THEOREM:  $\,\,\,-1>1$. 

\medbreak\noindent PROOF:  Define a function $f(x)$ by letting 
$$f(x)=1+x+x^2+x^3+x^4+\ldots \,\,\,\,{\rm for}\,\,\,x\in{\bf R}\,.
\leqno{(1)}$$
Then 
$$f(x)>1\,\,\,\,{\rm if}\,\,\,\,x>0\,,\leqno{(2)}$$
because $x+x^2+x^3+x^4+\ldots>0$. 
On the other hand, if $x$ is any real number,
$$xf(x)=x+x^2+x^3+x^4+\ldots\,\,\,\,.\leqno{(3)}$$
So, subtracting (3) from (1), we get
$$f(x)-xf(x)=1\,.\leqno{(4)}$$
So $(1-x)f(x)=1$. Therefore
$$f(x)={1\over 1-x}\,\,\,\,{\rm for}\,\,\,x\in{\bf R}\,.\leqno{(5)}$$
Now plug in $x=2$. We get
$$f(2)={1\over 1-2}=-1\,.\leqno{(6)}$$
But in (2) we already proved that $f(x)>1$ if $x>0$. In particular
$f(2)>1$, since $2>0$.  So $-1>1$.  END~OF~PROOF

\startexpage

\item{{\bf 11.}} ({\sl 21 points}.) Let $f$ be a real-valued function
with domain ${\bf R}$. Let $C$ be the graph of $f$. Prove that $f$ is
even if and only if $C$ is symmetric with respect to the $y$ axis.

The relevant definitions here are as follows:

\medbreak
\itemitem{i.} A function $g$ with domain ${\bf R}$ is {\it even} if
$g(-x)=g(x)$ for all $x\in{\bf R}$. 


\itemitem{ii.} The {\it graph} of a function $g$ is the set
$\{(x,y):y=g(x)\}$. 

\itemitem{iii.} A subset $S$ of ${\bf R}\times{\bf R}$ is {\it symmetric
with respect to the $y$ axis} if 
$$(\forall x\in{\bf R})(\forall y\in{\bf R})\Bigl((x,y)\in S\to (-x,y)\in S\Bigr)\,.$$

\vfill
\end



%\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\input mymacros
\def\rrr{{\bf R}}
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\mymagtwobf MATH 300}}

\medbreak
\centerline{{\mymagtwobf INTRODUCTION TO}}

\medbreak
\centerline{{\mymagtwobf MATHEMATICAL REASONING}}

\medbreak
\centerline{{\mymagtwobf SPRING 1997}}

\bigbreak
\centerline{{\mymagtworm SECOND MIDTERM EXAM.}}

\medbreak
\centerline{{\mymagtworm \ \  April 24, 1996 }}

\medbreak
\centerline{{\mymagtworm INSTRUCTOR: H\'ector J.  Sussmann}}

\bigbreak \noindent {\mymagtwosl Please make sure that your name 
and student I.D. number are clearly indicated  on every page, 
and that the pages are properly stapled.} {\mymagtwobf \ \ Please

\noindent write
clearly, in complete sentences with correct grammar 

\noindent and punctuation.}

\medbreak \noindent {\mymagtwoit If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure to tell us that
you are continuing in the back.} 

\medbreak \noindent {\mymagtwobf This test consists of {\mymagfourbf 6}
problems. \ The \ test \ will \ be \ graded 

\noindent on \ a \ scale \ from 
{\mymagfourbf \ 0 \ to \ 100}. \ \ \ The \ point
\ values \ of \ all 

\noindent the \ problems \ add \ up \ to {\mymagfivebf \ 133}. \ That \ means \ that
{\mymagfourbf \ you 

\noindent have some freedom to choose.  You do not need
to do everything right to get a perfect 100 score. It suffices to do 
75\%{} of the work right to get a perfect score.}}

\bigbreak \noindent {\mymagtworm
In Pages 8 and 9 you will find the complete list of rules of inference,

\noindent derived rules and axioms that appear in the March 7 handout.}

\vskip 0.2cm
\noindent {\mymagtworm In this test, {\mymagtwobf R}, {\mymagtwobf N}, 
denote, respectively, the
sets of real numbers and 

\noindent natural numbers.}

\vfill

\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vfill
\startexpage


\item{{\bf 1.}} ({\sl 23 points  total}.)  

\itemitem{(i)} ({\sl 8 points}.)  Define what an {\it interval} is.

\vfill

\itemitem{(ii)} Using the definition of ``interval'' given
in (i), prove or disprove each of the following:

\medbreak
{\parindent=30pt 
\itemitem{(a)} ({\sl 5 points}.)  If $a$ is a real number,
then the set $I=\{x\in{\bf R}: x\ge a\}$ is an interval.

\vfill

\itemitem{(b)} ({\sl 5 points}.)  The set ${\bf N}$ is an interval.

\vfill

\itemitem{(c)} ({\sl 5 points}.)  The union of two intervals is an interval.
}

\vfill

\startexpage

\item{{\bf 2.}} ({\sl 22 points}.) In this problem, we use a formal
language which is a predicate calculus with the usual logical connectives
---that is $\vee$, $\wedge$, $\sim$, $\to$, $\leftrightarrow$, $\exists$
and $\forall$--- as well as the comma, the left and right parentheses
---``('' and ``)''--- and the following predicate symbols: 
a binary (i.e. two-variable) predicate $p$, and two 1-variable
predicates $q$, $r$.  The following four formulas are
sentences of this predicate calculus:
$$(\forall y)(\exists x)p(x,y)\to(\exists x)(\forall y)p(x,y)\,;
\eqno{(1)}$$
$$(\exists x)(\forall y)p(x,y)\to(\forall y)(\exists x)p(x,y)
\,;\eqno{(2)}$$
$$(\exists x)(q(x)\to r(x))\to\Bigl((\exists x)q(x)\to (\exists x)r(x)\Bigr)
\,;\eqno{(3)}$$
$$\Bigl(\sim (\exists x)q(x)\Bigr)\to\Bigl((\forall x)(q(x)\to r(x))\Bigr)
\,.\eqno{(4)}$$
For each of the four formulas, decide whether the formula is a theorem 
of the predicate calculus.  If it is a theorem prove it. (In the proofs,
you are allowed to use the 8 rules of inference {\bf and nothing else}.)
If it is not a theorem, then establish this fact by exhibiting an
interpretation of the predicates  for which the sentence is false. 


\vfill

\startexpage

\item{{\bf 3.}} ({\sl 22 points.}) In this problem, we work with real
number arithmetic.  The axioms are the ones of the March 7 handout. (You
will find the complete list of axioms in Page 9.)
``2'' means ``$1+1$.''  ``$x\neq y$'' means ``$\sim x=y$.''
The expression ``$x-y$'' is defined to mean ``$x+(-y)$'', where $-y$ is
characterized by the fact that $(-y)+y=0$.  Then ``$x<y$'' means
``$y-x\in{\cal P}$,''  and ``$x\le y$'' means ``$x<y\vee x=y$.''
The following seven formulas
are sentences of real number arithmetic.  For each formula, there are four
possibilities:  (a)~the formula is a theorem (that is, it can be proved
from the axioms of Page 9); (b)~the negation of the formula is a
theorem; (c)~the formula is true as a statement about the real numbers but
cannot be proved from the axioms; (d)~the negation of the 
formula is true as a statement about the real numbers but
cannot be proved from the axioms.  For each of the formulas, (i)~translate 
the formula into plain English, without using letter variables or
mathematical symbols, (ii)~decide which of the above four possibilities 
is the case, (iii)~if (a) holds, prove the formula (using the rules and the
axioms; reasonable shortcuts are allowed);  
(iv)~if (b) holds, prove the negation of the formula (using the rules and the
axioms; reasonable shortcuts are allowed);  
(v)~if (c) or (d) holds, give a convincing reason.   
$$(\exists x)(\forall y)(x\le y)\,;\eqno{(1)}$$
$$(\forall y)(\exists x)(x<y)\,;\eqno{(2)}$$
$$(\forall x)(\forall y)(x<y\to x.x<y.y)\,;\eqno{(3)}$$
$$(\forall x)(0<x\to (\exists y)(y.y=x))\,;\eqno{(4)}$$
$$(\forall x)(\forall y)(\forall z)(x.y=x.z\to y=z)\,;\eqno{(5)}$$
$$(\forall x)(\forall y)(\forall z)\Bigl(x.y=x.z\to(x\neq 0\to
y=z)\Bigr)\,;\eqno{(6)}$$
$$(\forall x)(x.x\neq 2)\,.\eqno{(7)}$$
\vfill

\startexpage
\item{{\bf 4.}} ({\sl 22 points}.) Some of the following seven formulas are 
valid identities among sets, and others are not.  Determine which of the
formulas are valid set identities (i.e. true for all possible choices of
the universe of discourse $U$ and the sets $A$, $B$) and which ones are
not. For each of the seven formulas: (i)~If the formula is a valid set
identity, prove it.  (In your proof, you are allowed to use the axioms, the
eight rules of inference, and reasonable shortcuts.  Informal proofs such
as those that occur in the book are O.K.) (ii)~If the formula is not
valid, show it by giving an appropriate counterexample.
$$(A\cap B)'=A'\cup B'\,;\eqno{(1)}$$
$$(A\cap B)'\subseteq A'\cup B'\,;\eqno{(2)}$$
$$A\cap A'=\emptyset\,;\eqno{(3)}$$
$$A\cap B\subseteq A\,;\eqno{(4)}$$
$$A\cup B\subseteq A\,;\eqno{(5)}$$
$$A=(A\cup B)-B\,;\eqno{(6)}$$
$$A\cap B'=(A\cup B)-B\,.\eqno{(7)}$$

\vfill

\startexpage
\item{{\bf 5.}} ({\sl 22 points}.) Do either 5a or 5b below.  Indicate
clearly which of the two you are choosing.   
(An informal proof is O.K. here.) 

\medbreak
\itemitem{{\bf 5a.}} A set $I$ of real numbers is {\it
convex} if $(\forall x\in I)(\forall y\in I)(\forall t\in {\bf R})
(0<t<1\to t.x+(1-t)y\in I)$.  

\itemitem{} Prove: $I$ is convex $\leftrightarrow\,\,\,I$ is an interval.


\medbreak
\itemitem{{\bf 5b.}} A set $I$ of real numbers is {\it bounded} if $(\exists
x\in{\bf R})(\exists y\in{\bf R})(\forall z\in I)(x\le z\le y)$.  

\itemitem{} Prove: the union of two bounded sets is also
bounded.


\vfill

\startexpage
\item{{\bf 6.}} ({\sl 22 points}.) A set $S$ of points $(x,y)$ in
${\bf R}\times{\bf R}$ is {\it symmetric with respect to the origin}
if 
$$(\forall x)(\forall y)((x,y)\in S\to (-x,-y)\in S)\,.$$
(Here ``$(x,y)$'' means ``the ordered pair formed by $x$ and $y$.'' 
It does {\it not} mean ``the open interval from $x$ to $y$.'')

\item{} A real-valued function $f$ with domain ${\bf R}$ is {\it odd}
if $(\forall x)(f(-x)=-f(x))$.  The {\it graph} of a function $f$ is the
set $G(f)=\{(x,y):y=f(x)\}$. 

\item{} Prove or disprove each of the following 
statements (informal proofs are O.K.):

\itemitem{(i)} The set $\{(x,y):x^2+y^2\le 1\}$ is symmetric with respect
to the origin.

\itemitem{(ii)} The set $\{(x,y):(x-1)^2+y^2\le 1\}$ is symmetric with respect
to the origin. 

\itemitem{(iii)} The function $f$ given by $f(x)=x^2$ is odd.

\itemitem{(iv)} The function $f$ given by $f(x)=x+x^3$ is odd.

\itemitem{(v)} A function $f$ is odd if and only if $G(f)$ is symmetric 
with respect to the origin. 


\vfill\eject


\def\ste{\enspace\enspace\enspace\enspace}
\newcount{\statco}
\newcount{\quesco}
\newcount{\defco}
\newcount{\notco}
\newcount{\remco}
\newcount{\rulco}
\newcount{\drulco}
%\def\co#1{#1}
\def\rulo{\global\advance\rulco by 1
\medbreak\noindent{\bf  RULE
\the\rulco.}\ }
\def\drulo{\global\advance\drulco by 1
\medbreak\noindent {\bf DERIVED RULE
\the\drulco.}\ }
\def\justo{
\medbreak\noindent {\bf JUSTIFICATION$\,$:}\ \ }
\def\co#1{}
\def\ep{\varepsilon}
\def\lto{\leftrightarrow}
\def\clo{{\rm Clos}}
\defco=0
\statco=0
\remco=0
\rulco=0
\drulco=0
\notco=0
\quesco=1
\def\ino{{\rm Int}}

\centerline{{\bf INFERENCE RULES FOR PROOFS}}

\medbreak\noindent Here are the eight basic rules of inference 
that you are allowed to use in a formal proof.

\rulo You are allowed to bring in any statement which is
an instance of a tautology.

\rulo If you have $p$ and $p\to q$ then you can go to $q$. (This is called {\it
Modus Ponens}).

\rulo If you have proved $q$ under the assumption $p$ (that is, if you have
assumed $p$ and proved $q$), then you can go to $p\to q$.

\rulo If you have $(\forall x)p(x)$ and $s$ is anything, you can go 
to $p(s)$. (This is called {\it specialization}).

\rulo To prove a statement of the form $(\forall x)p(x)$, you start 
by saying ``Let $a$ be arbitrary,'' and then try to prove $p(a)$. 


\rulo If you have $(\exists x)p(x)$ then you can introduce a new
object and give it a letter name such as $a$ or $\alpha$ or anything you
want, provided that this name has not been used before, and stipulate that
$p(a)$.  (Here {\bf it's important that this new object should be given a
name that has not been used before.})

\rulo If you have an object $a$ such that $p(a)$,  then you can go to 
$(\exists x)p(x)$.  

\rulo You can substitute equals for equals.  That is, 
if you know that $a=b$, then you can go from a statement involving 
$a$ to the statement in which $b$ is substituted for $a$ everywhere,
and from a statement involving 
$b$ to the statement in which $a$ is substituted for $b$ everywhere.
(Example: After you have given the definition
$$[0,1]=\{x:0\le x\le 1\}\,$$
you can go from
$a\in [0,1]$ to $a\in\{x:0\le x\le 1\}$, and then you can use Set Axiom 2
below to get to $0\le a\le 1$.) 

\bigbreak

\centerline{{\bf DERIVED RULES (SHORTCUTS)}}

\drulo    If you have $(\forall x)(p(x)\to q(x))$ and $p(a)$ then
you can go to $q(a)$.

\drulo    If you have $p\wedge q$ then you can go to $p$. 

\drulo    If you have $p\wedge q$ then you can go to $q$. 

\drulo    If you have $p$ and $q$ then you can go to $p\wedge q$. 

\drulo    If you have $p$ then you can go to $p\vee q$. 

\drulo    If you have $q$ then you can go to $p\vee q$. 

\drulo    If you have $p\to q$ and $q\to p$ 
then you can go to $p\leftrightarrow q$. 

\drulo    If you have $p\leftrightarrow q$ 
then you can go to $p\to q$.

\drulo    If you have $p\leftrightarrow q$ then you can go to $q\to p$.

\drulo    If assuming $\sim q$ you get a contradiction, then 
you can go to $q$.

\drulo    From $\sim p$ and $p\vee q$ you can go to $q$. 

\drulo    From $\sim q$ and $p\vee q$ you can go to $p$.

\drulo    If assuming $p$ and $\sim q$ you get $\sim p$, then 
you can go to $p\to q$.

\drulo    If assuming $\sim p$ you get $q$, then 
you can go to $p\vee q$.

\drulo    If assuming $\sim p$ you get $q$, and assuming $p$ you also get $q$, 
then you can go to $q$ assuming nothing.

\drulo    If assuming $p$ you get $q$, and assuming $q$ you get $p$, 
then you can go to $p\leftrightarrow q$.

\drulo    If you have $(\forall x\in S)p(x)$ and you have
$s\in S$, you can go to $p(s)$. 

\drulo    To prove a statement of the form $(\forall x\in S)p(x)$, you
start by saying ``Let $a$ be an arbitrary element of $S$'' (that is,
``let $a$ be arbitrary; assume $a\in S$'') and then try to
prove $p(a)$. (Here, again, instead of ``$a$'' you could use ``$b$,'' or
``$z$,'' or ``$\alpha$,'' or ``$\beta$,'' or ``$\aleph$,'' or
``$\diamond$,'' or any symbol you want.  What is important that what you do
should apply to a {\it completely arbitrary} element of $S$.)

\drulo    If you have $(\exists x\in S)p(x)$ then you can introduce a
new object and give it a letter name such as $a$ or $\alpha$ or anything you
want, provided that this name has not been used before, and stipulate that
$a\in S$ and $p(a)$.  

\drulo     If you have $p(a)$ and $a\in S$ then you can go to $(\exists
x\in S)p(x)$. 

\bigbreak
\centerline{{\bf SET AXIOMS}}

\bigbreak\noindent SET AXIOM 1: If $S$, $T$ are sets, then
$$S=T\leftrightarrow (\forall x)(x\in S\leftrightarrow x\in T)\,.$$

\bigbreak\noindent SET AXIOM 2: If $S$ is a set and 
$p(x)$ is any formula containing the
variable $x$, no quantifiers involving $x$, and no other free variable,
then $\{x:x\in S\wedge p(x)\}$ (that is, 
$\{x\in S:p(x)\}$) is a set. 

\bigbreak\noindent SET AXIOM 3: If $p(x)$ is any formula containing the
variable $x$, and no quantifiers involving $x$,  and $p(x)$ does 
not contain the variable $y$, then 
$$(\forall y)(y\in\{x:p(x)\}\leftrightarrow p(y))\,.$$

\def\mitem{\item}
\def\rrbf{{\bf R}}
\bigbreak
\centerline{{\bf AXIOMS ABOUT EQUALITY}}

\item{Eq1.} $(\forall x)(x=x)$.

\mitem{Eq2.} $(\forall x)(\forall y)(x=y\to y=x)$.

\mitem{Eq3.} $(\forall x)(\forall y)(\forall z)((x=y\wedge y=z)\to x=z)$.

\bigbreak
\centerline{{\bf THE AXIOMS OF ARITHMETIC }}
\medbreak

\mitem{A0.} ${\bf R}$ is a set, ${\cal P}$ is a subset of ${\bf R}$,
$0\in{\bf R}$ and $1\in{\bf R}$.

\mitem{A1.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(x+y\in\rrbf)$.

\mitem{A2.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(x.y\in\rrbf)$.

\mitem{A3.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(x+y=y+x)$.

\mitem{A4.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(x.y=y.x)$.

\mitem{A5.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(\forall z\in\rrbf)\Bigl(
(x+y)+z=x+(y+z)\Bigr)$.

\mitem{A6.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(\forall
z\in\rrbf)\Bigl((x.y).z=x.(y.z)\Bigr)$.

\mitem{A7.} $(\forall x\in\rrbf)(\forall y\in\rrbf)(\forall z\in\rrbf)\Bigl(
(x+y).z=x.z+y.z\Bigr)$.

\mitem{A8.} $(\forall x\in\rrbf)(0+x=x)$.

\mitem{A9.} $(\forall x\in\rrbf)(\exists y\in\rrbf)(x+y=0)$.

\mitem{A10.} $(\forall x\in\rrbf)(1.x=x)$.

\mitem{A11.} $(\forall x\in\rrbf)\Bigl(x\neq 0\rightarrow(\exists
y\in\rrbf)(y.x=1)\Bigr)$. 

\mitem{A12.} $(\forall x\in\rrbf)(\forall y\in\rrbf)\Bigl((x\in{\cal P}\wedge
y\in{\cal P})\to x+y\in{\cal P}\Bigr)$.

\mitem{A13.} $(\forall x\in\rrbf)(\forall y\in\rrbf)\Bigl((x\in{\cal P}\wedge
y\in{\cal P})\to x.y\in{\cal P}\Bigr)$. 

\mitem{A14.} $(\forall x\in\rrbf)(x\in{\cal P}\vee x=0\vee
(\forall y\in\rrbf)(y+x=0\to y\in{\cal P}))$. 

\mitem{A15.} $0\notin{\cal P}$.

\mitem{A16.} $1\neq 0$.

\end

\magnification=\magstephalf
%\nopagenumbers
%\hsize 18true cm
\xdef\startexpage{
\eject
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}

\vskip 0.5cm
}
\def\boxoto{\enspace\lower5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{O}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}}
\def\plttr#1{\hbox to 9cm{#1\enspace\hrulefill} \boxoto\ True \ \boxoto\  False\ }
\def\tra{\vskip 0.05cm Translation:
\vskip 1.5cm}
\def\pltt#1{\hbox to 6.4cm{#1\hfill\enspace is$\,\,\,$a$\,\,\,$}
\boxoto$\,$Tautology \boxoto$\,$Contradiction\ \boxoto$\,$Contingency .}
\def\lr{\leftrightarrow}
\def\plcor{\boxoto\ Correct \ \ \ \boxoto\  Incorrect.}
\def\plhas{\boxoto\ Has a solution \ \ \ \boxoto\  Does not have a solution.}
\raggedbottom
\centerline{{\bf MATH 300 (INTRODUCTION TO MATHEMATICAL REASONING), FALL 1995}}

\medbreak
\centerline{FIRST MIDTERM EXAM.\ \  October 16, 1995}

\bigbreak \noindent {\sl Please make sure that your name 
and student I.D. number are clarly indicated  on every page, 
and that the pages are properly stapled.} {\bf Please write
clearly, in complete sentences with correct grammar and punctuation.}

\medbreak \noindent {\bf If you need extra space to write, you can use
the back of the exam sheets. In that case, make sure you tell us that
you are continuing in the back.} 

\medbreak \noindent {\bf This test will be graded on a scale from 0 to
100. The point values of all the problems add up to 140. That means that
you do not need to do everything right to get a perfect 100 score.}


\vskip 0.5cm
\centerline{\hbox to 6cm{\hrulefill}\ \ \ \hbox to 6cm{\hrulefill}}
\centerline{\hbox to 6cm{\hfill Student's name\hfill}\ \ \ \hbox to
6cm{\hfill Student's I.D. number\hfill}}


\vskip 0.2cm
\noindent In this test, ${\bf R}$, ${\bf N}$, denote, respectively, the
sets of real numbers and natural numbers.

\vskip 0.7cm

\item{{\bf 1.}} ({\sl 5 points}.) Assuming that $1=0$, prove that the
capital of New Jersey is Piscataway.

\vfill

\startexpage

\item{{\bf 2.}} ({\sl 18 points, but you get  $-1.5$ for each wrong answer, so
if you don't know you are better off not guessing.}) In each of the
following cases, check the appropriate box, and
{\it justify your answer}. (In most cases there is no need
to write a complete truth table.)

\medbreak
\item{1.} \pltt{$p\to(q\to (r\to (s\to (p\wedge q\wedge r\wedge s))))$}

\vfill
\item{2.} \pltt{$(p\wedge(\sim p))\to q$}

\vfill
\item{3.} \pltt{$p\to(\sim p)$}

\vfill
\item{4.} \pltt{$(p\wedge (\sim q))\to(p\to q)$}

\vfill
\item{5.} \pltt{$((\sim p)\vee(\sim q))\to(p\to(q\to r))$}

\vfill
\item{6.} \pltt{$((\sim p)\wedge(\sim q))\wedge(p\vee q)$}

\vfill

\startexpage
\item{{\bf 3.}} ({\sl 24 points}.) In this problem, ``$f^{(j)}$'' means 
``the $j$-th derivative of $f$.'' (For example, $f^{(0)}=f$, $f^{(1)}=f'$,
$f^{(2)}=f''$, $f^{(3)}=f'''$, etc.) Recall that 
${n\choose k}={n!\over k!(n-k)!}$ for $k,n\in{\bf N}$, 
$0\le k\le n$.

\medbreak\item{} Using the formula for the derivative of a product,
$$   (fg)'\,\,=\,\,f'g+fg'\,,$$
(or, if you prefer,
${d\over dx}\Bigl(f(x)g(x)\Bigr)\,={df\over dx}(x)g(x)
+f(x){dg\over dx}(x)$,
which is another way of saying exactly the same thing), prove the
following:

\medbreak\item{} For every natural number $n$, if $f$, $g$ are 
functions having $n$ derivatives, 
then
$$ (fg)^{(n)} = \sum_{k=0}^n{n\choose k}f^{(n-k)}g^{(k)}\,.$$

\medbreak\item{} You may use the formulas: 
${n\choose k}+{n\choose k-1}={n+1\choose k}\,\,\,{\rm for\,\,\,}
k,n\in{\bf N}\,,\,0<k\le n$, and also

\item{} $\sum_{j=m}^{\mu}x_j=\sum_{j=m+i}^{\mu+i}x_{j-i}$.

\vfill

\startexpage
\item{{\bf 4.}} ({\sl 20 points}.) In this problem, ``word'' means
``finite sequence of letters.''  For example, {\it mathematics}, {\it
qwertyuiop}, {\it asdfghjjkkl}, {\it zxcvbnm}, {\it babebibobu}, {\it
moooo}, {\it gfafxegf}, {\it ggggggggg}, {\it grrr} are all words.

\itemitem{(i)}  Determine how many six-letter words can be formed 
consisting of exactly three vowels and three consonants, with repetition
allowed for the vowels but not for the consonants.  (For example: {\it
arabap}, {\it benito}, {\it yaaexz} and {\it xaeiyz} are O.K., but {\it
baeiou}, {\it babeic} and {\it gugugu} are not.  Recall that in the English
alphabet there are 5 vowels and 21 consonants.)

\itemitem{(ii)}  Determine what percentage of the words described in (i)
contain the letter ``a''.    Is it higher or lower than 50{\%}? 

\item{} In both cases, {\it explain how you get your answer}, and {\it make
sure that you clearly indicate what your final answer is}, and that {\it
you carry out all arithmetic operations and produce a number}. (An answer
such as $41\times 38\times 32$ is {\it not} acceptable.)

\vfill

\startexpage
\item{{\bf 5.}} ({\sl 18 points}.) Prove or disprove:

\medbreak\item{} The equation $8x^5+5x^8=-3.01$ has a solution $x\in{\bf R}$.

\vfill
\startexpage

\item{{\bf 8.}} ({\sl 18 points}.) Prove or disprove each of the following:

\medbreak
\itemitem{1.} If every real number has a multiplicative inverse then 
every real number has a real square root. 

\vfill

\itemitem{2.} If a real number has a multiplicative inverse then it has a
real square root. 

\vfill

\itemitem{3.} If a real number has a real square root then it has a
multiplicative inverse.

\vfill

\itemitem{4.} If $A$, $B$ are sets then $(A\cup B)\cap A=A$.

\vfill

\itemitem{5.} If $A$, $B$, $C$ are sets then $(A\cup B)\times C=
(A\times C)\cup (B\times C)$.

\vfill
\vfill

\itemitem{6.} If $A$, $B$, $C$ are sets then $(A\cup B)\times(C\cup D)=
(A\times C)\cup (B\times D)$.

\vfill
\end





\subsection{Write clearly in complete sentences}
You should write so that you can be easily understood by a properly trained
English-speaking individual. In particular, this means that you must 

\medbreak
%{\parindent=1.71828
\bi
\item{} Use {\it complete English sentences}, that make 
clearly identifiable {\it statements} with a {\it clear meaning} that can

can be understood by anyone reading what you  wrote.
For example:


\bi
\item{} If you tell me that
``she is very smart,'' but you haven't told me who ``she'' is, then I don't
know what you are talking about, so you haven't made a statement with a
clear meaning.

\item{} If you write ``$x>0$,'' but you haven't told me who ``$x$'' is,
then I don't know what you are talking about, so you haven't made a
statement. 

\item{} If I ask you to state Pythagoras' theorem and your answer
only says ``$a^2+b^2=c^2$,'' then nobody will know what you are talking about,
because you have not said what ``$a$,''  ``$b$,'' and ``$c$'' are supposed 
to be.\footnote{Here is a correct statement of Pyhtagoras' theorem:
{\em Let $c$ be the lenght of the hypothenuse of a right triangle,
and let $a$, $b$ be the lengths of the other two sides. Then $a^2+b^2=c^2$.}}
\ei

\item{} Avoid exaggerated or incorrect use of cryptic 
mathematical notation. 

\item{} Explain what you are doing.

\item{} Make sure that letter ``variables'' are only used if
either: (i)~it has been said before what these letters stand for, or
(ii)~they are ``closed variables'' (or ``dummy variables,'' or ``bound
variables'') in the sense that will be discussed in detail in class, and 
is also explained in \S\ref{using-lets}.

\item{} Provide proper connectives between equations as well as
between ideas. 

\item{} Make sure that all the rules of English grammar (including
those of spelling and punctuation) are strictly obeyed.  

\item{} Try to say things correctly,  following all the rules,
but {\it in your own \mbox{words}.} Please {\it no rote learning.}  If
you have to memorize a definition or a \mbox{statement}, then that is 
not a good sign, because it indicates lack of understanding.

\item{} Please proofread carefully what you hand in. Ideally, you should
read and reread and revise almost any formal communication.  {\bf Neatness
and clarity count}, as you well know if you've tried to read any
complicated document.

\item{} {\Large{\em Do not assume that the people reading your paper can
read your mind. Do assume that they are intelligent, but also assume that
they are busy, and cannot and will not spend an excessive amount of time
puzzling out your meaning.  Communication is difficult, and written
technical communication is close to an art.}}

\ei

\begin{center}
\fbox{\fbox{
\begin{minipage}{12cm}
{\Large\bf Effective written exposition will be worth at least 50\% of your
grade. Conversely, bad or unclear exposition may be penalized as much as 50\%
of the grade or even more.}
\end{minipage}}}\end{center}

\bi
\item{} The best reference known to me on effective writing is {\sl
The Elements of Style} by Strunk and White, a very thin paperback
published by Macmillan. It isn't expensive, and it is easy to read.  I
recommend it.
\ei

\subsection{Your written work}
{\large{\bf You should pay attention to presentation, 
\mbox{especially} for the
homework}:}     

\vspace{-2mm}
\bi
\item{} A nicely typed homework (e.g., using a word processor)
is preferable to handwritten work. 
Handwritten work is acceptable too, but in that case:

\vspace{-3mm}
\bi 
\item{} If you have to cross out lots of words, then you should rewrite the
whole thing anew, {\bf cleanly and neatly}.  If you are not willing to
spend some of your time doing this; if what you hand in shows that you were
in a hurry and that you did not make the effort to write things neatly and
properly, then there is no reason for the instructor 
or the grader to spend any of our  time reading
what you wrote, and we will not do it.  

\vspace{-1mm}
\item{} You should use a pen. Never use a pencil. 

\vspace{-1mm}
\item{} Use any color other than red (for example, black, blue, or green),
but DO NOT USE RED. (Reason: The use of red is reserved for the
instructor's and grader's comments.) 
\ei

\vspace{-4mm}
\item{} Make sure that {\bf your name appears in every sheet 
of paper you hand in}, and that if you are handing in more than one sheet
then the sheets are {\bf stapled} and the pages are {\bf numbered}.

\vspace{-3mm}
\item{} If you tear off the sheets from a spiral notebook, please make sure
before you hand them in that there are none of those ugly hanging
shreds of paper
at the margins.  Use scissors, or a cutter, if necessary.
\ei
\begin{center} 
\fbox{\fbox{\fbox{\begin{minipage}{11cm} {\bf {\large
\null
\noindent If you hand in a homework assignment that has
at least one of the following flaws:

\vspace{-2mm}
\bi
\item{} it is written carelessly or in a hurry, 

\vspace{-3mm}
\item{} it has lots of words crossed out, 

\vspace{-3mm}
\item{} it has unreadable handwriting, 

\vspace{-3mm}
\item{} it has pages that are unstapled or unnumbered or fail to show your
name,  

\vspace{-3mm}
\item{} it has shreds of paper at the margins,

\vspace{-3mm}
\item{} it is written using pencil rather than a pen,

\vspace{-3mm}
\item{} it is written in red,
\ei

\vspace{-3mm}\noindent then }
{\huge the assignment will be marked ``unacceptable'' and 
\mbox{returned} unread.}}

$\quad$
\end{minipage}}}}\end{center}

\subssection{Answering questions in this course}\label{onanswering}

{\large{\bf In this course, 
whenever you are given a problem where you are asked to do something, your 
answer should be either:

\medbreak
\bi
\item[(a)] doing what you were asked to do, 
\item[or]
\item[(b)] showing---that is {\em proving}---that it cannot be done.
\ei
}}  

\medbreak\noindent (See, for example, Problems 2, 3, 4, 5, 8,
10, 12  below, where the
correct answer is ``what you asked me to do cannot be done.'')

Notice that, when the answer is that ``it cannot be done,'' it is not
enough for you to {\it say} that it cannot be done. {\it You have to tell
me why.} In other words, you have to {\it prove} that it cannot be
done. 

\medbreak\noindent {\large {\bf This remark is very important, and will
apply \mbox{throughout} the semester, not just during the first week. And it 
\mbox{applies} to all your work, to the homework, the quizzes, the 
\mbox{midterm}
exams, and the final exam. So please
read it and reread it until you are sure you got the point.}} \proofend


\subssection{Some examples of problems with correct answers}

\bigbreak\noindent 
Here are some examples of problems with correct solutions:

\medbreak\noindent PROBLEM 1: Express the number $26$ as a sum of two odd
natural numbers.\label{probwithcor}

\medbreak\noindent ANSWER: $26=3+23$.  \proofend

\medbreak\noindent REMARK: {\em There are lots of other solutions, of
course! For example, here are two solutions different from the one 
given above: \mbox{$26=7+19$}, $26=13+13$.}   

\bigbreak\noindent PROBLEM 2: How many solutions does 
Problem 1 have? \label{page-howmany}

\medbreak\noindent ANSWER: This question cannot be answered
in a unique way, because we have not told you exactly 
what is meant by a
``solution of Problem 1,'' and, depending on how you define
what is meant by a ``solution,'' the answers to our question may be different. 

Here are two equally legitimate ways to define ``solution of Problem 1'':

\medbreak\noindent{\bf First way:} A {\em solution of Problem 1} 
is an ordered pair
$(n,m)$ of odd natural numbers such that $n+m=26$. (What this means is
that a solution consists of specifying a first odd natural number $n$
and a second odd natural number $m$ such that $n+m=26$. So ``$n=3$ and
$m=23$'' is a solution, and ``$n=23$ and $m=3$'' is a {\em different}
solution.)

According to this definition of ``solution of Problem 1,''
there are exactly {\bf 13 solutions}, namely, the ordered pairs
$(1,25)$, $(3,23)$, $(5,21)$, $(7,19)$, $(9,17)$, $(11,15)$,
$(13,13)$, $(15,11)$, $(17,9)$, $(19,7)$, $(21,5)$,
$(23,3)$, $(25,1)$.   

\medbreak\noindent{\bf Second way:} A {\em solution of Problem 1} 
is an unordered 
pair $\{n,m\}$ of odd natural numbers such that $n+m=26$. 

According to this definition of ``solution of Problem 1,'' there are
exactly {\bf 7 solutions}, namely, the unordered pairs $\{1,25\}$, $\{3,23\}$,
$\{5,21\}$, $\{7,19\}$, $\{9,17\}$, $\{11,15\}$, $\{13,13\})$.  (What
this means is that a solution consists of specifying two odd natural
numbers whose sum is 26, without singling out one of them as the 
``first'' one. So in this case
``$n=3$ and $m=23$'' is a solution, and ``$n=23$ and $m=3$'' is  {\em
the same} solution.) \proofend

%\medbreak\noindent{\bf What do we learn from this example?}
%We learn that {\bf the meaning of a question is not clear 
%if you do not understand the meanings 

\bigbreak\noindent PROBLEM 3: Express the number $27$ as the sum of two odd
natural numbers.

\medbreak\noindent SOLUTION: This is impossible. REASON: the sum of two odd
numbers is always even. Since $27$ is odd, it cannot be the sum of two odd
numbers. \proofend

\bigbreak\noindent PROBLEM 4: Prove that $2+2=5$.

\medbreak\noindent SOLUTION: This cannot be done. REASON: 
The statement ``$2+2=5$'' is false, and false
statements cannot be proved. 

\bigbreak\noindent PROBLEM 5: Prove that for every positive integer $x$
there is another positive  integer $y$ such that $0<y<x$. 

\medbreak\noindent SOLUTION: This cannot be done. REASON: 
The statement that we are being asked to prove is false, and a false
statement cannot be proved. 

The reason that the statement is false is as
follows: $1$ is a positive integer, and there is no positive integer $y$ such
that $0<y<1$, because $1$ is the smallest positive integer. So if we take $x$
to be $1$ then the conclusion is not true. So it is not true that ``for {\em
every} positive integer $x$ 
there is another positive integer $y$ such that $0<y<x$.'' \proofend

\bigbreak\noindent PROBLEM 6: Express the number $12$ as the difference of 
the squares of two natural numbers. 

\medbreak\noindent SOLUTION: $12=16-4$, so $12=4^2-2^2$. \proofend 

\bigbreak\noindent PROBLEM 7: Express the number $20$ as the sum of 
the squares of two natural numbers. 

\medbreak\noindent SOLUTION: $20=4+16=2^2+4^2$. \proofend

\bigbreak\noindent PROBLEM 8: Express the number $4$ as the sum of 
the squares of two positive integers. 

\medbreak\noindent SOLUTION: This cannot be done. REASON:
here are all the ways to express $4$ as a sum of two positive integers:
$$
4=1+3\,,\,\,\,\quad 4=2+2\,.
$$
In neither of these two equalities are the two terms on the right-hand side
squares.

\bigbreak\noindent PROBLEM 9: Express the number $4$ as the sum of 
the squares of two natural numbers. 

\medbreak\noindent SOLUTION: $4=0+4=0^2+2^2$.\proofend

\bigbreak\noindent PROBLEM 10: Express the number $12$ as the sum of 
the squares of two natural numbers. 

\medbreak\noindent SOLUTION: This cannot be done. REASON:
here are all the ways to express $12$ as a sum of two natural numbers:
$$
12=0+12,\,12=1+11,\,12=2+10,\,12=3+9,\,
12=4+8,\,12=5+7,\,12=6+6.
$$
In none of these expressions are the two terms on the right-hand side
squares of natural numbers. \proofend

\bigbreak\noindent PROBLEM 11: Express the number $12$ as the sum of 
the squares of three natural numbers. 

\medbreak\noindent SOLUTION: $12=4+4+4$, so $12=2^2+2^2+2^2$.\proofend

\bigbreak\noindent PROBLEM 12: Express the number $15$ as the sum of
the squares of three natural numbers. 

\medbreak\noindent SOLUTION: This cannot be done. REASON:
If we are going to write $15$ as the sum of three squares, then
we should be able to write it as the sum of two natural numbers,
one of which is a square, while the other is the sum of two squares.

So let us look at all the ways to write $15$ as a sum of two natural
numbers, see in which cases one of these numbers is square,
and then see if the other number is the sum of two squares.

Here are all the ways to express $15$ as a sum of two natural numbers:
$$
\left.\begin{array}{llll}
15=0+15,&\quad 15=1+14,&\quad 15=2+13,&\quad 15=3+12,\\
15=4+11,&\quad 15=5+10,&\quad 15=6+9,&\quad 15=7+8\,.\end{array}\right.
$$
There are four expressions in the above list where one of the numbers
is a square:
$$
15=0+15,\quad 15=1+14,\quad 15=4+11,\quad 15=6+9\,.
$$
In each case, we must ask whether the other number is a sum of two squares.

For ``15=0+15,'' we have to see if $15$ is the sum of two squares.
The answer is ``no,'' because we already have the list of all (eight) ways
of writing $15$ as a sum of two natural numbers, and in no case
are both those numbers suqares. 

For ``15=1+14,'' we have to see if $14$ is the sum of two squares.
The answer is ``no.'' (Reason: the only squares not greater
than $14$ are $0$, $1$, $4$, and $9$. So if are going to write
$14$ as a sum of two squares, the only possibilities
are $14=0+14$, $14=1+13$, $14=4+10$, $14=9+6$. But in all these cases
the other number---that is, $14$, $13$, $10$, or $6$---is not a square.)

For ``15=4+11,'' we have to see if $11$ is the sum of two squares.
The answer is ``no.'' (Reason: the only squares not greater
than $11$ are $0$, $1$, $4$, and $9$. So if are going to write
$11$ as a sum of two squares, the only possibilities
are $11=0+11$, $11=1+10$, $11=4+7$, $14=9+5$. But in all these cases
the other number (that is, $11$, $10$, $7$, or $5$) is not a square.)

Finally, for ``15=6+9,'' we have to see if $6$ is the sum of two squares.
The answer is ``no.'' (Reason: the only squares not greater
than $6$ are $0$, $1$, and $4$. So if are going to write
$6$ as a sum of two squares, the only possibilities
are $6=0+6$, $6=1+5$, $6=4+2$. But in all these cases
the other number (that is, $6$, $5$, or $2$) is not a square.)\proofend

\bigbreak\noindent PROBLEM 13: Express the number $15$ as the sum of
the squares of four natural numbers. 

\medbreak\noindent SOLUTION: $15=1+1+4+9=1^2+1^2+2^2+3^2$. \proofend

\bigbreak\noindent PROBLEM 14: Express the number $674$ as the sum of
the squares of four natural numbers. 

\medbreak\noindent SOLUTION: $674=144+400+121+9=12^2+20^2+11^2+3^2$. \proofend

\bigbreak\noindent REMARK: You may be thinking {\em how on Earth 
did the author of these notes figure out how to solve problem 14?}
That's a good question. It turns out that there is a technique
for doing this, but it's not easy to explain. Maybe we will talk about
it later in the course. But keep in mind that the question
``how did I find a solution to problem 14?''  is quite different 
from the question 
``is $674=12^2+20^2+11^2+3^2$ a correct solution?'' 
This second question is easy to answer: you just compute 
$12^2+20^2+11^2+3^2$ and verify that what you get is $674$.\proofend

\bigbreak\noindent REMARK: You may have noticed that I gave you
examples of natural numbers that could be expressed as the sum of two
squares, and of natural numbers that could not. Then I gave you
examples of natural numbers that could be expressed as the sum of
three squares, and of natural numbers that could not.  But for {\em
four} squares what I have done is different: I gave you examples of
natural numbers that could be expressed as the sum of four squares,
but I did not give you examples of natural numbers that could not.  
Is there a reason for this? The answer is {\bf yes}. The mathematician 
Lagrange (1736-1813) proved that {\bf every natural number 
can be expressed as the sum of the squares of four natural numbers.}
This is a hard theorem, and I will try to say a few words about it later
in the course.\proofend

\newpage
\section{A first look at the world of mathematical objects: number systems}\label{worldofmath}

\begin{flushright}

\begin{minipage}{7cm}

\vspace{1.5mm}

{\it ``The time has come,'' the Walrus said, 

\ \ \ ``To talk of many things:

Of shoes---and ships---and sealing wax---

\ \ \ Of cabbages--and kings

And why the sea is boiling hot---

\ \ \ And whether pigs have wings.''}

\vspace{1.5mm} 

\end{minipage}

\end{flushright}

\vspace{4mm}




\begin{center}
\fbox{\fbox{\begin{minipage}{11cm}
\boxon{SUMMARY OF CHAPTER \ref{worldofmath}\label{sum-chap-things}}
\medbreak\noindent There are many different types of mathematical objects.
Among them, there are {\bf numbers} of various kinds,
and also lots of other things that are not numbers.

{\bf Numbers} belong to {\bf number systems}. The most 
important number systems are:

\vspace{-2.1mm}
\begin{itemize}
\item{} $\nn$, the set of {\em natural numbers},

\vspace{-3.1mm}
\item{} $\zz$, the set of {\em integers},

\vspace{-3.1mm}
\item{} $\qq$, the set of {\em rational numbers},

\vspace{-3.1mm}
\item{} $\rr$, the set of {\em real numbers},

\vspace{-3.1mm}
\item{} $\cc$, the set of {\em complex numbers},

\vspace{-3.1mm}
\item{} $\kk$, the set of {\em quaternions},

\vspace{-3.1mm}
\item{} $\zz_2$, $\zz_3$, $\zz_4$, $\zz_5$, $\zz_6$, and,
more generally, $\zz_n$---the set of {\em integers modulo $n$}---for 
every natural number $n$ such that $n\ge 2$.
\end{itemize}

\vspace{-2.1mm}
In addition to numbers, there are all kinds of other mathematical 
\mbox{objects}, such as {\em sets}, {\em lists}, {\em operations}, 
{\em functions},
{\em relations}, {\em arrays} of various kinds and, in particular, 
{\em matrices}, {\em linear spaces} (such as $\rr^2$, $\rr^3$, $\rr^4$, 
etc.), {\em geometrical figures} such as {\em curves}
of different kinds (for example, lines, parabolas, ellipses, hyperbolas,
cycloids, exponential curves), {\em \mbox{surfaces}} 
(such as planes, spheres, cylinders,
cones, ellipsoids, paraboloids, \mbox{hyperboloids}), and other
figures (e.g., triangles, squares, rectangles, discs,
balls), and many, many more things.

We will talk about those other things later. For the moment,
we just concentrate on {\bf number systems}.
\end{minipage}}}\end{center}
\medbreak

\noindent Our first task is to take a look at the rich and
varied collection of things that populate the mathematical world.
Some of them should be familiar to you, others may be less so or not
at all.  Do not worry if you find in our list things that you have
never heard of before: we will be coming back to the list later, and
discussing all the items in much greater detail.

Mathematicians like to give names to things so that they can talk about
complicated objects without repeating their long descriptions every
time. So let us start by agreeing to use \wmo{} as our name 
for the world of mathematical objects.   

\subsection{Numbers: $\zz_{+}$, $\nn$, $\zz$, $\qq$ and $\rr$.}

\noindent The best known inhabitants of \wmo{} are {\bf numbers}
and {\bf number systems}.  There are several different kinds of
numbers, and they can be organized into {\sl number systems}. A number
can belong to different number systems, in the same way as, say, a
person can belong to different associations.  (For example, somebody
could be a member, say, of the American Association of University
Professors, the Rutgers Alumni Association, and the Sierra Club.
Similarly, the number $3$ belongs to lots of
different number systems, such as, for example, 
$\zz$, $\nn$, $\qq$, $\rr$, $\zz_4$, $\zz_5$, $\zz_6$, 
$\ldots\,$, $\zz_{397}$, $\ldots\,\,$.)

It is convenient to give the number systems
{\bf names}, and to introduce mathematical symbols 
to represent them. Here are some examples:  

\medbreak
\bi
\item[1.]  The set $\nn$ of {\bf natural numbers}. (Natural numbers are
also called {\bf positive integers}, or---sometimes---{\bf whole numbers}.) 
The members of this set are the
integers $1,2,3\ldots$. So, for example,

\begin{itemize}
\item[]  $1$ is a member of $\nn$ (or ``$1$ belongs to $\nn$''), 
\item[]  $2$ is a member of $\nn$ (or ``$2$ belongs to $\nn$''), 
\item[]  $\ldots\ldots$ 
\item[]  $385$ is a member of $\nn$ (or ``$385$ belongs to $\nn$''), 
\item[]  $\ldots\ldots$ 
\item[]  $10,891,032$ is a member of $\nn$ (or ``$10,891,032$ 
belongs to $\nn$''), 
\item[]  $\ldots\ldots$ 
\item[]  $0$ is not a member of $\nn$ (or ``$0$ does not belong to $\nn$''), 
\item[]  $-1$ is not a member of $\nn$ (or ``$-1$ does not belong
to $\nn$''), 
\item[]  $3.5$ is not a member of $\nn$ (or ``$3.5$ does not belong
to $\nn$''), 
\end{itemize}
and so on.

\medbreak
\medbreak\item[2.] The set $\zz$ of all {\bf integers}. The members
of $\zz$ are the natural numbers as well as $0$ and the negatives 
of natural numbers, i.e., the numbers $-1$, $-2$, $-3$, etc. 

\medbreak
\item[3.] The set $\zz_+$ of all {\bf nonnegative 
integers}.
The members of this set are exactly the same as the members 
of $\nn$, except only for the fact that $0$, which is not a member
of $\nn$, counts as a member of $\zz_+$.
\end{itemize}

\noindent
\fbox{\begin{minipage}{12cm}
\boxon{THE SYMBOL $\quad$ ``$\in$'' $\quad$ (``BELONGS TO'')}

\noindent We use the symbol $\in$ (read ``belongs to,'' or ``is a member 
of,'' or ``belonging to,'' or ``in'') to indicate {\em membership in a set}:

\medbreak
\begin{center}
\fbox{\begin{minipage}{10.5cm}
If $A$ is a set and $x$ is an object, then 
``$x\in A$'' means ``$x$ belongs to $A$,'' or ``$x$ is a member of $A$.''
\end{minipage}}\end{center}

\medbreak\noindent The symbol 
$\notin$ (read ``does not belong to,'' or ``is not a member of,''
or ``not belonging to,'' or ``not in'') indicates non-membership: 

\medbreak
\begin{center}
\fbox{\begin{minipage}{10.5cm}
If $A$ is a set and $x$ is an object, then 
``$x\notin A$'' means ``$x$ does not belong to $A$,'' or 
``$x$ is not a member of $A$.''
\end{minipage}}\end{center}

\medbreak\noindent For example the following statements are true:
$$
\left.\begin{array}{rrrr} 
1\in\nn\,,\phantom{_{+}}&
2\in\nn\,,\phantom{_{+}}&
385\in\nn\,,\phantom{_{+}}&
10,891,032\in\nn\,,\phantom{_{+}}\\
0\notin\nn\,,\phantom{_{+}}&
-1\notin\nn\,,\phantom{_{+}}&
3.5\notin\nn\,,\phantom{_{+}}&
\pi\notin\nn\,,\phantom{_{+}}\\
\frac{8}{5}\notin\nn\,,\phantom{_{+}}&
\frac{10}{5}\in\nn\,,\phantom{_{+}}&
\frac{382,403,313}{9}\in\nn\,,\phantom{_{+}}&
\pi+7\notin\nn\,,\phantom{_{+}}\\
3+7\in\nn\,,\phantom{_{+}}&
8-1\in\nn\,,\phantom{_{+}}&
1-8\notin\nn\,,\phantom{_{+}}&
\pi-(2-\pi)\notin\nn\,,\phantom{_{+}}\\
1\in\zz\,,\phantom{_{+}}& 2\in\zz\,,\phantom{_{+}}&
385\in\zz\,,\phantom{_{+}}&
10,891,032\in\zz\,,\phantom{_{+}}\\
0\in\zz\,,\phantom{_{+}}&
-1\in\zz\,,\phantom{_{+}}&
3.5\notin\zz\,,\phantom{_{+}}&
\pi\notin\zz\,,\phantom{_{+}}\\
\frac{8}{5}\notin\zz\,,\phantom{_{+}}&
\frac{10}{5}\in\zz\,,\phantom{_{+}}&
\frac{382,403,313}{9}\in\zz\,,\phantom{_{+}}&
\pi+7\notin\zz\,,\phantom{_{+}}\\
3+7\in\zz\,,\phantom{_{+}}&
8-1\in\zz\,,\phantom{_{+}}&
1-8\in\zz\,,\phantom{_{+}}&
\pi^2-(2-\pi)^2-4\pi\in\zz\,,\phantom{_{+}}\\
0\in\zz_+\,,&
1\in\zz_{+}\,,&
4\in\zz_{+}\,,&
-3\notin\zz_{+}\,.
\end{array}\right.
$$
\end{minipage}}
\medbreak


\begin{itemize}
\item[4.] The set $\rr$ of all {\bf real numbers}. Every 
natural number is a member of $\rr$. Every integer is a member of $\rr$. 
In addition, there are many other real numbers besides the integers.  

For example the following statements are true: 
\end{itemize}
$$
\begin{array}{rrrr} 
\qquad 1\in\rr\,,\quad\quad&
2\in\rr\,,\quad\quad&
385\in\rr\,, \quad\quad&
10,891,032\in\rr\,,\\
\qquad 0\in\rr\,,\quad\quad&
-1\in\rr\,,\quad\quad&
3.5\in\rr\,,\quad\quad&
\pi\in\rr\,,\\
\qquad\frac{8}{5}\in\rr\,,\quad\quad&
\frac{10}{5}\in\rr\,,\quad\quad&
\frac{382,403,313}{9}\in\rr\,,\quad\quad&
\pi+7\in\rr\,,\\
\qquad 3+7\in\rr\,,\quad\quad&
8-1\in\rr\,,\quad\quad&
1-8\in\rr\,,\quad\quad&
\pi-(2-\pi)\in\rr\,.
\end{array}
$$

\begin{itemize}
\item[] A real number (and, in particular, an integer) is 
``positive'' if it is $>0$.   A number is  ``nonnegative'' if it is $\ge 0$.  

In particular, {\bf $0$ is not positive}, but {\bf $0$ is nonnegative.} 
{\bf $0$ is a nonnegative integer, but $0$ is not a positive integer.}
\end{itemize}

\begin{itemize}
\item[5.] The set $\qq$ of all {\bf rational numbers}. (Rational numbers
are often called ``fractions''). By \mbox{definition},
a {\sl rational number} is a real number that can be expressed as the quotient 
(or ``ratio\footnote{``Ratio'' is an old word for ``quotient.''}) of two
integers. So, for example, $\frac{8}{3}\in\qq$,  
$6\in\qq$, $\frac{9}{3}\in\qq$, $-6\in\qq$, $-\frac{9}{3}\in\qq$. 
\ei


\begin{remark}
Do not confuse the symbols $\nn$, $\zz$, $\rr$ with the letters
$N$, $Z$, $R$. They are different! (This is very convenient.
If we agreed to use $N$ as the name for the set of natural numbers,
then the meaning of the letter $N$ would be fixed for ever, and we
would not be able to use $N$ for other things.) \proofend
\end{remark}

\medbreak\noindent Another way of saying that 
``$\nn$ is the set consisting of $1$, $2$, $3$, $4$, and so on,''
is to write 
\begin{equation}\label{equ-1}
\nn=\{\,1,2,3,4,\ldots\,\}\,.
\end{equation}
Similarly, 
$$
\zz=\{\,0,1,-1,2,-2,3,-3,4,-4,\ldots\,\}\,.
$$

\subsection{The complex numbers}

\noindent For a long time, people used the word ``number'' to mean
``real number.''  But then the ``complex numbers'' were invented, and
mathematicians decided, after long discussions on whether this was a
good thing to do, that these ``complex numbers'' were objects worth
admitting as {\sl bona fide} numbers as well. 
So we can add a sixth example to our list of number systems:

\medbreak
\bi
\item[6.] The set $\cc$ of all {\sl complex numbers}. 
Here are some examples of these ``numbers'': 
$$
3\,,\qquad i\,,\qquad 3+i\,,\qquad 4-i\,,\qquad 35-67i\,,\qquad
\frac{8}{5}\,,\qquad \frac{8}{5}-\pi i\,.
$$
Complex numbers can be {\bf added} and {\bf multiplied}. To add two complex
numbers, you just add separately the ``$i$-parts'' and the
``non-$i$-parts.'' So, for example, 
$$
3+i+(4-2i)=7-i\,,\,\,\,3+(8+5i)=11+5i\,,\,\,\,\frac{8}{5}+i+
\Big(\frac{12}{5}+83i\Big)=4+84i\,.
$$
To multiply two complex numbers, you just multiply as you would real
numbers, except that every time you run into $i.i$ you put $-1$ instead. 
For example:  

\begin{eqnarray*}
(23 +7i).(4+2i)&=&23.4+7.4.i+23.2.i+7.2.i.i\\
&=&92+28i+46i+14.i.i\\
&=&92-14+28i+46i\\
&=&78+72i\,.
\end{eqnarray*}

\begin{remark}
{\rm The invention, or discovery, of the
complex numbers was a long process, driven by the effort to prove the
``fundamental theorem of algebra.'' This theorem ought to say that an
$n$-th order equation has $n$ solutions, so for example the equation
$x^2+1=0$ should have two solutions, the equation $x^7+3x^5+4x^2+5=0$
should have seven solutions, and so on.  But the only way for $x^2+1=0$ to
have two solutions is to stipulate that there is a number $x$ such that
$x^2=-1$, that is, that a ``square root of $-1$'' exists.  This, however,
leads to all kinds of difficulties, if you think that this square
root---let us call it $i$---is an ordinary real number. For example, every
real number is either positive or negative or zero. Now, $i$ of course
cannot be zero, so $i$ is either positive or negative. Since the product of
two positive numbers is positive, if $i$ was positive then $i^2$ would have
to be positive.  Since the product of two negative numbers is positive, if
$i$ was negative then $i^2$ would have to be positive. So in either
case, whether $i$ is positive or negative, the number $i^2$ would have to be
positive. But $i^2=-1$, so $-1$ would have to be positive, and this is not
possible. The way out of this was to think of $i$ as not being an ordinary
number, really, but only a symbol to be manipulated according to certain
rules. So, for example, following what {\bf Descartes} wrote in 1637, one
can ``imagine'' that the equation $x^2+1=0$ has two roots ($i$ and
$-i$) but these imagined roots are not ``real'' numbers.  Earlier, in
1572, {\bf Bombelli} had given precise rules for the manipulation of these
imaginary numbers. {\bf Argand} in 1814 had the idea of representing
complex numbers as points in a two-dimensional plane. The term ``complex
number'' was actually introduced by {\bf Gauss} in 1831. The term
``conjugate'' had been introduced by {\bf Cauchy} in 1821.}\proofend
\end{remark}

\noindent\fbox{\begin{minipage}{11cm}

\boxon{THE SYMBOL $\quad$ ``$\subseteq$'' $\quad$ (``IS A SUBSET 
OF'')\label{subsetbox}}

\noindent We use the symbol $\subseteq$ (read ``is a subset of,''
or ``is contained in'') to indicate
{\em set inclusion}, and the symbol $\not\subseteq$ (read ``is not 
a subset of'') to indicate that the inclusion does not hold.

If $A$ and $B$ are sets, the expression ``$A\subseteq B$ 
(read ``$A$ is a subset of $B$'') means that {\em every member
of $A$ belongs to $B$}. For example: if $A$ is the set of 
all U.S. Senators, and $B$ is the set of all citizens of
the U.S., then ``$A\subseteq B$'' and ``$B\not\subseteq A$''
are true statements.

The following statements about $\zz_{>0}$, $\nn$,
$\zz$, $\qq$, $\rr$,  are true:
$$
\left.\begin{array}{rrrr} 
\zz_+\subseteq\zz_+\,,& 
\qquad\zz_+\not\subseteq\nn\,,\phantom{_{+}}&
\qquad\nn\subseteq\zz_+\,,&
\qquad\nn\subseteq\nn\,,\phantom{_{+}}\\
\nn\subseteq\zz\,,\phantom{_{+}}&
\zz\not\subseteq\nn\,,\phantom{_{+}}&
\zz\subseteq\zz\,,\phantom{_{+}}&
\zz_{+}\subseteq\zz\,,\phantom{_{+}}\\
\zz\subseteq\qq\,,\phantom{_{+}}&
\qq\not\subseteq\zz\,,\phantom{_{+}}&
\zz\subseteq\rr\,,\phantom{_{+}}&
\nn\subseteq\qq\,,\phantom{_{+}}\\
\qq\not\subseteq\nn\,,\phantom{_{+}}&
\qq\not\subseteq\zz\,,\phantom{_{+}}&
\rr\not\subseteq\qq\,,\phantom{_{+}}&
\qq\subseteq\rr\,.\phantom{_{+}}
\end{array}\right.
$$
\end{minipage}}
\medbreak

\begin{center}
\fbox{\begin{minipage}{11cm}
\null
\boxon{NOT ALL REAL NUMBERS ARE RATIONAL\label{notratbox}}

\noindent In Box No. \ref{subsetbox} we asserted that
$\rr\not\subseteq\qq$.   How do we know that?

For ``$\rr\subseteq\qq$'' to be true, it
would have to be the case that every real number is rational.
Yet, this is {\em not} the case, because there are real 
numbers that are not rational. A famous example
is $\sqrt{2}$. We will prove later that

\medbreak
\begin{center}
\fbox{The number $\sqrt{2}$ is not rational.}
\end{center}

Actually, many other real numbers are not rational,
although in lots of interesting cases this is very hard to prove. 
For example, the numbers $\pi$ and $e$ are not rational. 
\end{minipage}}\end{center}
\medbreak

\subsection{The integers modulo $n$}

Here are some more examples of important ``number systems'' of a rather
different kind. These systems will be very important to us later.

\bi
\item[7.a.] The set $\zz_{12}$ (read this as ``z-twelve'') 
of all {\sl integers modulo $12$}.
The members of $\zz_{12}$ are the following $12$ integers:
$$
0\,,\,\,1\,,\,\,2\,,\,\,3\,,\,\,4\,,\,\,5\,,\,\,6\,,\,\,7\,,\,\,
8\,,\,\,9\,,\,\,10\,,\,\,11\,.
$$
Addition and multiplication are carried out in the usual way, except that
any time you get $12$ or something larger, you must ``reduce modulo
$12$,'' that is, subtract $12$ and keep subtracting $12$ until you get a
member of $\zz_{12}$. For example, the following equalities are true 
in $\zz_{12}$:\label{inz12}
\ei
$$
\begin{array}{rrrrr}
\!\!3+5=8\,,\quad&6+9=3\,,\quad&8+9=5\,,\quad&7+1=8\,,\quad&11+11=10\,,\\
4.2=8\,,\quad&6.9=6\,,\quad&8.7=8\,,\quad&7.11=5\,,\quad&7.7=1\,\,\,\,.
\end{array}
$$

\begin{remark} {\rm The numbers modulo $12$ are very natural objects, 
and you are very much used to working with them.  For example, 
suppose you start at $10$ o'clock and work $8$ hours.  What time is it when
you finish?  The answer is ``$10+8$ modulo $12$,'' that is, 
$6$ o'clock. If a movie theatre starts its show
at $8$ o'clock and shows a 2-hour movie $8$ times, without intermission, 
what time is it at the end of the show?  The answer is 
``$10+2.8$ modulo $12$,'' that is, $2$ o'clock.   

Naturally, if you wanted to keep track of ``am'' vs. ``pm'' then you would
want to use so-called ``military time,'' that is, work in $\zz_{24}$, the
numbers modulo $24$. (In $\zz_{24}$,   $16+13=5$, and 
$8.5=16$, and so on.)}\proofend
\end{remark}

\bi\item[7.b.] The set $\zz_{7}$ of all {\sl integers modulo $7$}.


The members of $\zz_{7}$ are the following $7$ numbers:
$$
0\,,\,\,1\,,\,\,2\,,\,\,3\,,\,\,4\,,\,\,5\,,\,\,6\,.
$$
Addition and multiplication are carried out in the usual way, except that
any time you get $7$ or something larger, you must ``reduce modulo
$7$,'' that is, subtract $7$ and keep subtracting $7$ until you get a
member of $\zz_{7}$. For example, the following equalities are true in 
$\zz_7$:
\ei
$$
\begin{array}{rrrrr}
3+2=5\,,\quad&6+5=11\,,\quad&4+3=0\,,\quad&2+6=1\,,\quad&4+2=6\,,\\
4.2=1\,,\quad&6.5=2\,,\quad&3.4=5\,,\quad&3.3=2\,,\quad&3.5=1\,\,\,\,.
\end{array}
$$
\begin{remark}{\rm  \ Like the members of $\zz_{12}$, the numbers modulo $7$ 
are also very natural objects that you are very much used to working with.
For example,  let us think of the days of the week as numbered, 
starting with Sunday as Day No. 0.  In other words, think of 

\bigbreak
$\qquad$\begin{minipage}{4cm}
Sunday\\
Monday\\
Tuesday\\
Wednesday\\
Thursday\\
Friday\\
Saturday
\end{minipage}
\begin{minipage}{4cm}
as day No.\\
as day No.\\
as day No.\\
as day No.\\
as day No.\\
as day No.\\
as day No.
\end{minipage}
\begin{minipage}{2cm}
0\\
1\\
2\\
3\\
4\\
5\\
6
\end{minipage}

\bigbreak
Suppose today is Thursday. What day of the week is it going to be 
$5$ days from today? The answer is {\bf\em Tuesday}, of course. Why? 
Because Thursday is Day 4, so $5$ days from today is Day $4+5$,
that is, Day 2, modulo $7$.  

Suppose today, January 5, is a Wednesday. Whay day of the week 
is February 11 going to be?  ANSWER: today is Day 3; to go from 
January 5 to February 11 you must add 36 days (that is, 31 days to go to
February 5, and 6 more to go to February 11), so you'll end up in Day 
3+36=39, which is equal to 4 modulo $7$. So February 11 is going to be a 
Day 4, that is, {\bf\em a Thursday}.

Suppose today, January 5, is a Wednesday. Whay day of the week 
is July 5 going to be, assuming that we are NOT in a leap year?
ANSWER: today is Day 3; to go from January 5 to July 5 we must add
31 days (to get to February 5), then 28 days 
(to get to March 5, using the fact that we are not in a leap year), 
then 31 days (to get to April  5), 
then 30 days (to get to May 5), 
then 31 days (to get to June 5), and, finally, 
30 days (to get to July 5). So we end up in Day 
$3+31+28+30+31+30$ which, modulo $7$, is the same as
$3+3+0+2+3+2$, that is, $13$, that is, $6$ (modulo $7$). 
So {\bf\em July 5 is going to be a Saturday}. }\proofend
\end{remark}

\medbreak\bi
\item[7.c.] Naturally, one could consider integers modulo any other 
number, not just $12$ or $24$.  For example, $\zz_3$ is the set of 
{\bf integers modulo $3$}.  So $\zz_3$ has exactly three members, 
namely, 
$$
0\,,\qquad\qquad 1\,,\qquad\qquad{\rm and}\qquad 2\,.
$$
Arithmetic modulo $3$ is very simple. The addition table is as follows:  
\begin{eqnarray*}
0+0&=&0\,,\\
0+1&=&1+0=1\,,\\
0+2&=&2+0=2\,,\\
1+1&=&2\,,\\
1+2&=&2+1=0\,,\\
2+2&=&1\,,
\end{eqnarray*}
\item[] and the multiplication table is
\begin{eqnarray*}
0.0&=&0.1=1.0=0.2=2.0=0\,,\\
1.1&=&2.2=1\,,\\ 
1.2&=&2.1=2\,.
\end{eqnarray*}
\ei

\begin{remark}
To say that an equality holds in $\zz_n$, we can say just that,
or we can say that the equality holds (or is true) ``modulo $n$.''

For example, we can say that
$$
5+8=2\qquad{\rm in}\quad\zz_{11}\,,
$$
or that
$$
5+8=2\qquad{\rm modulo}\quad 11\,.
$$
Also, if it is clear from the context that we are working in $\zz_n$, then
there is no need to say ``in $\zz_n$'' or ``modulo $n$.'' (Example:
after we said in Page \pageref{inz12} that ``the following equalities
are true in $\zz_{12}$,'' we then just went ahead and wrote things
like $8.7=8$, without explaining again that this was meant to be
``in $\zz_{12}$,'' or ``modulo $12$.''  There was no need, because 
we had already stated clearly that we were working in $\zz_{12}$.\proofend
\end{remark}


\begin{remark}
If you look at the way we defined the integers modulo $n$, you may think 
that the integers that are larger than $n$ do not belong to $\zz_n$.
(For example, the integer $8$ does not belong to $\zz_7$,
because $\zz_7$ consists of the integers $0$, $1$, $2$, $3$, 
$4$, $5$, and $6$, and nothing else.) We can, however, assign a value
to a symbolic expression such as $8$ in $\zz_7$, by letting
$$
8=1+1+1+1+1+1+1+1\,,
$$
where the sum is understood to be in $\zz_7$. Then, in $\zz_7$, the expression
``$8$'' denotes the number $1$, and we can write
$$
8=1\qquad{\rm in}\qquad\zz_7\,.
$$
Notice that 
$$
8\neq 1\qquad{\rm in}\qquad\zz\,.
$$
So the symbol $8$, when we are working in 
$\zz_7$,  is the name of a thing which is not the integer
$8$. (If it was the same, then ``$8\neq 1$'' would also be true
in $\zz_7$.)

One can also talk about {\em negatives} of integers modulo $n$. 
What do we mean by ``the negative'' of an integer modulo 
$n$? We just mean ``the integer modulo $n$ that added in $\zz_n$ 
to our given integer modulo $n$ yields $0$.''

For example, what is the negative of $3$ in $\zz_7$? Well, in $\zz_7$ 
the sum of $3$ and $4$ is $0$. Therefore 
{\bf the negative of $3$ in $\zz_7$ is $4$}. We can also write 
this using the symbol ``$-$'':
$$
-3=4\qquad{\rm in}\qquad\zz_7\,,
$$
or 
$$
-3=4\qquad{\rm modulo }\quad 7\,.
$$

\proofend
\end{remark}

An important application of arithmetic modulo $3$ is that it enables us
to find a simple criterion to check if a given integer is divisible
by $3$. The key fact here is that $10=1$ modulo $3$. This means that
if we are working modulo $3$ then a big number such as $4,687$ 
is just the same (in $\zz_3$) as $4+6+8+7$, i.e. as $25$, 
which is equal to $1$ in $\zz_3$. So $4,687$ is not divisible by $3$.

\medbreak
\noindent\fbox{

\begin{minipage}{4.2in}

\boxonnop{When is a natural number divisible by $3$?}

\vspace{1mm}
\noindent Take some natural number like $16,547$. Is it divisible by $3$? 
Naturally, you can do the division and see what happens. But
can you tell quickly and directly?  

When we say that a number is ``divisible by $3$'' what we mean
is that if you start subtracting $3$s  from it until you cannot do that any
more (that is, until you get to $0$, $1$ or $2$), then you end up with
$0$.  In other words, {\bf\em our number is equal to $0$ modulo $3$.} 

So in our example we need to find out whether  
$16,547$ is equal to $0$, $1$ or $2$ modulo $3$. Now:
$$
16,547=1\times 10,000+6\times1,000+5\times 100+4\times 10+7\times 1\,.
$$
Also, $10$ is just equal to $1$ modulo $3$. So $100=10\times 10=1\times
1=1$
modulo $3$. And $1,000=10\times 10\times 10=1\times 1\times 1=1$ modulo
$3$. And so on. Therefore, the following is true modulo $3$:
$$
16,547=1+6+5+4+7=23=2\,.
$$
So {\bf\em $16,547$ is not divisible by $3$.}

The point of the above argument is that

$$
\left.\begin{array}{rclc}
10,000&=&1&{\rm modulo}\,\,\,3\,,\\ 
6,000&=&6&{\rm modulo}\,\,\,3\,,\\
500&=&5&{\rm modulo}\,\,\,3\,,\\
40&=&4&{\rm modulo}\,\,\,3\,,\\
7&=&7&{\rm modulo}\,\,\,3\,,
\end{array}\right.
$$
so 
$$
16,547=1+6+5+4+7=23=2\,\qquad{\rm modulo}\,\,3.
$$
\vspace{1mm}
The general criterion is this: {\bf to find out if an integer
is divisble by $3$, just add the digits in its decimal 
expression. If the sum of the digits is divisible by 
$3$, then the given number is divisible by $3$. Otherwise it is not.}
\end{minipage}
}
 
\medbreak\noindent{\bf QUESTION \quesno}. Which of the following three natural
numbers are divisible by $3$? 
$$
\begin{array}{r}
999,111,999,111\\
543,043,664,987\\
6,915,571,375,016,461,671,444,102,534
\end{array}
$$


\medbreak
\noindent\fbox{

\begin{minipage}{4.2in}

\boxonnop{When is a natural number divisible by $11$?}

\vspace{1mm}
\noindent The situation here is very similar to what we found
when we looked at divisibility by $3$, but with an important 
twist. Now $10$ is not equal to $1$ modulo $11$, but to $-1$. 
So if we look at a natural number like $16,547$, the $7$ should
count as a $7$ modulo $11$, the $4$ as a $-4$, 
the $5$ as $5$, the $6$ as a $-6$, and the $1$ as a $1$.

(Reason: $16,547=1\times 10,000+6\times 1,000+5\times 100
+4\times 10+7$. Modulo $11$, we have 
$10=-1$, $100=10^2=1$, $1,000=-1$, $10,000=1$.
So, $16,547=1-6+5-4+7$ modulo $11$.) Therefore 
$16,547=3$ modulo $11$, and then 
{\bf\em $16,547$ is not divisible by $11$.}

\vspace{1mm}
The general criterion is this: {\bf\em to find out if an integer
is divisble by $11$, just add the digits in its decimal 
expression with alternating signs (plus, minus, plus, minus, and so on). 
If the alternating sum of the digits is divisible by 
$11$, then the given number is divisible by $11$. Otherwise it is not.}
\end{minipage}
} 
 
\medbreak\noindent{\bf QUESTION \quesno}. Which of the three natural
numbers of the previous question are divisible by $11$? 

\medbreak\noindent{\bf QUESTION \quesno}. Formulate and justify
a criterion for testing if a number is divisible by $9$.

\medbreak\noindent{\bf QUESTION \quesno}. Is the following true or false?
If you take a natural number, reverse the order of its decimal figures,
and subtract the smaller of the numbers you got from the larger, then what you
get is divisible by $3$.  (Here are two examples: 
(1)~start with $925$, then get $529$ by reversing the order of the figures,
then subtract, and you get $925-529$, which is equal to \fbox{$396$};
(2)~start with $386$, then get $683$ by reversing the order of the figures,
then subtract, and you get $683-386$, which is equal to \fbox{$297$}. In
both cases, we got numbers that are divisible by $3$. The question for you
is: is this going to to happen no matter which natural number we take?)

\medbreak\noindent{\bf QUESTION \quesno}. Is the following true or false?
If you take a natural number, reverse the order of its decimal figures,
and subtract the smaller of the numbers you got from the larger, then what you
get is divisible by $11$.  (Here are two examples: 
(1)~start with $925$, then get $529$ by reversing the order of the figures,
then subtract, and you get $925-529$, which is equal to \fbox{$396$};
(2)~start with $386$, then get $683$ by reversing the order of the figures,
then subtract, and you get $683-386$, which is equal to \fbox{$297$}. In
both cases, we got numbers that are divisible by $11$. The question for you
is: is this going to to happen no matter which natural number we take?)

\medbreak\noindent{\bf QUESTION \quesno}. Is the following true or false?
If you take a natural number, reverse the order of its decimal figures,
and add the two numbers, then what you
get is divisible by $3$.  (Here are two examples: 
(1)~start with $321$, then get $123$ by reversing the order of the figures,
then add, and you get $321+123$, which is equal to \fbox{$444$};
(2)~start with $801$, then get $108$ by reversing the order of the figures,
then add, and you get $801+108$, which is equal to \fbox{$909$}. In
both cases, we got numbers that are divisible by $3$. The question for you
is: is this going to to happen no matter which natural number we take?)

\medbreak\item[7.d.] Even simpler is $\zz_2$, the set of 
{\bf integers modulo $2$}.  Clearly, $\zz_2$ has exactly two members, 
namely, 
$$
0\qquad\qquad\qquad\qquad{\rm and}\qquad\qquad\qquad 1\,.
$$
Arithmetic modulo $2$ is so ridiculously simple that we can fit the 
complete addition and multiplication tables together in one line: 
$$
0+0=0\,,\,\,0+1=1+0=1\,,\,\,1+1=0\,,\,\,0.0=0.1=1.0=0\,,\,\,1.1=1\,.
$$
\ei

\medbreak\noindent{\bf QUESTION \quesno}. Does the equation $x^2+1=0$ have
a solution  in $\zz_7$?  

\medbreak\noindent{\bf ANSWER}. Let us write down explicitly the squares of 
all members of $\zz_7$:
$$
0^2=0\,,\,\,\,1^2=1\,,\,\,\,2^2=4\,,\,\,\,3^2=2\,,\,\,\,
4^2=2\,,\,\,\,5^2=4\,,\,\,\,6^2=1\,.
$$
The only way for $x^2+1$ to be equal to zero would be for $x^2$ to be equal
to $6$. But in the above list we see clearly that there is no member of
$\zz_7$ whose square is $6$.  So the answer is {\bf NO}. 

\medbreak\noindent{\bf QUESTION \quesno}. Does the equation $x^2+1=0$
have a solution  in $\zz_{17}$?  

\medbreak\noindent{\bf ANSWER}. In $\zz_{17}$, $4.4=16$, so 
$4.4+1=16+1=0$. So the answer is {\bf YES}. 

\medbreak\noindent{\bf QUESTION \quesno}. Does the equation $x^2+1=0$
have a solution  in $\zz_{101}$?  

\medbreak\noindent{\bf ANSWER}. In $\zz_{101}$, $10^2+1=101=0$, so 
the answer is {\bf YES}. 

\medbreak\noindent{\bf QUESTION \quesno}. Does the equation $x^2+1=0$
have a solution  in $\zz_{103}$?  

\medbreak\noindent{\bf COMMENT}. This is not easy. 
It turns out that the answer is ``no.'' Of course, you could 
verify this by listing 
the squares of all $103$ members of $\zz_{103}$ 
and checking that you never get $102$ (which is $-1$ in $\zz_{103}$).
But this would be a lot of work. There ought to be a smarter 
way. Maybe we can {\bf prove} that the answer is ``no'' 
without having to do lengthy boring computations? 
Try to think about it. We will come back to this problem later.

\medbreak\noindent{\bf QUESTION \quesno}. In $\zz_{24}$, are there two
nonzero numbers  whose product is equal to zero?

\medbreak\noindent{\bf ANSWER}. In $\zz_{24}$, $8.3=0$. 
So the answer is {\bf YES}. 

\medbreak\noindent{\bf QUESTION \quesno}.  In $\zz_{4}$, does the equations 
$a^2+b^2=3$ have a solution?

\medbreak\noindent{\bf ANSWER}.  Here are the values
of $a^2+b^2$ for all possible choices of $a,b\in\zz_4$,
\begin{eqnarray*}
0^2+0^2&=&0\,,\\
0^2+1^2=1^2+0^2&=&1\,,\\
0^2+2^2=2^2+0^2&=&1\,,\\
1^2+1^2&=&2\,,\\
1^2+2^2=2^2+1^2&=&2\,,\\
2^2+2^2=&=&0\,.
\end{eqnarray*}
We see that the right-hand side is never equal to $3$.
So the answer is {\bf NO}. 

\medbreak\noindent{\bf QUESTION \quesno}. In $\zz_{17}$, are there two
nonzero numbers  whose product is equal to zero?

\medbreak\noindent{\bf ANSWER}. If you multiply two numbers in $\zz_{17}$, the only
way the product can be equal to zero in $\zz_{17}$ is if, when you reduce 
modulo $17$, you get zero. That is, the product of these two numbers must 
be divisible by $17$. But $17$ is a prime number, so there is no way 
to multiply two nonzero numbers smaller than $17$ and get a product which
is divisible by $17$.   So the answer is {\bf NO}. 

\bigbreak
\begin{remark} {\rm Clearly, the class of number systems
presented in this subsection consists of {\sl infinitely many} ``number
systems,'' namely, the systems }
$$
\qquad\qquad\qquad
\zz_2,\quad\zz_3,\quad\zz_4,\quad\zz_5,\ldots,\quad\zz_{397}\,,\ldots\,,
{\rm \quad and\,\,\,so\,\,\,on}\,.\qquad\qquad
\diamondsuit
$$
\end{remark}

\subsection{The quaternions}\label{sec-quat}

\noindent In 1843 {\bf W. R. Hamilton} (1805-1865)
invented the {\bf quaternions}, which we include as the eighth
item in our list.

\bi
\item[8.] The set $\kk$ of all {\bf quaternions}.  Quaternions are
like complex numbers, but with four components rather than two.
So for example 
$3+4i-7j+25k$ and $-1+2i+9j+k$ are quaternions.  To multiply quaternions,
you just multiply as you would ordinary real or complex numbers, except
that when you run into products such as $i.j$ or $k.i$ you rewrite them
using the following ``multiplication table'':  
\begin{eqnarray*}
&\,&i.j=k\,,\qquad\qquad\qquad\qquad\qquad j.i=-k\,,\\
&\,&j.k=i\,,\qquad\qquad\qquad\qquad\qquad k.j=-i\,,\\
&\,&k.i=j\,,\qquad\qquad\qquad\qquad\qquad i.k=-j\,,\\
&\,&i.i=j.j\,=k.k=-1\,.
\end{eqnarray*}
For example, suppose that we want to multiply
$$
(3+4i-7j+25k).(-1+2i+9j+k)\,.
$$ 
Here is the calculation in full:
\ei
\begin{eqnarray*}
\!\!\!&\!\!&\!\!\!\!\!\!\!\!\!\!(3+4i-7j+25k).(-1+2i+9j+k)\\
\!\!\!&\!\!&=3.(-1)+3.2.i+3.9.j+3.k+4.(-1).i+4.2.i.i+4.9.i.j+4.i.k\\
&\,&\,\,\,\,\,\,\,+(-7).(-1).j+(-7).2.j.i+(-7).9.j.j+(-7).j.k+25.(-1).k\\
&\,&\,\,\,\,\,\,\,+25.2.k.i+25.9.k.j+25.k.k\\
\!\!\!&\!\!&=-3+6i+27j+3k-4i+8i.i+36i.j+4i.k+7j\\
\!\!\!&\!\!&\,\,\,\,\,\,-14j.i-63j.j-7j.k-25k+50k.i+225k.j+25k.k\\
\!\!\!&\!\!&=-3+6i+27j+3k-4i-8+36k-4j+7j\\
\!\!\!&\!\!&\,\,\,\,\,\,\,+14k+63-7i-25k+50j-225i-25\\
\!\!\!&\!\!&=(\!-\!3\!-\!8\!+\!63\!-\!25)\!+\!(6\!-\!4\!-\!7\!-\!225)i
\!+\!(27\!-\!4\!+\!7\!+\!50)j\\
\!\!\!&\!\!&\,\,\,\,\,\,\,+(3\!+\!36\!+\!14\!-\!25)k\\
\!\!\!&\!\!&=\,\,\,\,27-230i+80j+28k\,.
\end{eqnarray*}

\begin{center}
\fbox{
\begin{minipage}{5.1in}
\boxon{Hamilton discovers the quaternions}
\vspace{1mm}\noindent Hamilton was knighted in 1835 and that year his 
second son, Archibald Henry, was born. But the next few years did not 
bring him much
happiness. After the discovery of algebraic couples, he tried to extend the
theory to triplets, and this became an obsession that plagued him for many
years. The following autumn he went to Bristol for a meeting of the British
Association, and Helen took the children with her to Bayly Farm for ten
months. His cousin Arthur died, and not long after Helen returned from her
mother's she went away again to England this time leaving the children
behind after the birth of a daughter, Helen Eliza Amelia. At this point,
William became depressed and started to have problems with alcohol so his
sister came back to live at Dunsink.

Helen returned in 1842 when Hamilton was so preoccupied with the triplets
that even his children were aware of it. Every morning they would inquire:

\begin{center}
       {\sl Well, Papa can you multiply triplets?}
\end{center}

but he had to admit that he could still only add and subtract them.

On 16 October 1843 (a Monday) Hamilton was walking in along the Royal Canal
with his wife to preside at a Council meeting of the Royal Irish
Academy. Although his wife talked to him now and again Hamilton hardly
heard, for the discovery of the quaternions, the first noncommutative
algebra to be studied, was taking shape in his mind:

\begin{center}

\begin{minipage}{4.4in}


{\it And here there dawned on me the notion that we must admit, in some
sense, a fourth dimension of space for the purpose of calculating with
triples ... An electric circuit seemed to close, and a spark flashed forth.}


\end{minipage}

\end{center} 

\noindent He could not resist the impulse to carve the formulae for the
quaternions  

$$
i = j = k = i j k = -1
$$

in the stone of Brougham Bridge as he and his wife passed it. Hamilton felt
this discovery would revolutionise mathematical physics and he spent the
rest of his life working on quaternions. He wrote:-

\begin{center}

\begin{minipage}{4.4in}


{\it I still must assert that this discovery appears to me to be as
important for the middle of the nineteenth century as the discovery of
fluxions [the calculus] was for the close of the seventeenth.}

\end{minipage}

\end{center} 

\vspace{1mm}

\end{minipage}

}

\end{center} 

\newpage\section{Reading formulas precisely} 

It is important that you acquire
from the very beginning the habit of reading mathematical formulas aloud
precisely and  correctly.  For example, the correct way to read
the formula 
\begin{equation}\label{equ-2}
3\in\nn
\end{equation}
aloud is

\begin{center}
$3$ belongs to $\nn$
\end{center}
or
\begin{center}
$3$ is a member of $\nn$
\end{center}
or 
\begin{center}
$3$ is in  $\nn$
\end{center}
or 
\begin{center}
$3$ is a natural number.
\end{center}
But you {\bf cannot} read (\ref{equ-2}) as, for example,
``$3$ is contained in $\nn$.'' ``Is contained'' means something else!

Here is another example: Formula (\ref{equ-1}) is read as
\begin{center}
$\nn$ is the set consisting of $1$, $2$, $3$, $4$, etc.
\end{center}
It would be {\bf wrong} ({\bf very} wrong!) to read it as ``$\nn$ is 
$1$, $2$, $3$, $4$, etc.''  Do you see why this is bad?
The best way to understand it is to think of more familiar examples.

For example, you cannot say that ``the U.S. Supreme Court 
is Chief Justice Rehnquist and Justice Scalia and Justice Thomas
and Justice O'Connor and Justice Souter and Justice Kennedy
and Justice Ginsburg and Justice Breyer and Justice Stevens.''
Why not? Because the U.S. Supreme Court is {\bf one single thing,
not nine things!} What you {\em can} say is:

\begin{center}
\begin{minipage}{10cm} The U.S. Supreme Court is a court whose members are
Chief \mbox{Justice} Rehnquist and Justices Scalia, Thomas,
O'Connor, Souter, Kennedy, Ginsburg, Breyer, and Stevens.
\end{minipage}
\end{center}
Similarly, the set $\nn$ of natural numbers is {\bf one single object,
not many objects}, so you can say 
``$\nn$ is the set consisting of $1$, $2$, $3$, $4$, etc.''
but you cannot say ``$\nn$ is $1$, $2$, $3$, $4$, etc.'' 

Also, you cannot say ``$\nn$ is the natural numbers,'' or ``$\nn$ is
natural numbers.'' (Actually, ``$\nn$ is the natural numbers'' is
bad, but ``$\nn$ is natural numbers'' is even worse. Why?)  What you should
say is ``$\nn$ is the set of natural numbers,'' or ``$\nn$ is the set
of all natural numbers.''\footnote{Recommended reading: the short story 
``Pigs is pigs,'' by Ellis Parker Butler. You can find it in the Web site
{\em www.bookvalley.com/cgibin/bv?b=162}. It's funny, and one of the things
that make it funny is precisely the misuse of the verb ``is'' in
``Pigs is pigs.''}

Similarly, you cannot say ``$\zz$ is the integers.'' (And 
``$\zz$ is integers'' would be even worse!!!)  The correct
ways to say it are 
\begin{itemize}
\item[$\bullet$] $\zz$ is the set of all integers,
\item[$\bullet$] $\zz$ is the set whose members are all the integers,
\item[$\bullet$] $\zz$ is the set whose members are the integers,
\item[$\bullet$] $\zz$ is the set consisting of all the integers.
\end{itemize}

\medbreak\noindent FOOD FOR THOUGHT. Can you say 
``$\zz$ is a set of integers''?  (Hint: The answer depends very much 
on what you are
trying to say or what question you are trying to answer. To answer this 
question, you should give examples of (a)~a question for which
``$\zz$ is a set of integers'' would be a correct answer, and 
(b)~a question for which
``$\zz$ is a set of integers'' would not be a correct answer.)



\section{Some remarks on mathematical writing}\label{using-lets}

\noindent When we talk about mathematical objects, we often use 
{\em letters} (or other special symbols such as
$\rr$, $\nn$, $\zz$, $0$, $1$, $2$, etc.).
The way letters are used in mathematical discourse
may often appear mysterious to the students. The first thing you 
need to do is to realize that it is not mysterious at all. It is,
in fact, quite simple!

\medbreak
\begin{center}
\fbox{\begin{minipage}{11.5cm}
{\Large\bf Letters are only used in mathematics in \mbox{order}
to shorten certain statements that would \mbox{otherwise} be too long.}
\end{minipage}}\end{center}
\medbreak

\vbox{

\subssection{Some examples}

\begin{example} {\rm Suppose you want to define the word ``cousin.''
You could say:  

\begin{equation}
\mbox{\begin{minipage}{10cm}
A person is said to be a {\em cousin} of another person if
one of the first person's parents is a sibling of one of the second person's
parents.
\end{minipage}}\end{equation}

\noindent This is O.K., but having to talk about the ``first person'' and
the ``second person'' and having to repeat the word ``person''
so many times  makes it awkward.  So what we can do is give names to those
persons, for example $x$ and $y$, and say the following:
\begin{equation}
\mbox{\begin{minipage}{10cm}
Let $x$ and $y$ be persons. Then $x$ is said to be a {\em cousin} of $y$
if one of $x$'s parents is a sibling of one of $y$'s
parents.
\end{minipage}}\end{equation}}
\end{example}
}

\medbreak

\begin{example} {\rm Suppose you want to say 

\begin{equation}
\mbox{\begin{minipage}{10cm}\label{youcan}
If you take two real numbers, add them, and square the sum,
then you get the same result as if you had squared each number separately,
added the squares, and added to the result twice the product of the two
numbers.
\end{minipage}}\end{equation}

\medbreak\noindent This is awfully complicated, awkward,  and ugly.
You can make it much shorter by saying 

\begin{equation}
\mbox{\begin{minipage}{10cm}
Let $x$, $y$ be two real numbers. Then 
$(x+y)^2=x^2+y^2+2xy\,$.
\end{minipage}}\end{equation}
or
\begin{equation}\mbox{\begin{minipage}{10cm}\label{evenm-a}
For all real numbers $x$, $y$, $(x+y)^2=x^2+y^2+2xy\,$.
\end{minipage}}\end{equation}

\medbreak\noindent We can say (\ref{evenm-a}) even more
concisely, by writing   

\begin{equation}\label{evenm}
\forall x\in\rr,\forall y\in\rr,\Bigl((x+y)^2=x^2+y^2+2xy\Bigr)\,.
\end{equation}  

\medbreak\noindent  Here the symbol $\forall$ is the {\bf universal
quantifier}. Statement (\ref{evenm}) is read as follows:

\begin{equation}\mbox{\begin{minipage}{10cm}
For all real numbers $x$ and for all real numbers $y$,
the square of $x$ plus $y$ is equal to the sum of the square of $x$,
the square of $y$, and twice the product of $x$ and $y$.
\end{minipage}}\end{equation}

\noindent Actually {\bf you can even read (\ref{evenm}) without any
letter variables at all.}   How? Well, that's precisely what
(\ref{youcan}) does!}\proofend
\end{example}


\medbreak
\begin{example}\label{sqrat} Suppose you want to say 
\begin{equation}\label{squa}\mbox{\begin{minipage}{10cm}
If you take the number $\frac{\sqrt{5}-1}{2}$, and add it to its square,
you get $1$.
\end{minipage}}\end{equation}
You could say this as follows:
\begin{equation}
\Big(\frac{\sqrt{5}-1}{2}\Big)^2+\frac{\sqrt{5}-1}{2}=1\,.
\end{equation}
Or you could give a name to the number, say $a$, and write
\begin{equation}\label{squalo}\mbox{\begin{minipage}{10cm}
Let $a=\frac{\sqrt{5}-1}{2}$. Then $a^2+a=1$.
\end{minipage}}\end{equation}

\end{example}


\medbreak
\begin{example}  Suppose you want to say 

\begin{equation}\label{ifyoupr}\mbox{\begin{minipage}{10cm}
It is possible to find a real number whose cube, added to its 
square, is equal to the number itself.  
\end{minipage}}\end{equation}

\medbreak\noindent Again, this looks horribly complicated and ugly, and you 
can make it much shorter by saying 

\begin{equation}\mbox{\begin{minipage}{10cm}
There exists a real number $x$ such that $x^2+x^3=x\,.$
\end{minipage}}\end{equation}

\noindent or 

\begin{equation}\mbox{\begin{minipage}{10cm}
There is a real number $x$ such that $x^2+x^3=x\,.$
\end{minipage}}\end{equation}

\noindent or 

\begin{equation}\mbox{\begin{minipage}{10cm}
$x^2+x^3=x$ for some real number $x$.  
\end{minipage}}\end{equation}

\medbreak\noindent We can say this thing even more concisely, by writing   

\begin{equation}\label{evenmex}\mbox{\begin{minipage}{10cm}
$\exists x\in\rr,(x^2+x^3=x)\,.$
\end{minipage}}\end{equation}

\medbreak\noindent  Here the symbol $\exists$ is the {\bf existential
quantifier}. Statement (\ref{evenmex}) is read as follows:

\begin{equation}\mbox{\begin{minipage}{10cm}
There exists a real number $x$ such that $x^2+x^3=x$.
\end{minipage}}\end{equation}

\begin{equation}\mbox{\begin{minipage}{10cm}
There exists a real number $x$ such that the sum of the square of
$x$ and the cube of $x$ cube equals $x$.
\end{minipage}}\end{equation}

or even, if you don't want to use any letter variables at all

\begin{equation}\mbox{\begin{minipage}{10cm}
There exists a real number such that the sum of the square of the number
plus the cube of the number equals the number itself.
\end{minipage}}\end{equation}

or, if you prefer, you can read it as statement 
(\ref{ifyoupr}) above.\proofend
\end{example}

\bigskip\begin{center}\fbox{\fbox{\begin{minipage}{12cm}
\begin{center}{\Large\bf QUANTIFIERS}\end{center} 

The symbols $\exists$ and $\forall$ are the {\bf quantifiers}.

\medbreak
``$\,\,\exists\,\,$'' is the {\bf existential quantifier}.

\medbreak
``$\,\,\forall\,\,$'' is the {\bf universal quantifier}.

\medbreak
A statement of the form ``$\exists x$,blahblahblah-about-$x$'' is 
an {\bf existential statement}. We read it 
``There exists $x$ such that blahblahblah-about-$x$.'' 

\begin{center}\fbox{\fbox{\begin{minipage}{11.5cm}
{\bf IMPORTANT!}  The 3-symbol string ``$\exists x$,'' is read
``there exists $x$ such that,'' or ``there exists an
$x$ such that,'' or ``there is and $x$ such that.''   
{\bf Do not forget the ``such that.''} Students sometimes read
``$\exists x$,'' as ``there exists $x$ and,'' This is wrong
and could be very confusing. (Think about this, and
ask the intructor if you don't see why ``there exists $x$ and''
is bad.) 
\end{minipage}}}\end{center}

\medbreak
A statement of the form ``$\forall x$,blahblahblah-about-$x$'' is 
a {\bf universal statement}. We read it 
``For all $x$ blahblahbla-about-$x$.'' 

\medbreak
EXAMPLE: What kind of statement is ``$\forall x,\exists y,y<x$''?

ANSWER: It is a universal statement, because it says that ``for all , 
$x$ something is true.'' The ``something'' that is true about $x$ is 
an {\em existential statement}, namely, that ``there exists
a $y$ such that $y<x$.'' 

Let us be more precise. Let us give a name to our statement, say
$S$. So from now until the end of this box the letter $S$ stands
for\footnote{Pay attention to
what we have just done! Does it look familiar? If it does not, 
read \S\ref{constantvar} and then come back to this box. 
You will see that 
what we have done is this: we introduced a new letter, namely, $S$,
and declared it to be a constant, by declaring its value to be the statement
``$\forall x,\exists y,y<x\,$.''  This is exactly the same kind of thing
we did in Example \ref{sqrat} when we introduced the letter 
$a$ and declared it to be a constant with value 
$\frac{\sqrt{5}-1}{2}$. In Example \ref{sqrat} the letter 
$a$ is used as a {\em numerical constant}, that is, a 
constant whose value is a number. Here we are using 
the letter $S$ as a {\em statement constant}, that is, a 
constant whose value is a statement. In both cases, the letters 
are thing-symbols (that is, they stand for a thing, which is a number
in one case and a statement in the other case) and are constants
(that is, they stand for a fixed thing).} the statement ``$\forall
x,\exists y,y<x$.'' Then what we just said is that {\em $S$ is a
universal statement} and, in fact, $S$ is of the form ``$\forall x,T$,'' 
where {\em $T$ is an existential statement}, namely,
``$\exists y,y<x$.''  So $T$ is of the form ``$\exists y,W$,'' where
$W$ is the statement ``$y<x$.''

\medbreak
QUESTION: Suppose we are talking about integers. What does
``$\forall x,\exists y,y<x$'' say in plain English? Is it true?

ANSWER: it says ``for every integer there is a smaller integer.''
This is true, because given any integer $x$ we can let
$y=x-1$, and then $y<x$.

\medbreak
QUESTION: Suppose we are talking about natural numbers. What does
``$\forall x,\exists y,y<x$'' say in plain English? Is it true?

ANSWER: it says ``for every natural number there is a smaller natural
\mbox{number}.'' This is false, because if we take $x$ to be $0$ then there is 
no natural number $y$ such that $y<x$.

\end{minipage}}}\end{center}




\subssection{A $\,$$\,$very $\,$$\,$important $\,$$\,$remark $\,$$\,$on 
$\,$$\,$reading $\,$$\,$existential \mbox{statements}} 

An {\bf existential statement} is a statement that asserts that
there exists an object of a certain kind. For example:
\begin{equation}\label{polokil}\mbox{\begin{minipage}{10cm}
Somebody killed Polonius
\end{minipage}}\end{equation}
is an existential statement, because it says that there is
a person who killed Polonius. Using the symbolism of existential
quantifiers, we could rewrite (\ref{polokil}) as 
\begin{equation}\label{polokil-2}\mbox{\begin{minipage}{10cm}
$\exists x,x$ killed Polonius. 
\end{minipage}}\end{equation}

The correct way to read the three-character expression ``$\exists x,$'' 
(including the comma) is ``there exists an $x$ such that''.  For example:
\bi
\item{} you should  read (\ref{polokil-2}) as ``there exists
$x$ such that $x$ killed Polonius,'' or 
``there exists $x$ who killed Polonius.'' 
\item{} a statement of the form 
``$\exists x\in\rr,{\rm something\,\,about\,\,}x$'' is read as
\begin{equation}\mbox{\begin{minipage}{10cm}
There exists a real number $x$ such that ${\rm something\,\,about\,\,}x$.
\end{minipage}}\end{equation}
or 
\begin{equation}\mbox{\begin{minipage}{10cm}
There exists an $x$ belonging to $\rr$ such that 
${\rm something\,\,about\,\,}x$.
\end{minipage}}\end{equation}
\ei
\medbreak\noindent It is {\bf wrong} ({\bf very} wrong!)  to read 
(\ref{polokil-2}) as ``there exists $x$ and  $x$ killed Polonius.'' 

Similarly, you should {\bf never} read
``$\exists x\in\rr,{\rm something\,\,about\,\,}x$'' as
\begin{equation}\label{itswrong}\mbox{\begin{minipage}{10cm}
There exists a real number $x$ and 
${\rm something\,\,about\,\,}x$.
\end{minipage}}\end{equation}

\medbreak\noindent {\bf If you read 
(\ref{polokil-2}) as ``there exists $x$ and  $x$ killed Polonius,'' 
or ``\mbox{$\exists x\in\rr$,} something about $x$'' as in (\ref{itswrong})
then you are completely changing the meaning of the statement,
and you are not conveying what it actually means.}

\medbreak\noindent{\bf EXAMPLE:} Suppose you want to explain
what it means for two people to be ``siblings or half-siblings''
using an existential quantifier. You could say:

\begin{equation}\label{sibl}\mbox{\begin{minipage}{10cm}
Let $x$ and $y$ be persons. We say that 
{\em $x$ and $y$ are siblings or half-siblings} if $x\neq y$ and
$\exists z,z$ is a parent of $x$ and a parent of $y$.
\end{minipage}}\end{equation}

How would you read this in plain English? You would certainly {\bf not}
say ``there exists a $z$ and $z$ is a parent of $x$ and a parent of $y$.''!

What you would say (somewhat awkwardly, I know!) is
``there exists a $z$ who is parent of $x$ and a parent of $y$,''
or (even more awkwardly, but still correctly)
``there exists a $z$ such that $z$ is a parent of $x$ and a parent of $y$.''

{\bf It is important that you acquire as soon as possible the habit
of using ``such that'' correctly. Please start right now!}
\proofend


\medbreak\noindent{\bf  RECOMMENDED QUESTION FOR YOU TO THINK ABOUT:}
If I were to say ``There exists a real number such that $x^2+x^3=x$,''
this would {\bf not} be acceptable. Why?  What is missing?

\medbreak\noindent {\bf ANOTHER RECOMMENDED QUESTION FOR YOU TO THINK ABOUT:} 
Is (\ref{evenmex}) true? Why? 

\subssection{More examples}
\begin{example}
Suppose you want to say 

\begin{equation}\label{sibl-u}\mbox{\begin{minipage}{10cm}
For each integer between 2 and 25, take that integer, 
square it, add 5, and compute the multiplicative inverse of the result
(that is, $1$ over the result). 
Having done this for all the integers between 2 and 25 (including 2 and
25 themselves), add all the resulting numbers. Then the number 
you get is smaller than 20.   
\end{minipage}}\end{equation}

\medbreak\noindent Once again, this is awful, so you can shorten the whole
thing and just say 

\begin{equation}\label{sibl-v}\mbox{\begin{minipage}{10cm}
Let $N=\sum_{k=2}^{25}{1\over k^2+5}\,.$  Then $N<20$. 
\end{minipage}}\end{equation}

\medbreak\noindent Or, if you prefer, you could just say: 

\begin{equation}\label{sibl-w}\mbox{\begin{minipage}{10cm}
$\sum_{k=2}^{25}{1\over k^2+5}<20\,.$  
\end{minipage}}\end{equation}

\medbreak\noindent{\bf RECOMMENDED QUESTIONS FOR YOU TO THINK ABOUT:}
Is it true or false that $\sum_{k=2}^{25}{1\over k^2+5}<20$?  How about 
$\sum_{k=2}^{25}{1\over k^2}<20$? And $\sum_{k=2}^{25}{1}<20$? And 
what would you say if I ask you whether it's true that 
$\sum_{n=2}^{25}{1\over k^2+5}<20$? 
\end{example}

\medbreak
\begin{example} Suppose you want to say

\begin{equation}\label{subl-w}\mbox{\begin{minipage}{10cm}
If you take the function that to each real number assigns its 
cube, and integrate it from $0$ to $4$, then the result is 64. 
\end{minipage}}\end{equation}

\medbreak\noindent Again, this can be made much shorter 
by saying 

\begin{equation}\label{subl-a}\mbox{\begin{minipage}{10cm}
Let $N=\int_{0}^4x^3\,dx\,.$  Then $N=64$.
\end{minipage}}\end{equation}

\medbreak\noindent Or you could just say: 

\begin{equation}\label{subl-b}\mbox{\begin{minipage}{10cm}
$\int_{0}^4x^3\,dx=64\,.$ 
\end{minipage}}\end{equation}

\medbreak\noindent{\bf RECOMMENDED QUESTIONS FOR YOU TO THINK ABOUT:}
Is it true or false that $\int_{0}^4x^3\,dx=64$?  
Is it true or false that $\int_{0}^4y^3\,dx=64$? 
\end{example}

\medbreak
\begin{example} Suppose you want to say:  
\begin{equation}\label{sabl-b}\mbox{\begin{minipage}{10cm}
The set of all real numbers whose square is 
larger than the cube of the number has at least two different members.
\end{minipage}}\end{equation}

\medbreak\noindent We can make this shorter by saying

\begin{equation}\label{sabl-a}\mbox{\begin{minipage}{10cm}
Let $S$ be the set of all real numbers $x$ such that
$x^2>x^3$. Then $S$ has at least two members.   
\end{minipage}}\end{equation}

\medbreak\noindent or

\begin{equation}\label{sabl-aa}\mbox{\begin{minipage}{10cm}
Let $S=\{x\in\rr:x^2>x^3\}\,.$ Then $S$ has at least two members.   
\end{minipage}}\end{equation}


\medbreak\noindent Or you could just say: 

\begin{equation}\label{sabl-c}\mbox{\begin{minipage}{10.5cm}
Let $S=\{x\in\rr:x^2>x^3\}\,.$ Then $\exists x\exists y,
(x\in S, y\in S, {\rm \,\,and\,\,}x\neq y)$.
\end{minipage}}\end{equation}
\end{example}

\medbreak
\begin{example} Suppose you want to say that 
\begin{equation}\label{sobl-b}\mbox{\begin{minipage}{10cm}
I am older than 20 and less than 70, and my brother is younger than I
am, but his age is larger than half my age.
\end{minipage}}\end{equation}

\medbreak\noindent Here too we can be more concise, and say

\begin{equation}\label{sobl-ba}\mbox{\begin{minipage}{10cm}
Let $x$ be my age and let $y$ be my brother's age. Then
$20<x<70$ and ${x\over 2}<y<x$.
\end{minipage}}\end{equation} 
\end{example}

\subssection{Constants and variables}\label{constantvar}
In all the examples we have discussed so far, 
we have used {\bf special symbols} 
(most commonly letters, such as $x,y,a,k,N,n,S$) 
as labels, so that we can avoid  writing clumsy phrases 
such as ``the number that we have been talking about.'' 
Sometimes we use other special symbols, such as Greek letters
($\alpha$, $\beta$, $\eps$, etc.), letters 
modified in a particular way (for example $\rr$, $\nn$, $\zz$,
$\cc$), numerals ($0$, $1$, $2\,,\ldots\,$)

These special symbols stand for {\em objects}, or {\em things}
of various kinds (numbers, human beings, sets, functions, 
wolfes, stars), so we are going to call
them {\em individual symbols}, or {\em thing-symbols}. 
(There are other kinds of mathematical symbols that 
do not stand for things. For example: the universal quantifier
$\forall$, the existential quantifier $\exists$, the symbol
$+$ for the addition operation.  We will talk about many of these symbols
later.)

All the thing-symbols we have used are either

\bigbreak

\fbox{\noindent {\bf constants}$\qquad$ or $\qquad$
{\bf variables} $\quad$(also called {\bf dummy variables}).} 

\medbreak\noindent A {\em constant} is a symbol that stands for a
thing that stays fixed throughout an explanation or argument.

For instance, in Example \ref{sqrat}, the letter $a$ stands
for the number $\frac{\sqrt{5}-1}{2}$.  We made this clear
by  {\bf introducing} the letter $a$ by means of the 
{\bf declaration} 

\begin{center}
Let $a=\frac{\sqrt{5}-1}{2}$. 
\end{center}
That is, we {\bf declared} the value of $a$ to be 
$\frac{\sqrt{5}-1}{2}$, so from that moment on we know 
that $a$ stands for ``$\frac{\sqrt{5}-1}{2}$.''

\medbreak{\bf Variables} (also called ``dummy variables'') 
are letters used as an {\em internal label}  within a statement,
but which do not have a value outside this statement.

This may look complicated, but it is in fact quite
simple. Actually, it is \mbox{something} you are probably used to 
if you write computer programs in a \mbox{language} like Pascal.
You often introduce a variable to keep track of some \mbox{number}
with which you perform several operations. For example, you 
may want to keep track of 
a list of 20 numbers, called 
$\quad{\rm list}[1]$, ${\rm list}[2]$, $\ldots$, up to ${\rm list}[20]$.
Suppose you want to add the 20 numbers, and end up with a number
called ``sum'' as the result.  You could write the following 
program:

\begin{itemize}
\item[] begin
\begin{itemize}
\item[] ${\rm sum}:=0$;
\item[] For $i:=1$ to $20$ do ${\rm sum}:={\rm sum}+{\rm list}[i]$;
\end{itemize}
\item[] end.
\end{itemize}

\medbreak\noindent
In this program, the letter $i$ is just a label used to keep track of 
the variable index that keeps changing all the time.  You could have written
instead  

\begin{itemize}
\item[] begin
\begin{itemize}
\item[] ${\rm sum}:=0$;
\item[] For $k:=1$ to $20$ do ${\rm sum}:={\rm sum}+{\rm list}[k]$;
\end{itemize}
\item[] end.
\end{itemize}

\medbreak\noindent and you would have achieved the same thing.
{\em Within the program}, the letter $i$ (or $k$ in our second version)
has a value that keeps changing all the time. 
{\em Outside the program}, it does not make sense to ask ``what is 
the value of $i$?''  

Notice that what we are doing with this program is computing the sum
$$
{\rm Sum}=\sum_{i=1}^{20}{\rm list}[i]\,,
$$
which is exactly the same as saying 
$$
{\rm Sum}=\sum_{k=1}^{20}{\rm list}[k]\,,
$$
or 
$$
{\rm Sum}=\sum_{\alpha=1}^{20}{\rm list}[\alpha]\,,
$$
or 
$$
{\rm Sum}=\sum_{\clubsuit=1}^{20}{\rm list}[\clubsuit]\,.
$$
{\bf The distinctive feature of a dummy variable is that you could have 
used any other symbol instead, and you would be saying exactly the same 
thing.}  For example, if you write


\begin{equation}\label{frol-a}\mbox{\begin{minipage}{10cm}
For all real numbers $x$, $y$, $(x+y)^2=x^2+y^2+2xy\,$, 
\end{minipage}}\end{equation} 
or 
\begin{equation}\label{frol-b}
\forall x\in\rr,\forall y\in\rr,((x+y)^2=x^2+y^2+2xy)\,, 
\end{equation} 
then that has exactly the same meaning as if you had written 
\begin{equation}\label{frol-c}\mbox{\begin{minipage}{10cm}
For all real numbers $a$, $b$, $(a+b)^2=a^2+b^2+2ab\,$, 
\end{minipage}}\end{equation} 
or 
\begin{equation}\label{frol-d}
\forall a\in\rr,\forall b\in\rr,((a+b)^2=a^2+b^2+2ab)\,, 
\end{equation} 
or
\begin{equation}\label{frol-e}\mbox{\begin{minipage}{10cm}
For all real numbers $\alpha$, $\beta$, $(\alpha+\beta)^2
=\alpha^2+\beta^2+2\alpha\beta\,$, 
\end{minipage}}\end{equation} 
or 
\begin{equation}\label{frol-f}
\forall\alpha\in\rr,\forall\beta\in\rr,((\alpha+\beta)^2=\alpha^2+
\beta^2+2\alpha\beta)\,, 
\end{equation} 
or 
\begin{equation}\label{frol-g}\mbox{\begin{minipage}{10cm}
For all real numbers $m$, $n$, $(m+n)^2=m^2+n^2+2mn\,$, 
\end{minipage}}\end{equation} 
or 
\begin{equation}\label{frol-h}
\forall m\in\rr,\forall n\in\rr,((m+n)^2=m^2+n^2+2mn)\,, 
\end{equation} 
or 
\begin{equation}\label{frol-i}\mbox{\begin{minipage}{10cm}
For all real numbers $\diamond$, $\#$,
$(\diamond+\#)^2=\diamond^2+\#^2+2\diamond\#\,$. 
\end{minipage}}\end{equation} 
or 
\begin{equation}\label{frol-j}
\forall\diamond\in\rr,\forall\#\in\rr,
((\diamond+\#)^2=\diamond^2+\#^2+2\diamond\#)\,. 
\end{equation} 

\bigbreak\noindent Similarly, instead of 
$\,\qquad\qquad\exists x\in\rr,\Bigl(x^2+x^3=x\Bigr)\,$,

\nin you could have said $\,\,\qquad\qquad\exists z\in\rr,
\Bigl(z^2+z^3=z\Bigr)\,,$

\nin or $\qquad\qquad\qquad\qquad\qquad\qquad\exists a\in\rr,
\Bigl(a^2+a^3=a\Bigr)\,,$

\nin or $\qquad\qquad\qquad\qquad\qquad\qquad\exists \omega\in\rr,
\Bigl(\omega^2+\omega^3=\omega\Bigr)\,,$

\nin or $\qquad\qquad\qquad\qquad\qquad\qquad\exists \aleph\in\rr,
\Bigl(\aleph^2+\aleph^3=\aleph\Bigr)\,,$

\nin or $\qquad\qquad\qquad\qquad\qquad\qquad\exists\diamond\in\rr,
\Bigl(\diamond^2+\diamond^3=\diamond\Bigr)\,.$

\bigbreak\noindent Just one more example:  

$$
{\rm Instead\,\,of}\qquad\qquad \sum_{k=2}^{25}{1\over k^2+5}<20\,,
\quad{\rm you\,\,could\,\,have\,\,said\,\,} 
\sum_{n=2}^{25}{1\over n^2+5}<20\,,
$$
$$
{\rm or}\qquad\sum_{\rho=2}^{25}{1\over \rho^2+5}<20\,,
\qquad\qquad\qquad
{\rm or}\qquad\qquad\sum_{GNU=2}^{25}{1\over GNU^2+5}<20\,.
$$


\bnin  The ``dummy variables'' in the above examples (that is, 
$i$, $k$, $\alpha$, $\clubsuit$, $x$, $y$, $a$, $b$, $\beta$,
$m$, $n$, $\diamond$, $\#$, $z$, $\omega$, $\aleph$,
$\rho$, $GNU$) are really just
{\bf internal tags}, using  to keep track of numbers {\bf within a particular
statement}.  Outside this text, they variables do not have any meaning. For
example, after you have written 

$$\sum_{k=2}^{25}{1\over k^2+5}<20\,,$$  

\mnin it would not make sense for me to ask you ``what is the value of
$k$?''   The letter $k$ is just a tag to specify a {\it range of values},
as part of a set of instructions on how to compute a sum.  Any other letter
could be used instead, so in fact ``$k$ isn't really there.''   You can
persuade yourself of this in another way, by remembering that 

$$\sum_{k=2}^{25}{1\over k^2+5}<20$$  

\mnin is just another way of saying 

\medbreak
\bi
\item{} For each integer between 2 and 25, take that integer, 
square it, add 5, and compute the multiplicative inverse of the result. 
Having done this for all the integers between 2 and 25 (including 2 and
25 themselves), add all the resulting numbers. Then the number 
you get is smaller than 20.   
\ei

\mnin As you can see, there is no value of $k$ there!

\bnin Similarly, when we say $\int_0^4x^3\,dx=64$, this is just a way
of saying ``if you take the function that to each real number assigns
its cube, and integrate it from $0$ to $4$, you get $64$.''
Once again, there is no value of $x$ there! You could equally well
have said $\int_0^4y^3\,dy=64$, or $\int_0^4u^3\,du=64$, or 
$\int_0^4\aleph^3\,d\aleph=64$.

\bnin And when we say $\exists x\in\rr,x^2=2$, 
this is just a way of saying ``there is a real number whose square 
is $2$'' (or, even better, ``$2$ has a square root''). 
This statement does not talk about $x$.

\bnin  This way of speaking and writing, using dummy variables, 
is very useful, especially when what we
are trying to say is quite long.  But {\bf you have to be careful} when
writing this way, and have to make sure that you are using the letters
correctly.

\vskip 0.3cm

\begin{center}\fbox{\fbox{\begin{minipage}{9.5cm}
\baselineskip=21pt
{\huge\bf
\mnin In a statement, a
thing-symbol can only be used if either its value has been
declared before or it is a dummy variable.}
\end{minipage}}}\end{center}

\subsection{Examples of things you should NOT write}

\nin Suppose you write 

$$x^2+x=90\,$$

\nin just like that, without any words before or after explaining 
what $x$ is, or making it a dummy variable. Then what you wrote is {\bf
not a statement}\footnote{It's more or less the same situation as if somebody
walked into our classroom and said ``she has arrived.'' 
If we already know who
``she'' is supposed to be then this makes sense. But if we don't then our 
reaction would probably be to ask ``who are you talking about?''
The word ``she'' is a {\em pronoun}. It stands for somebody,
and a sentence using ``she'' is meaningless if we do not know
which person ``she'' stands for. Similarly, the letter
$x$ in ``$x^2+x=90\,$'' functions like a pronoun. The sentence
``$x^2+x=90\,$'' is meaningless if you do not know who---or what---
$x$ stands for.}.  We have no way to tell if what you wrote is true or false,
because we have no idea what you are talking about.     

If you write something like this in your homework, then I will 
\begin{itemize}
\item[(a)] Give you a 
grade of {\bf x}. (Then you will come and ask me ``what is this $x$?'' 
And I will tell you ``well, if you explain to me what your $x$ is, 
then I will tell you what my $x$ is.'' Then you will probably answer
``$x$ is a number.'' And then I will reply: ``O.K., the $x$ I have written
in your homework is a grade.'') 
\end{itemize}
or
\begin{itemize}
\item[(b)] Give a special assignment to you: ``read page $x$ of the book 
and tell me what it says next week.''
\end{itemize}

\medbreak\noindent
For comparison, here are some perfectly meaningful statements you could have
written:

\medbreak
\bi
\item[1.] Let $x=9$. Then $x^2+x=90$.  

\medbreak
\item[2.] Let $x=6$. Then $x^2+x=90$.  

\medbreak
\item[3.] Let $x$ be the number of planets of the solar system. 
Then $x^2+x=90$.  

\medbreak
\item[4.] $x^2+x=90$ for every real number $x$.

\medbreak

\item[5.] $x^2+x=90$ for some real number $x$.

\medbreak
\item[6.] $\forall x\in\rr,(x^2+x=90)$.  

\medbreak
\item[7.] $\exists x\in\rr,(x^2+x=90)$.  

\medbreak
\item[8.] The equation $x^2+x=90$ has a solution $x\in\rr$.
(Or ``the equation $x^2+x=90$ has a real solution.'')
\ei


\bnin Some of these are true, while the others are false. (Precisely:
1,3,5,7,8 are true; 2,4,6 are false. Moreover:  6 is just another way of
saying the same thing as 4, and 7 and 8 are just alternative ways of
stating 5.) 

Notice that in 1, 2 and 3 the letter $x$ is a constant: its
value is declared before the letter is used to make a statement.    
In 4, 5, 6, 7 and 8 the letter $x$ is a dummy variable. 

Again for comparison, let us now write some things that do {\bf not} have a
completely precise meaning:

\medbreak
\bi
\item[A.] The equation $x^2=2$ has a solution.

\medbreak
\item[B.] The equation $x^2+1=0$ has a solution.

\medbreak
\item[C.] The equation $3x=5$ has a solution.
\ei


\bnin Why are these things not good enough?   Because they do not specify 
the {\bf range} of values within which the solution $x$ is to be sought. 
For example, A is true if we look for solutions $x\in\rr$, but false 
if we look for solutions $x\in\qq$.  B is true if we look for solutions 
$x\in{\bf C}$ (the set of complex numbers) but false if we look for
solutions $x\in\rr$.   C is true if we look for solutions $x\in\qq$
(or $x\in\rr$) but false if we look for solutions $x\in\zz$.   


On the other hand, the following are perfectly meaningful statements: 

\medbreak
\noindent A1. The equation $x^2=2$ has a solution $x\in\rr$.  (That is:
$\exists x\in\rr,(x^2=2)$.) 

\noindent A2. The equation $x^2=2$ has a solution $x\in\qq$.  (That is:
$\exists x\in\qq,(x^2=2)$.) 

\noindent B1. The equation $x^2+1=0$ has a solution $x\in\zz$.  (That is:
\mbox{$\exists x\in\zz,(x^2+1=0)$}.) 

\noindent B2. The equation $x^2+1=0$ has a solution $x\in\rr$.  (That is:
\mbox{$\exists x\in\rr,(x^2+1=0)$}.) 

\noindent B3. The equation $x^2+1=0$ has a solution $x\in{\bf C}$.  (That is:
\mbox{$\exists x\in{\bf C},(x^2+1=0)$}.) 

\noindent C1. The equation $3x=5$ has a solution $x\in\zz$.  (That is:
$\exists x\in\zz,(3x=5)$.) 

\noindent C2. The equation $3x=5$ has a solution $x\in\qq$.  (That is:
$\exists x\in\qq,(3x=5)$.) 

\noindent C3. The equation $3x=5$ has a solution $x\in\rr$.  (That is:
$\exists x\in\rr,(3x=5)$.) 

\noindent C4. The equation $3x=5$ has a solution $x\in{\bf C}$.  (That is:
$\exists x\in{\bf C},(3x=5)$.) 


\bnin (Actually, A1, B3, C2, C3 and C4 are true, whereas 
A2, B1, B2 and C1 are false. Of course, in each case you could say the same
thing in lots of different ways.  For example, instead of 
C2 you could say 
``the equation $3x=5$ has a rational solution,''   or 
``the equation $3x=5$ can be solved in the set of rational numbers,''  
or ``there is a rational number that satisfies the equation $3x=5$,''
or ``$3x=5$ is solvable in the set of rational numbers.'')


\subssection{Words are important}

It's the {\bf words} that glue the formulas together and give a structure and a
meaning to mathematical writing.  If you just write a bunch of formulas
then you probably do not know what you are saying, and if you don't know
then you should not expect that the instructor or the grader will know.

Look, for example,  at Statements 1, 2, 4 and 5 above. In all cases, 
the ``formula'' $x^2+x=90$ is the same, but the statements are quite
different and have totally different meanings.   That's because the words
are different!   

\subssection{Avoid ambiguous sentences}

\noindent A sentence whose meaning is imprecise because it can be 
interpreted
in more than one way is said to be {\it ambiguous}.   For example, the
sentences 


\bnin\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Flying planes can be dangerous

\noindent\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ College students are turning to
vegetables

\noindent\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ The University president promised
to stop drinking on campus

\noindent\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I shot an elephant in my 
pajamas\footnote{``Last night I shot an elephant in my pajamas.
How the elephant got into my pajamas I have no idea.'' (Groucho Marx)}

\noindent\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Mary had a little 
lamb\footnote{
Mary had a little lamb,\\% $\backslash\backslash$
\phantom{a}\ \ \ \ Its fleece was white as snow,\\% $\backslash\backslash$
\phantom{a}\ \ \ \ Everywhere that Mary went,\\% $\backslash\backslash$
\phantom{a}\ \ \ \ The lamb was sure to go.\\% $\qquad\qquad$
\phantom{a}\ \ \ \ {\em But also:}$\qquad$\\
\phantom{a}\ \ \ \ Mary had a little lamb,\\%$\backslash\backslash$
\phantom{a}\ \ \ \ A loaf of bread, a pot of jam,\\% $\backslash\backslash$
\phantom{a}\ \ \ \ An ice cream soda topped with fizz,\\%$\backslash\backslash$
\phantom{a}\ \ \ \ And oh! - How sick our Mary is!}

\noindent\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ George W. Bush
and Hillary R. Clinton are married

\bnin are ambiguous. Why? Think of other examples of ambiguous sentences.

\mnin Similarly, A, B, C above are ambiguous.  Another example of ambiguous 
\mbox{sentence} is a
sentence where a letter does not have a declared value and is not a 
\mbox{dummy}
variable either. For example, if I tell you to ``read chapter $x$,'' then
that's \mbox{ambiguous}---unless I have told you before what ``$x$'' 
stands for,
or given you clues so you can find out, or perhaps I have 
said ``read Chapter $x$ for all $x$'' (that is ``read all the chapters'')
or ``read Chapter $x$ for some $x$'' (that is ``read one chapter'')--- 
because you don't know what $x$  is supposed to be.


%\vspace{-6cm}
\end{document}




