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\begin{document}

\psol{1}{September 14, 1995}

\bigskip

\begin{problems}

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\problem 

\bparts
\ppart $r \Rightarrow p.$
\ppart $p\ \bigwedge \  \neg q \bigwedge r.$
\ppart $(p\ \bigwedge \   q) \Rightarrow r.$
\ppart $r \Leftrightarrow (p \bigvee q).$
\eparts

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\problem 
$(p \bigwedge q \bigwedge \neg r) \bigvee (p \bigwedge \neg q \bigwedge r) (\neg p \bigwedge q \bigwedge r).$

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\problem 

\bparts
\ppart $\exists x \ (P(x) \bigwedge Q(x)).$
\ppart $\exists x \ (P(x) \bigwedge \neg Q(x)).$
\ppart $\forall x \ (P(x) \bigvee  Q(x)).$
\ppart $\forall x \ \neg (P(x) \bigvee Q(x)).$
\ppart $\forall x \ P(x) \Rightarrow \neg Q(x).$
\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem 

\bparts
\ppart $\forall x \exists y \ L(x,y).$
\ppart $\exists y \forall x \ L(x,y).$
\ppart $\forall x \exists y \ \neg L(x,y).$
\ppart $\exists x (\forall y \ L(y,x) \bigwedge \forall z ((\forall w \ L(w,z)) \Rightarrow z = x)).$
\ppart $\exists x \exists y (x \not= y \bigwedge L(Lynn,x) \bigwedge L(Lynn,y) \bigwedge \forall z (L(Lynn,z) \Rightarrow (z = x \bigvee z = y))).$
\ppart $\exists x \forall y (L(x,y) \Rightarrow x = y).$
\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem 

\bparts
\ppart $\forall x (P(x) \Rightarrow \neg Q(x)).$
\ppart $\forall x (Q(x) \Rightarrow R(x)).$
\ppart $\forall x (P(x) \Rightarrow \neg R(x)).$
\ppart The conclusion does not follow. There could be other vain people 
besides ignorant ones and hence there may be vain professors.
\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem $\forall x P(x) \bigvee \forall x Q(x) \ \Rightarrow \forall x \forall y (P(x) \bigvee Q(y) ).$ For   $\forall x P(x) \bigvee \forall x Q(x)$ to be true it must be the case that  $\forall x P(x)$ or $\forall x Q(x)$ is true. In either case $\forall x \forall y (P(x) \bigvee Q(y) )$ becomes true.

 $\forall x P(x) \bigvee \forall x Q(x) \ \Leftarrow \forall x \forall y (P(x) \bigvee Q(y) ).$ For $\forall x \forall y (P(x) \bigvee Q(y) )$ to be true it must be the case that  $\forall x P(x)$ or $\forall x Q(x)$ is true. Because if not, then there exists an $x'$ and a $y'$ such that $ (P(x) \bigvee Q(y) )$ is false. 

Hence  $\forall x P(x) \bigvee \forall x Q(x) \ \Leftrightarrow \forall x \forall y (P(x) \bigvee Q(y) ).$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem 

\bparts
\ppart Fallacy of affirming the conclusion.
\ppart Fallacy of begging the question.
\ppart Valid argument using modus tollens.
\ppart Fallacy of denying the hypothesis.
\eparts

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem 
Suppose that $r$ is rational and $i$ is irrational and $s = r + i$ is
rational. Then $s + (-r) = i$ is rational, a contradiction.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem There are four cases.
\begin{enumerate}
\item $x \geq 0$ and $y \geq 0.$ Equality holds.
\item $x < 0$ and $y < 0.$ Equality holds.
\item  $x \geq 0$ and $y < 0.$ If $x \geq -y$ then $|x| + |y| = x + (-y) > x + y = |x+y|.$ If $x < -y$ then  $|x| + |y| = x + (-y) \geq -x + (-y) = |x+y|.$
\item  $x < 0$ and $y \geq 0.$ Same as the previous case with the roles of $x$
and $y$ reversed.
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem Let the squares of the checkerboard be numbered with Cartesian coordinates so that the square in the lower lefthand corner gets the number $(1,1)$. Then the square in the upper righthand corner gets numbered $(8,8)$. Consider the the three squares $(1,1), (5,1)$ and $(3,3)$. These three squares are pairwise a distance of $4$ apart. Since two of these three squares must get the same color we see that there are two squares at distance exactly $4$ apart that have the same color.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\problem

\begin{theorem}
\[
\forall m,n,p \in \naturals, (m+p=n+p) \implies m=n
\]
\end{theorem}

\begin{proof}
Fix $m,n$.
Let $S= \{ p \in \naturals \mid (m+p=n+p) \implies m=n \}$.
We now prove $S=\naturals$ by induction.

\begin{itemize}

\item  {\bf Base case:}  $m+0= n+0 \implies m=n$ by defn of +.

\item {\bf Inductive step:}  Suppose $p\in S$.  Then, 
$(m+p=n+p) \implies m=n$.

\begin{eqnarray*}
m+p' = n+p' & \implies & (m+p)'=(n+p)' \hspace{.2in} \mbox{(defn of +)} \\
& \implies & m+p = n+p \hspace{.2in} \mbox{(Third Peano axiom)} \\
& \implies & m = n \hspace{.2in} \mbox{(true for $p$)}
\end{eqnarray*}
\end{itemize}

Therefore, $S=\naturals$ by induction axiom.
By universal generalization, 
$\forall m,n,p \in \naturals, (m+p=n+p) \implies m=n$.
\end{proof}
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\problem
The function 
\[f(x) = \overbrace{x + x + \ldots + x}^x\]

is defined only for nonnegative integers $x$. It is defined only on discrete points and is not continuous. Hence its derivative does not make sense.

\end{problems}

\end{document}





