\documentclass{article}

\input{exer-defs}

%\renewcommand{\thepage}{}

\newcommand{\Zero}{\mathsf{zero}}
\newcommand{\Succ}{\mathsf{succ}}
\newcommand{\NATS}{\mathsf{Nats}}
\newcommand{\PLUS}{\mathsf{Plus}}

\begin{document}
\begin{center}\large\bf
CIS 705 --- Programming Languages --- Spring 2009
\end{center}
\begin{center}\Large\bf
Assignment 1
\end{center}
\begin{center}\large\bf
Model Answers
\end{center}

\section*{Exercise 1}
\renewcommand{\thesection}{1}
\theoremnumber{1}

Suppose $t_2\in\NATS$.  It will suffice to show that, for all
$t_1\in\NATS$, there is a $t_3\in\NATS$ such that
$(t_1,t_2,t_3)\in\PLUS$.  Let
\begin{displaymath}
P=\setof{t_1\in\NATS}{\eqtxtr{there is
    a} t_3\in\NATS\eqtxt{such that}(t_1,t_2,t_3)\in\PLUS} .
\end{displaymath}
It will suffice to show that, for all $t_1\in\NATS$, $P(t_1)$, and we show
this by structural induction on $\NATS$.
\begin{description}
\item[\quad(Zero)] We must show $P(\Zero)$, i.e., that there is
  a $t_3\in\NATS$ such that $(\Zero,t_2,t_3)\in\PLUS$.  By (Plus-Zero),
  we have that $t_2\in\NATS$ and $(\Zero,t_2,t_2)\in\PLUS$.

\item[\quad(Succ)] Suppose $t_1\in\NATS$, and assume the inductive
  hypothesis: $P(t_1)$, i.e., that there is a $t_3\in\NATS$ such that
  $(t_1,t_2,t_3)\in\PLUS$.  We must show $P(\Succ\;t_1)$, i.e., that
  there is a $t'_3\in\NATS$ such that $(\Succ\;t_1,t_2,t'_3)\in\PLUS$.
  Since $(t_1,t_2,t_3)\in\PLUS$ and by (Plus-Succ), we have that
  $\Succ\;t_3\in\NATS$ and $(\Succ\;t_1,t_2,\Succ\;t_3)\in\PLUS$.
\end{description}

\section*{Exercise 2}
\renewcommand{\thesection}{2}
\theoremnumber{1}

Let
\begin{displaymath}
P =
\setof{(t_1,t_2,t_3)\in\NATS\times\NATS\times\NATS}{\eqtxtr{for all}
  t'_3\in\NATS,\eqtxt{if}
  (t_1,t_2,t'_3)\in\PLUS,\eqtxt{then}t_3=t'_3} .
\end{displaymath}
It will suffice to show that,
\begin{displaymath}
  \eqtxtr{for all}t_1,t_2,t_3\in\NATS,\eqtxt{if}(t_1,t_2,t_3)\in\PLUS,
  \eqtxt{then}P(t_1,t_2,t_3) .
\end{displaymath}
(Given this result, we would proceed as follows.
Suppose $t_1,t_2,t_3,t'_3\in\NATS$, $(t_1,t_2,t_3)\in\PLUS$ and
$(t_1,t_2,t'_3)\in\PLUS$.  We must show that $t_3=t'_3$.  By
the result, we have that $P(t_1,t_2,t_3)$, i.e., for all
$t'_3\in\NATS$, if $(t_1,t_2,t'_3)\in\PLUS$, then $t_3=t'_3$.  But
$t'_3\in\NATS$ and $(t_1,t_2,t'_3)\in\PLUS$, so that $t_3=t'_3$.)

We proceed by induction on $\PLUS$.
\begin{description}
\item[\quad(Plus-Zero)] Suppose $t\in\NATS$.  We must show that
  $P(\Zero, t, t)$, i.e., for all $t'_3\in\NATS$, if $(\Zero, t,
  t'_3)\in\PLUS$, then $t=t'_3$.  Suppose $t'_3\in\NATS$ and $(\Zero,
  t, t'_3)\in\PLUS$.  We must show that $t=t'_3$, and this follows by
  the inversion lemma for $\PLUS$.

\item[\quad(Plus-Succ)] Suppose $t_1,t_2,t_3\in\NATS$ and
  $(t_1,t_2,t_3)\in\PLUS$, and assume the inductive hypothesis:
  $P(t_1,t_2,t_3)$, i.e., for all $t'_3\in\NATS$, if
  $(t_1,t_2,t'_3)\in\PLUS$, then $t_3=t'_3$.  We must show that
  $P(\Succ\;t_1,t_2,\Succ\;t_3)$, i.e., for all $t'_3\in\NATS$, if
  $(\Succ\;t_1,t_2,t'_3)\in\PLUS$, then $\Succ\;t_3=t'_3$.
  Suppose $t'_3\in\NATS$ and $(\Succ\;t_1,t_2,t'_3)\in\PLUS$.  We must
  show that $\Succ\;t_3=t'_3$.  By the inversion lemma for $\PLUS$,
  we have that there is a $t''_3\in\NATS$ such that $t'_3=\Succ\;t''_3$
  and $(t_1,t_2,t''_3)\in\PLUS$.  By the inductive hypothesis, we
  have that $t_3=t''_3$.  Thus $\Succ\;t_3=\Succ\;t''_3=t'_3$.
\end{description}

\section*{Exercise 3}
\renewcommand{\thesection}{3}
\theoremnumber{1}

Let $P$ be the set of all $(t_1,t_2,s_1)\in\NATS\times\NATS\times\NATS$
such that,
\begin{displaymath}
  \eqtxtr{for all}
  t_3,s_2,s_3\in\NATS, \eqtxt{if} (s_1,t_3,s_2)\in\PLUS \eqtxt{and}
  (t_2,t_3,s_3)\in\PLUS, \eqtxt{then} (t_1,s_3,s_2)\in\PLUS .
\end{displaymath}
It will suffice to show that,
\begin{displaymath}
  \eqtxtr{for all}t_1,t_2,s_1\in\NATS,\eqtxt{if}(t_1,t_2,s_1)\in\PLUS,
  \eqtxt{then}P(t_1,t_2,s_1) .
\end{displaymath}
(Given this result, we would proceed as follows.  Suppose
$t_1,t_2,t_3,s_1,s_2,s_3\in\NATS$, $(t_1,t_2,s_1)\in\PLUS$,
$(s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$.  We must show
that $(t_1,s_3,s_2)\in\PLUS$.  By the result, we have that
$P(t_1,t_2,s_1)$, i.e., for all $t_3,s_2,s_3\in\NATS$, if
$(s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, then
$(t_1,s_3,s_2)\in\PLUS$.  But $t_3,s_2,s_3\in\NATS$,
$(s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, and thus we can
conclude that $(t_1,s_3,s_2)\in\PLUS$.)

We proceed by induction on $\PLUS$.
\begin{description}
\item[\quad(Plus-Zero)] Suppose $t\in\NATS$.  We must show that
  $P(\Zero, t, t)$, i.e., for all $t_3,s_2,s_3\in\NATS$, if
  $(t,t_3,s_2)\in\PLUS$ and $(t,t_3,s_3)\in\PLUS$, then
  $(\Zero,s_3,s_2)\in\PLUS$.  Suppose $t_3,s_2,s_3\in\NATS$,
  $(t,t_3,s_2)\in\PLUS$ and $(t,t_3,s_3)\in\PLUS$.  We must show that
  $(\Zero,s_3,s_2)\in\PLUS$.  Since 
  $(t,t_3,s_2)\in\PLUS$ and $(t,t_3,s_3)\in\PLUS$, Exercise~2 tells
  us that $s_2=s_3$.  Thus, by (Plus-Zero), we have
  that $(\Zero,s_3,s_2)=(\Zero,s_2,s_2)\in\PLUS$.

\item[\quad(Plus-Succ)] Suppose $t_1,t_2,s_1\in\NATS$ and
  $(t_1,t_2,s_1)\in\PLUS$, and assume the inductive hypothesis:
  $P(t_1,t_2,s_1)$, i.e., for all $t_3,s_2,s_3\in\NATS$, if
  $(s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, then
  $(t_1,s_3,s_2)\in\PLUS$.  We must show that
  $P(\Succ\;t_1,t_2,\Succ\;s_1)$, i.e., for all $t_3,s_2,s_3\in\NATS$, if
  $(\Succ\;s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, then
  $(\Succ\;t_1,s_3,s_2)\in\PLUS$.  Suppose $t_3,s_2,s_3\in\NATS$,
  $(\Succ\;s_1,t_3,s_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$.  We must
  show that $(\Succ\;t_1,s_3,s_2)\in\PLUS$.  Because
  $(\Succ\;s_1,t_3,s_2)\in\PLUS$, the inversion lemma for $\PLUS$
  tells us that there is an $s'_2\in\NATS$ such that $s_2=\Succ\;s'_2$
  and $(s_1,t_3,s'_2)\in\PLUS$.  By the inductive hypothesis, substituting
  $s'_2$ for $s_2$, we have
  that if $(s_1,t_3,s'_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, then
  $(t_1,s_3,s'_2)\in\PLUS$.  But 
  $(s_1,t_3,s'_2)\in\PLUS$ and $(t_2,t_3,s_3)\in\PLUS$, and thus
  $(t_1,s_3,s'_2)\in\PLUS$.  Hence
  $(\Succ\;t_1,s_3,s_2)=(\Succ\;t_1,s_3,\Succ\;s'_2)\in\PLUS$, by
  (Plus-Succ).
\end{description}

\section*{Exercise 4}
\renewcommand{\thesection}{4}
\theoremnumber{1}

First, we prove two lemmas.

\begin{lemma}
\label{Ex4Lem1}
For all $t\in\NATS$, $(t,\Zero,t)\in\PLUS$.
\end{lemma}

\begin{proof}
Let
\begin{displaymath}
P = \setof{t\in\NATS}{(t,\Zero,t)\in\PLUS}.
\end{displaymath}
It will suffice to show that, for all $t\in\NATS$, $P(t)$, and
we show this by structural induction on $\NATS$.
\begin{description}
\item[\quad(Zero)] We must show $P(\Zero)$, i.e., that
  $(\Zero,\Zero,\Zero)\in\PLUS$.  And this follows by
  (Plus-Zero).

\item[\quad(Succ)] Suppose $t\in\NATS$, and assume the inductive
  hypothesis: $P(t)$, i.e., that $(t,\Zero,t)\in\PLUS$.
  We must show $P(\Succ\;t)$, i.e., $(\Succ\;t,\Zero,\Succ\;t)\in\PLUS$,
  and this follows from the inductive hypothesis by (Plus-Succ).
\end{description}
\end{proof}

\begin{lemma}
\label{Ex4Lem2}
For all $t_1,t_2,t_3\in\NATS$, if $(t_1,t_2,t_3)\in\PLUS$,
then $(t_1,\Succ\;t_2,\Succ\;t_3)\in\PLUS$.
\end{lemma}

\begin{proof}
Let
\begin{displaymath}
P =
\setof{(t_1,t_2,t_3)\in\NATS\times\NATS\times\NATS}%
{(t_1,\Succ\;t_2,\Succ\;t_3)\in\PLUS} .
\end{displaymath}
It will suffice to show that,
\begin{displaymath}
  \eqtxtr{for all}t_1,t_2,t_3\in\NATS,\eqtxt{if}(t_1,t_2,t_3)\in\PLUS,
  \eqtxt{then}P(t_1,t_2,t_3) .
\end{displaymath}
(Given this result, we would proceed as follows.  Suppose
$t_1,t_2,t_3\in\NATS$ and $(t_1,t_2,t_3)\in\PLUS$.  We must show that
$(t_1,\Succ\;t_2,\Succ\;t_3)\in\PLUS$.  By the result, we have that
$P(t_1,t_2,t_3)$, i.e., $(t_1,\Succ\;t_2,\Succ\;t_3)\in\PLUS$.)

We proceed by induction on $\PLUS$.
\begin{description}
\item[\quad(Plus-Zero)] Suppose $t\in\NATS$.  We must show that
  $P(\Zero, t, t)$, i.e., $(\Zero, \Succ\;t, \Succ\;t)\in\PLUS$,
  and this follows by (Plus-Zero).

\item[\quad(Plus-Succ)] Suppose $t_1,t_2,t_3\in\NATS$ and
  $(t_1,t_2,t_3)\in\PLUS$, and assume the inductive hypothesis:
  $P(t_1,t_2,t_3)$, i.e., $(t_1, \Succ\;t_2, \Succ\;t_3)\in\PLUS$.  We must
  show that $P(\Succ\;t_1,t_2,\Succ\;t_3)$, i.e.,
  $(\Succ\;t_1, \Succ\;t_2, \Succ(\Succ\;t_3))\in\PLUS$, which follows
  from the inductive hypothesis by (Plus-Succ).
\end{description}
\end{proof}

Now, we use our lemmas to prove the main result.  Let
\begin{displaymath}
P = \setof{(t_1,t_2,t_3)\in\NATS\times\NATS\times\NATS}%
{(t_2,t_1,t_3)\in\PLUS} .
\end{displaymath}
It will suffice to show that,
\begin{displaymath}
  \eqtxtr{for all}t_1,t_2,t_3\in\NATS,\eqtxt{if}(t_1,t_2,t_3)\in\PLUS,
  \eqtxt{then}P(t_1,t_2,t_3) .
\end{displaymath}
(Given this result, we would proceed as follows.
Suppose $t_1,t_2,t_3\in\NATS$ and $(t_1,t_2,t_3)\in\PLUS$.
We must show that $(t_2,t_1,t_3)\in\PLUS$.  By the result,
we have that $P(t_1,t_2,t_3)$, i.e., $(t_2,t_1,t_3)\in\PLUS$.)

We proceed by induction on $\PLUS$.
\begin{description}
\item[\quad(Plus-Zero)] Suppose $t\in\NATS$.  We must show that
  $P(\Zero, t, t)$, i.e., $(t, \Zero, t)\in\PLUS$, which
  follows by Lemma~\ref{Ex4Lem1}.

\item[\quad(Plus-Succ)] Suppose $t_1,t_2,t_3\in\NATS$ and
  $(t_1,t_2,t_3)\in\PLUS$, and assume the inductive hypothesis:
  $P(t_1,t_2,t_3)$, i.e., $(t_2, t_1, t_3)\in\PLUS$.  We must show
  that $P(\Succ\;t_1,t_2,\Succ\;t_3)$, i.e., $(t_2, \Succ\;t_1,
  \Succ\;t_3)\in\PLUS$.
  Since $(t_2, t_1, t_3)\in\PLUS$, Lemma~\ref{Ex4Lem2} tells
  us that $(t_2, \Succ\;t_1, \Succ\;t_3)\in\PLUS$.
\end{description}
\end{document}