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\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 (100 Points)
\end{center}
\begin{center}\large\bf
Due by 2:30 p.m.\ on Thursday, February 12
\end{center}
\begin{center}
The context for this assignment is Chapters 2-3 of \emph{TAPL}.
\end{center}

\section*{Definition of $\NATS$}

Let $\NATS$ be the smallest set such that:
\begin{description}
\item[\quad(Zero)]
  \begin{displaymath}
    \Zero\in\NATS ;
  \end{displaymath}
\item[\quad(Succ)] for all $t$,
  \begin{displaymath}
    \frac{t\in\NATS}%
         {\Succ\;t\in\NATS.}
  \end{displaymath}
\end{description}
(Here, $\Zero$ and $\Succ$ are constructors, so that, for all $t$,
$\Zero\neq\Succ\;t$, and, for all $t_1,t_2$, $\Succ\;t_1=\Succ\;t_2$
iff $t_1=t_2$.)

The \emph{inversion lemma} for $\NATS$ says that, for all $s\in\NATS$,
either:
\begin{description}
\item[\quad(Zero)] $s = \Zero$; or

\item[\quad(Succ)] $s = \Succ\;t$, for some $t\in\NATS$.
\end{description}

Suppose $P\sub\NATS$.  The \emph{principle of structural induction on
  $\NATS$} says that,
\begin{displaymath}
\eqtxtr{for all}t\in\NATS, P(t)  
\end{displaymath}
follows from showing
\begin{description}
\item[\quad(Zero)] $P(\Zero)$;

\item[\quad(Succ)] for all $t\in\NATS$, if (\dag) $P(t)$, then
  $P(\Succ\;t)$.
\end{description}
We refer to (\dag) as the \emph{inductive hypothesis}.

\section*{Definition of $\PLUS$}

Let $\PLUS$ be the smallest subset of $\NATS\times\NATS\times\NATS$
such that:
\begin{description}
\item[\quad(Plus-Zero)] for all $t\in\NATS$,
  \begin{displaymath}
    (\Zero, t, t)\in\PLUS ;
  \end{displaymath}
\item[\quad(Plus-Succ)] for all $t_1,t_2,t_3\in\NATS$,
  \begin{displaymath}
    \frac{(t_1,t_2,t_3)\in\PLUS}%
         {(\Succ\;t_1,t_2,\Succ\;t_3)\in\PLUS.}
  \end{displaymath}
\end{description}

The \emph{inversion lemma} for $\PLUS$ says that, for all $s_1,s_2,s_3\in\NATS$,
if $(s_1,s_2,s_3)\in\PLUS$, then either:
\begin{description}
\item[\quad(Plus-Zero)] there is a $t\in\NATS$ such that
  $s_1=\Zero$, $s_2 = t$ and $s_3 = t$ (so that $s_2=s_3$); or

\item[\quad(Plus-Succ)] there are $t_1,t_2,t_3\in\NATS$ such
  that $s_1=\Succ\;t_1$, $s_2=t_2$, $s_3=\Succ\;t_3$ and
  $(t_1,t_2,t_3)\in\PLUS$ (so that $(t_1,s_2,t_3)\in\PLUS$).
\end{description}

Suppose $P\sub\NATS\times\NATS\times\NATS$.
We sometimes write ``$P(t_1,t_2,t_3)$'' for ``$(t_1,t_2,t_3)\in P$''.
The \emph{principle of induction on $\PLUS$} says that,
\begin{center}
for all $t_1,t_2,t_3\in\NATS$, if $(t_1,t_2,t_3)\in\PLUS$, then
$P(t_1,t_2,t_3)$,
\end{center}
follows from showing
\begin{description}
\item[\quad(Plus-Zero)] for all $t\in\NATS$,
    $P(\Zero, t, t)$;

\item[\quad(Plus-Succ)] for all $t_1,t_2,t_3\in\NATS$,
  if $(t_1,t_2,t_3)\in\PLUS$ and (\dag) $P(t_1,t_2,t_3)$,
  then $P(\Succ\;t_1,t_2,\Succ\;t_3)$.
\end{description}
We refer to (\dag) as the \emph{inductive hypothesis}.

\section*{Exercise 1 (25 Points)}

Prove that, for all $t_1,t_2\in\NATS$, there is a $t_3\in\NATS$
such that $(t_1,t_2,t_3)\in\PLUS$.

\section*{Exercise 2 (25 Points)}

Prove that, for all $t_1,t_2,t_3,t'_3\in\NATS$, if
$(t_1,t_2,t_3)\in\PLUS$ and $(t_1,t_2,t'_3)\in\PLUS$, then
$t_3=t'_3$.
(This completes the proof that, for all $t_1,t_2\in\NATS$, there
is a unique $t_3\in\NATS$ such that $(t_1,t_2,t_3)\in\PLUS$.)

\section*{Exercise 3 (25 Points)}

Prove that, for all $t_1,t_2,t_3,s_1,s_2,s_3\in\NATS$,
if $(t_1,t_2,s_1)\in\PLUS$, $(s_1,t_3,s_2)\in\PLUS$ and
$(t_2,t_3,s_3)\in\PLUS$, then $(t_1,s_3,s_2)\in\PLUS$.

\section*{Exercise 4 (25 Points)}

Prove that, for all $t_1,t_2,t_3\in\NATS$,
if $(t_1,t_2,t_3)\in\PLUS$, then $(t_2,t_1,t_3)\in\PLUS$.
\end{document}