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\begin{document}
\begin{center}\large\bf
CIS 705 --- Programming Languages --- Spring 2009
\end{center}
\begin{center}\Large\bf
Assignment 4
\end{center}
\begin{center}\large\bf
Model Answers to Exercise 1
\end{center}

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

Note that, if $v$ is a closed value, then $\overline{v} = v$.
Thus, if $t$ is a closed term that converges, then
$\overline{\overline{t}} = \overline{t}$.

\subsection*{Proposition D}

\begin{lemma}
\label{PropDLem}
For all closed terms $t$ and $t'$, closed values $v$ and $n,m\in\nats$,
if $t\fun^n v$ and $t\fun^m t'$, then $m\leq n$ and $t'\fun^{n-m}v$.
\end{lemma}

\begin{proof}
Suppose, toward a contradiction, that $m > n$.  Thus
$m-n>0$ and $n + (m-n) = m$.  Hence, because
$t\fun^m t'$, we have that $t\fun^{n+(m-n)}t'$, so that
there is a closed term $u$ such that $t\fun^n u\fun^{m-n}t'$.
Since $t\fun^n v$ and $t\fun^n u$, we have that
$v=u\fun^{m-n}t'$.  But $m-n>0$, so that there is
a $u'$ such that $v\fun u'$, contradicting the fact that $v$
is a value.  Thus $m\leq n$.

Since $m\leq n$, it follows that $n-m\geq 0$ and $n = m + (n-m)$.
Because $t\fun^n v$, we have that $t\fun^{m + (n-m)}v$,
so that there is a closed term $u$ such that $t\fun^m u\fun^{n-m}v$.
But $t\fun^m t'$, so that $t'=u\fun^{n-m}v$.
\end{proof}

Suppose $t_1$ and $t_2$ are closed terms and $t_1\,t_2$ converges.  We
must show that $t_1$, $t_2$ and $\overline{t_1}\;\overline{t_2}$
converge and $\overline{t_1t_2} =
\overline{\overline{t_1}\;\overline{t_2}}$.  Because $t_1\,t_2$
converges and is closed, there is a closed value $v$ such that
$t_1\,t_2\fun^*v$.  Thus $\overline{t_1\,t_2}=v$, and there is an
$n\in\nats$ such that $t_1\,t_2\fun^n v$.

Suppose, toward a contraction, that $t_1$ diverges.  Thus, by
Proposition~C, there is a closed term $t'_1$ such that
$t_1\fun^{n+1}t'_1$.  Thus $t_1\,t_2\fun^{n+1}t'_1\,t_2$, by
Proposition~A(1).  But $t_1\,t_2\fun^n v$, so that $n+1\leq n$, by
Lemma~\ref{PropDLem}---contradiction.  Thus $t_1$ converges, so that
there is a closed value $v_1$ such that $t_1\fun^*v_1$ and
$\overline{t_1}=v_1$.  Thus $t_1\fun^{n_1}v_1$ for some $n_1\in\nats$,
so that $t_1\,t_2\fun^{n_1}v_1\,t_2$, by Proposition~A(1).  But
$t_1\,t_2\fun^n v$, so that $n_1\leq n$ and $v_1\,t_2\fun^{n-n_1}v$,
by Lemma~\ref{PropDLem}.

Suppose, toward a contradiction, that $t_2$ diverges.  Thus, by
Proposition~C, there is a closed term $t'_2$ such that
$t_2\fun^{n-n_1+1}t'_2$, so that $v_1\,t_2\fun^{n-n_1+1}v_1\,t'_2$, by
Proposition~A(2).  But $v_1\,t_2\fun^{n-n_1}v$, so that $n-n_1+1\leq
n-n_1$, by Lemma~\ref{PropDLem}---contradiction.  Thus $t_2$
converges, so that there is a closed value $v_2$ such that
$t_2\fun^*v_2$ and $\overline{t_2}=v_2$.  Hence $t_2\fun^{n_2}v_2$ for
some $n_2\in\nats$, so that $v_1\,t_2\fun^{n_2}v_1\,v_2$, by
Proposition~A(2).  But $v_1\,t_2\fun^{n-n_1}v$, so that $n_2\leq n-n_1$ and
$v_1\,v_2\fun^{n-n_1-n_2}v$, by Lemma~\ref{PropDLem}.  Thus
$v_1\,v_2\fun^*v$.

We showed above that $t_1$ and $t_2$ converge, $\overline{t_1}=v_1$
and $\overline{t_2}=v_2$.  And $\overline{t_1}\;\overline{t_2} =
v_1\,v_2\fun^*v$, showing that $\overline{t_1}\;\overline{t_2}$
converges and $\overline{\overline{t_1}\;\overline{t_2}} = v$.
Finally, we have that $\overline{t_1\,t_2} = v =
\overline{\overline{t_1}\;\overline{t_2}}$.

\subsection*{Proposition E}

Suppose $t_1$ and $t_2$ are closed, and $t_1$, $t_2$ and
$\overline{t_1}\;\overline{t_2}$ converge.  We must show that
$t_1\,t_2$ converges and $\overline{t_1t_2} =
\overline{\overline{t_1}\;\overline{t_2}}$.  Because $t_1$ converges,
there is a closed value $v_1$ such that $t_1\fun^*v_1$, so that
$\overline{t_1} = v_1$.  Because $t_2$ converges, there is a closed
value $v_2$ such that $t_2\fun^*v_2$, so that $\overline{t_2}=v_2$.
Thus, $v_1\,v_2=\overline{t_1}\;\overline{t_2}$ converges,
so that there is a closed value $v$ such that $v_1\,v_2\fun^* v$.
By Proposition~B(1), because $t_1\fun^*v_1$, we have that
$t_1\,t_2\fun^*v_1\,t_2$.  And, by Proposition~B(2),
because $v_1$ is a value and $t_2\fun^*v_2$, we have that
$v_1\,t_2\fun^*v_1\,v_2$.  Finally, because
$t_1\,t_2\fun^*v_1\,t_2\fun^*v_1\,v_2\fun^*v$, we have that
$t_1\,t_2\fun^*v$.  Thus $t_1\,t_2$ converges
and, since $t_1\,t_2\fun^*v$ and $v_1\,v_2\fun^* v$, it follows that
\begin{displaymath}
\overline{t_1\,t_2} = v = \overline{v_1\;v_2} =
\overline{\overline{t_1}\;\overline{t_2}} .
\end{displaymath}

\subsection*{Proposition F}

Let $P$ be the set of all $n\in\nats$ such that, if $n\geq 2$, then,
for all closed terms $t_1$, \ldots, $t_n$, if $t_1\,\cdots\,t_n$
converges, then $t_1$, \ldots, $t_n$ and
$\overline{t_1}\;\cdots\;\overline{t_n}$ converge and
$\overline{t_1\,\cdots\,t_n} =
\overline{\overline{t_1}\;\cdots\;\overline{t_n}}$.  It will suffice
to use ordinary induction to prove that, for all $n\in\nats$, $P(n)$.
\begin{itemize}
\item (Basis Step)\quad $P(0)$ holds vacuously, because $0\geq 2$ is
  false.

\item (Inductive Step)\quad Suppose $n\in\nats$, and assume the
  inductive hypothesis, $P(n)$.  We must show that $P(n+1)$.  Suppose
  $n+1\geq 2$, $t_1$, \ldots, $t_{n+1}$ are closed terms, and
  $t_1\,\cdots\,t_{n+1}$ converges.  We must show that $t_1$, \ldots,
  $t_{n+1}$ and $\overline{t_1}\;\cdots\;\overline{t_{n+1}}$ converge
  and $\overline{t_1\,\cdots\,t_{n+1}} =
  \overline{\overline{t_1}\;\cdots\;\overline{t_{n+1}}}$.
  Since $n+1\geq 2$, we have that $n\geq 1$.  Hence there are
  two cases to consider.
  \begin{itemize}
  \item Suppose $n=1$.  Thus $n+1=2$, so that $t_1$ and $t_2$ are
    closed terms, and $t_1\,t_2$ converges.  We must show that $t_1$,
    $t_2$ and $\overline{t_1}\;\overline{t_2}$ converge, and that
    $\overline{t_1\,t_2} = \overline{\overline{t_1}\;\overline{t_2}}$,
    and this follows by Proposition~D.

  \item Suppose $n\geq 2$.  Thus $t_1$, \ldots, $t_n$, $t_{n+1}$ are
    closed terms, and $t_1\,\cdots\,t_n\,t_{n+1}$ converges.  We must
    show that $t_1$, \ldots, $t_n$, $t_{n+1}$ and
    $\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}$
    converge and $\overline{t_1\,\cdots\,t_n\,t_{n+1}} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}}$.
    Because $(t_1\,\cdots\,t_n)t_{n+1} = t_1\,\cdots\,t_n\,t_{n+1}$
    converges, Proposition~D tells us that $t_1\,\cdots\,t_n$,
    $t_{n+1}$ and $\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}$
    converge and
    \begin{displaymath}
      \overline{t_1\,\cdots\,t_n\,t_{n+1}} =
      \overline{(t_1\,\cdots\,t_n)t_{n+1}} =
      \overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}} .
    \end{displaymath}
    Because $n\geq 2$ and $t_1\,\cdots\,t_n$ converges, the inductive
    hypothesis tells us that $t_1$, \ldots, $t_n$ and
    $\overline{t_1}\;\cdots\;\overline{t_n}$ converge and
    $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$.  So, we have
    that $t_1$, \ldots, $t_n$ and $t_{n+1}$ converge.  Because
    $\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}$ converges,
    $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$ and
    $\overline{t_{n+1}} = \overline{\overline{t_{n+1}}}$, it follows
    that
    $\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;
    \overline{\overline{t_{n+1}}}$ converges.  Since
    $\overline{t_1}\;\cdots\;\overline{t_n}$, $\overline{t_{n+1}}$ and
    $\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;
    \overline{\overline{t_{n+1}}}$ converge, Proposition~E tells
    us that $\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}} =
    (\overline{t_1}\;\cdots\;\overline{t_n})\overline{t_{n+1}}$
    converges and
    \begin{displaymath}
      \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}}
      = \overline{(\overline{t_1}\;\cdots\;\overline{t_n})\overline{t_{n+1}}}
      = \overline{\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;\overline{\overline{t_{n+1}}}} .
    \end{displaymath}
    Since $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$ and
    $\overline{t_{n+1}} = \overline{\overline{t_{n+1}}}$, it follows that
    \begin{displaymath}
      \overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}} =
      \overline{\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;\overline{\overline{t_{n+1}}}} =
      \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}} .
    \end{displaymath}
    Finally,
    \begin{displaymath}
      \overline{t_1\,\cdots\,t_n\,t_{n+1}}
      = \overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}}
      = \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}} .
    \end{displaymath}
  \end{itemize}
\end{itemize}

\subsection*{Proposition G}

Let $P$ be the set of all $n\in\nats$ such that, if $n\geq 2$, then,
for all closed terms $t_1$, \ldots, $t_n$, if $t_1$, \ldots, $t_n$ and
$\overline{t_1}\;\cdots\;\overline{t_n}$ converge, then
$t_1\,\cdots\,t_n$ converges and $\overline{t_1\,\cdots\,t_n} =
\overline{\overline{t_1}\;\cdots\;\overline{t_n}}$.  It will suffice
to use ordinary induction to prove that, for all $n\in\nats$, $P(n)$.
\begin{itemize}
\item (Basis Step)\quad $P(0)$ holds vacuously, because $0\geq 2$ is
  false.

\item (Inductive Step)\quad Suppose $n\in\nats$ and assume the
  inductive hypothesis, $P(n)$.  We must show that $P(n+1)$.  Suppose
  $n+1\geq 2$, $t_1$, \ldots, $t_{n+1}$ are closed terms, and $t_1$,
  \ldots, $t_{n+1}$ and $\overline{t_1}\;\cdots\;\overline{t_{n+1}}$ converge.
  We must show that $t_1\,\cdots\,t_{n+1}$ converges and
  $\overline{t_1\,\cdots\,t_{n+1}} =
  \overline{\overline{t_1}\;\cdots\;\overline{t_{n+1}}}$.
  Since $n+1\geq 2$, we have that $n\geq 1$.  Hence there are two
  cases to consider.
  \begin{itemize}
  \item Suppose $n=1$.  Thus $n+1=2$, so that $t_1$ and $t_2$ are
    closed terms, $t_1$, $t_2$ and $\overline{t_1}\;\overline{t_2}$
    converge.  We must show that $t_1\,t_2$ converges and
    $\overline{t_1\,t_2} = \overline{\overline{t_1}\;\overline{t_2}}$,
    and this follows by Proposition~E.

  \item Suppose $n\geq 2$.  Thus $t_1$, \ldots, $t_n$, $t_{n+1}$ are
    closed terms, and $t_1$, \ldots, $t_n$, $t_{n+1}$ and
    $\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}$
    converge.  We must show that $t_1\,\cdots\,t_n\,t_{n+1}$ converges
    and $\overline{t_1\,\cdots\,t_n\,t_{n+1}} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}}$.
    Because
    $(\overline{t_1}\;\cdots\;\overline{t_n})\overline{t_{n+1}} =
    \overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}$
    converges, Proposition~D tells us that
    $\overline{t_1}\;\cdots\;\overline{t_n}$, $\overline{t_{n+1}}$ and
    $\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;
    \overline{\overline{t_{n+1}}}$ converge, and
    \begin{displaymath}
      \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}} =
      \overline{(\overline{t_1}\;\cdots\;\overline{t_n})\overline{t_{n+1}}} =
      \overline{\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;
        \overline{\overline{t_{n+1}}}} .
    \end{displaymath}
    Since $n\geq 2$, and $t_1$, \ldots, $t_n$ and
    $\overline{t_1}\;\cdots\;\overline{t_n}$ converge, the inductive
    hypothesis tells us that $t_1\,\cdots\,t_n$ converges and
    $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$.  So, we have
    that $t_1\,\cdots\,t_n$ and $t_{n+1}$ converge.  Because
    $\overline{\overline{t_1}\;\cdots\;\overline{t_n}}\;
    \overline{\overline{t_{n+1}}}$ converges,
    $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$ and
    $\overline{t_{n+1}} = \overline{\overline{t_{n+1}}}$, it follows
    that $\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}$ converges.
    Since $t_1\,\cdots\,t_n$, $t_{n+1}$ and
    $\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}$ converge,
    Proposition~E tells us that $t_1\,\cdots\,t_n\,t_{n+1}
    = (t_1\,\cdots\,t_n)t_{n+1}$ converges and
    \begin{displaymath}
      \overline{t_1\,\cdots\,t_n\,t_{n+1}} =
      \overline{(t_1\,\cdots\,t_n)t_{n+1}} =
      \overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}} .
    \end{displaymath}
    Since $\overline{t_1\,\cdots\,t_n} =
    \overline{\overline{t_1}\;\cdots\;\overline{t_n}}$ and
    $\overline{t_{n+1}} = \overline{\overline{t_{n+1}}}$, we have that
    $\overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}} =
    \overline{\overline{\overline{t_1}\;\cdots\overline{t_n}}\;
      \overline{\overline{t_{n+1}}}}$.  Hence, we have that
    \begin{displaymath}
      \overline{t_1\,\cdots\,t_n\,t_{n+1}} =
      \overline{\overline{t_1\,\cdots\,t_n}\;\overline{t_{n+1}}}  =
      \overline{\overline{\overline{t_1}\;\cdots\overline{t_n}}\;
        \overline{\overline{t_{n+1}}}} =
      \overline{\overline{t_1}\;\cdots\;\overline{t_n}\;\overline{t_{n+1}}} .
    \end{displaymath}
  \end{itemize}
\end{itemize}
\end{document}