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\begin{document}
\begin{center}\large\bf
CIS 705 --- Programming Languages --- Spring 2009
\end{center}
\begin{center}\Large\bf
Assignment 5
\end{center}
\begin{center}\large\bf
Due by 2:30 p.m.\ on Tuesday, April 21
\end{center}
\begin{center}
The context for this assignment is Chapters~9 and 11 of \emph{TAPL}.
\end{center}

\section*{Re-formulating the Simply Typed Lambda Calculus}

If $f$ is a function and $x,y$ are elements of our universe, we define
the function $f[x\mapsto y]$ from $\dom(f)\cup\{x\}$ to $\ran(f)\cup\{y\}$
by, for all $z\in\dom(f)\cup\{x\}$,
\begin{displaymath}
f[x\mapsto y](z) =
\left\{ \begin{array}{ll}
y,    & \eqtxtr{if} z = x, \\
f(z), & \eqtxtr{if} z\neq x .
\end{array} \right.
\end{displaymath}

We reformulate the syntax of the simply typed lambda calculus with
$\Unit$ as follows, where variables $x$ are as usual.  Our
\emph{types} are defined by:
\begin{alignat*}{2}
T &\goesto &\qquad \eqtxt{types:}\\
&\quad\Unit &\qquad \eqtxt{$\Unit$ type}\\
&\quad T\fun T &\qquad \eqtxt{type of functions}
\end{alignat*}
As usual, $\fun$ associates to the right.
Our \emph{terms} are defined by:
\begin{alignat*}{2}
t &\goesto &\qquad \eqtxt{terms:}\\
&\quad \unit &\qquad \eqtxt{constant $\unit$}\\
&\quad x &\qquad \eqtxt{variables}\\
&\quad \lambda x.\,t &\qquad \eqtxt{abstraction}\\
&\quad t\,t &\qquad \eqtxt{application}
\end{alignat*}
And our \emph{values} are defined by:
\begin{alignat*}{2}
v &\goesto &\qquad \eqtxt{values:}\\
&\quad\unit &\qquad \eqtxt{constant $\unit$}\\
&\quad \lambda x.\,t &\qquad \eqtxt{abstraction value}
\end{alignat*}
As usual, application associates to the left and abstractions
extend as far as possible.

In contrast to TAPL's approach, we do \emph{not} identify abstractions
up to the renaming of bound variables, so that, e.g.,
$\lambda x.x = \lambda y.y$ iff $x=y$.
The \emph{free variables} of a term $t$ ($\FV(t)$)
is defined recursively as follows:
\begin{align*}
\FV(\unit) &= \emptyset , \\
\FV(x) &= \{x\} , \\
\FV(\lambda x.\, t) &= \FV(t)\setminus \{x\} , \\
\FV(t_1\,t_2) &= \FV(t_1)\cup\FV(t_2) .
\end{align*}
A term is \emph{closed} iff it has no free variables; otherwise it is
\emph{open}.  The \emph{substitution of} a \emph{closed} term $s$
\emph{for the free occurrences of} a variable $x$ \emph{in} a term $t$
($[x\mapsto s]t$) is defined recursively by:
\begin{align*}
[x\mapsto s]\unit &= \unit , \\
[x\mapsto s]y &=
\left\{ \begin{array}{ll}
s, & \eqtxtr{if} y = x, \\
y, & \eqtxtr{if} y\neq x ,
\end{array} \right. \\
[x\mapsto s](\lambda y.\,t) &=
\left\{ \begin{array}{ll}
\lambda y.\,t, & \eqtxtr{if} y = x, \\
\lambda y.\,[x\mapsto s]t, & \eqtxtr{if} y\neq x ,
\end{array} \right. \\
[x\mapsto s](t_1\,t_2) &= [x\mapsto s]t_1\,[x\mapsto s]t_2 .
\end{align*}

The \emph{evaluation relation} \fbox{$t\fun t'$} between \emph{closed}
terms is defined inductively by:
\begin{alignat*}{2}
&\frac{t_1\fun t'_1}%
{t_1\,t_2 \fun t'_1\,t_2} & \quad\eqtxt{(E-App1)} \\[.2cm]
&\frac{t_2\fun t'_2}%
{v_1\,t_2 \fun v_1\,t'_2} & \quad\eqtxt{(E-App2)} \\[.2cm]
&(\lambda x.\,t)v\fun [x\mapsto v]t & \quad\eqtxt{(E-AppAbs)}
\end{alignat*}
So, in the above, $t_1$, $t'_1$, $t_2$, $t'_2$, $v$ and $v_1$ range over
\emph{closed} terms, whereas $\FV(t)\sub\{x\}$.
A closed term $t$ is a \emph{normal form} iff there is no
closed term $t'$ such that $t\fun t'$.  A closed term $t$ is
\emph{stuck} iff $t$ is a normal form but $t$ is not a value.

A \emph{typing context} (or just \emph{context}) $\Gamma$ is a function
such that $\dom(\Gamma)$ is a finite subset of the variables,
and $\ran(\Gamma)$ is a subset of the types.
The \emph{typing relation} \fbox{$\Gamma\vdash t\typrel T$} between
typing contexts, terms and types is defined inductively by:
\begin{alignat*}{2}
&\Gamma\vdash\unit:\Unit & \quad\eqtxt{(T-Unit)} \\[.2cm]
&\frac{(x, T)\in\Gamma}%
{\Gamma\vdash x \typrel T} & \quad\eqtxt{(T-Var)} \\[.2cm]
&\frac{\Gamma[x\mapsto T_1]\vdash t\typrel T_2}
{\Gamma\vdash\lambda x.\,t\typrel T_1\fun T_2} & \quad\eqtxt{(T-Abs)} \\[.2cm]
&\frac{\Gamma\vdash t_1\typrel T_1\fun T_2\qquad
\Gamma\vdash t_2\typrel T_1}%
{\Gamma\vdash t_1\,t_2 \typrel T_2} & \quad\eqtxt{(T-App)}
\end{alignat*}
We say that a closed term $t$ is \emph{well-typed} iff
$\emptyset\vdash t\typrel T$ for some type $T$.

\section*{Exercise 1 (5 Points)}

State the Inversion Lemma for the Evaluation Relation.
(You don't have to prove that it's correct.)

A consequence of this lemma will be that closed values are normal
forms, and you may use this fact without proof in what follows.

\section*{Exercise 2 (5 Points)}

State the Principle of Induction on the Evaluation Relation.
(You don't have to prove that it's correct.)

\section*{Exercise 3 (5 Points)}

State the Inversion Lemma for the Typing Relation.
(You don't have to prove that it's correct.)

\section*{Exercise 4 (5 Points)}

State the Principle of Induction on the Typing Relation.
(You don't have to prove that it's correct.)

\section*{Exercise 5 (10 Points)}

Prove the Free Variables Lemma:
\begin{quotation}
\noindent For all contexts $\Gamma$, terms $t$ and types $T$,
if $\Gamma\vdash t\typrel T$, then $\FV(t)\sub\dom(\Gamma)$.
\end{quotation}

\section*{Exercise 6 (5 Points)}

Prove the Canonical Forms Lemma:
\begin{quotation}
\noindent For all closed values $v$:
\begin{itemize}
\item if $\emptyset\vdash v\typrel\Unit$, then $v=\unit$.

\item if $\emptyset\vdash v\typrel T_1\fun T_2$, for some types $T_1$ and
  $T_2$, then $v$ is an abstraction.
\end{itemize}
\end{quotation}

\section*{Exercise 7 (10 Points)}

Either prove or disprove the Determinacy of Evaluation Proposition:
\begin{quotation}
  \noindent For all closed terms $t$, $t'$ and $t''$, if $t\fun t'$
  and $t\fun t''$, then $t'=t''$.
\end{quotation}

\section*{Exercise 8 (10 Points)}

Either prove or disprove the Uniqueness of Typing Proposition:
\begin{quotation}
  \noindent For all contexts $\Gamma$, terms $t$ and types $T$ and
  $T'$, if $\Gamma\vdash t\typrel T$ and $\Gamma\vdash t\typrel T'$,
  then $T=T'$.
\end{quotation}

\section*{Exercise 9 (20 Points)}

Either prove or disprove the Progress Theorem:
\begin{quotation}
\noindent For all closed terms $t$, if $t$ is well-typed, then
$t$ is not stuck.
\end{quotation}

\section*{Exercise 10 (25 Points)}

Either prove or disprove the Preservation Theorem:
\begin{quotation}
  \noindent For all closed terms $t$ and $t'$ and types $T$, if
  $\emptyset\vdash t\typrel T$ and $t\fun t'$, then $\emptyset\vdash
  t'\typrel T$.
\end{quotation}
\end{document}