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\newcommand{\dom}{\mathsf{dom}}
\newcommand{\ran}{\mathsf{ran}}
\newcommand{\goesto}{\mathrel{::=}}
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\newcommand{\Unit}{\mathsf{Unit}}
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\newcommand{\FV}{\mathsf{FV}}

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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
Model Answers
\end{center}

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

For all closed terms $s$ and $s'$, if $s\fun s'$, then one of the
following conditions holds:
\begin{description}
\item[\quad(E-App1)] there are closed terms $t_1$, $t'_1$ and $t_2$
  such that $s=t_1\,t_2$, $s'=t'_1\,t_2$ and $t_1\fun t'_1$;

\item[\quad(E-App2)] there are a closed value $v_1$ and closed terms
  $t_2$ and $t'_2$ such that $s=v_1\,t_2$, $s'=v_1\,t'_2$ and $t_2\fun
  t'_2$; or

\item[\quad(E-AppAbs)] there are a variable $x$, a term $t$ such that
  $\FV(t)\sub\{x\}$, and a closed value $v$ such that $s=(\lambda
  x.\,t)v$ and $s'=[x\mapsto v]t$.
\end{description}

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

Suppose $P$ is a binary relation on closed terms.
We sometimes write ``$P(t,t')$'' for ``$(t,t')\in P$''.
The \emph{principle of induction on the evaluation relation} says that,
\begin{center}
  for all closed terms $t$ and $t'$, if $t\fun t'$, then $P(t,t')$,
\end{center}
follows from showing
\begin{description}
\item[\quad(E-App1)] for all closed terms $t_1$, $t'_1$ and $t_2$, if
  $t_1\fun t'_1$ and $P(t_1,t'_1)$, then $P(t_1\,t_2, t'_1\,t_2)$;

\item[\quad(E-App2)] for all closed values $v_1$ and closed terms
  $t_2$ and $t'_2$, if $t_2\fun t'_2$ and $P(t_2,t'_2)$, then
  $P(v_1\,t_2, v_1\,t'_2)$;

\item[\quad(E-AppAbs)] for all variables $x$, terms $t$ such that
  $\FV(t)\sub\{x\}$, and closed values $v$, $P((\lambda x.\,t)v,
  [x\mapsto v]t)$.
\end{description}

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

For all contexts $\Gamma$, terms $s$ and types $T$,
if $\Gamma\vdash s\typrel T$, then one of the following conditions
holds:
\begin{description}
\item[\quad(T-Unit)] $s=\unit$ and $T=\Unit$;

\item[\quad(T-Var)] there is a variable $x$ such that $s=x$ and $(x,
  T)\in\Gamma$;

\item[\quad(T-Abs)] there are a variable $x$, a term $t$ and types
  $T_1$ and $T_2$ such that $s=\lambda x.\,t$, $T=T_1\fun T_2$ and
  $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$; or

\item[\quad(T-App)] there are terms $t_1$ and $t_2$ and types $T_1$
  and $T_2$ such that $s=t_1\,t_2$, $T=T_2$, $\Gamma\vdash t_1\typrel
  T_1\fun T_2$ and $\Gamma\vdash t_2\typrel T_1$.
\end{description}

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

Suppose $P$ is a ternary relation between contexts, terms and types.
We sometimes write ``$P(\Gamma,t,T)$'' for ``$(\Gamma,t,T)\in P$''.
The \emph{principle of induction on the typing relation} says that,
\begin{center}
  for all contexts $\Gamma$, terms $t$ and types $T$, if $\Gamma\vdash
  t\typrel T$, then $P(\Gamma,t,T)$
\end{center}
follows from showing
\begin{description}
\item[\quad(T-Unit)] for all contexts $\Gamma$, $P(\Gamma,\unit,\Unit)$;

\item[\quad(T-Var)] for all contexts $\Gamma$, variables $x$ and
  types $T$, if $(x,T)\in\Gamma$, then $P(\Gamma,x,T)$;

\item[\quad(T-Abs)] for all contexts $\Gamma$, variables $x$,
  terms $t$, and types $T_1$ and $T_2$, if
  $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$ and
  $P(\Gamma[x\mapsto T_1], t, T_2)$, then
  $P(\Gamma,\lambda x.\,t,T_1\fun T_2)$;

\item[\quad(T-App)] for all contexts $\Gamma$, terms $t_1$ and $t_2$,
  and types $T_1$ and $T_2$, if $\Gamma\vdash t_1\typrel T_1\fun T_2$,
  $\Gamma\vdash t_2\typrel T_1$, $P(\Gamma,t_1,T_1\fun T_2)$ and
  $P(\Gamma,t_2,T_1)$, then $P(\Gamma, t_1\,t_2, T_2)$.
\end{description}

\section*{Exercise 5}
\renewcommand{\thesection}{5}
\theoremnumber{1}

Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is a context}, t\eqtxtl{is
    a term}, T \eqtxt{is a type and} \FV(t)\sub\dom(\Gamma)}$.  We
use induction on the typing relation to prove that, for all contexts
$\Gamma$, terms $t$ and types $T$, if $\Gamma\vdash t\typrel T$, then
$P(\Gamma,t,T)$.
\begin{description}
\item[\quad(T-Unit)] Suppose $\Gamma$ is a context.  We must show that
  $P(\Gamma,\unit,\Unit)$, and this follows since $\FV(\unit) =
  \emptyset \sub \dom(\Gamma)$.

\item[\quad(T-Var)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $T$ is a type and $(x,T)\in\Gamma$.  We must show that
  $P(\Gamma,x,T)$, and this follows since
  $\FV(x)=\{x\}\sub\dom(\Gamma)$.

\item[\quad(T-Abs)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $t$ is a term, $T_1$ and $T_2$ are types, and $\Gamma[x\mapsto
  T_1]\vdash t\typrel T_2$, and assume the inductive hypothesis,
  $P(\Gamma[x\mapsto T_1], t, T_2)$.  We must show that
  $P(\Gamma,\lambda x.\,t,T_1\fun T_2)$.
  By the inductive hypothesis, we have that $\FV(t)\sub
  \dom(\Gamma[x\mapsto T_1]) = \dom(\Gamma)\cup\{x\}$.  Suppose
  $y\in\FV(\lambda x.\,t)$.  Since $\FV(\lambda x.\,t) =
  \FV(t)\setminus \{x\}$, it follows that $y\in\FV(t)$ and $y\neq x$.
  Thus $y\in\FV(t)\sub\dom(\Gamma)\cup\{x\}$.  But $y\neq x$,
  and thus $y\in\dom(\Gamma)$, as required.

\item[\quad(T-App)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$ are
  terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1\fun
  T_2$ and $\Gamma\vdash t_2\typrel T_1$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1\fun T_2)$ and $P(\Gamma,t_2,T_1)$.  We
  must show that $P(\Gamma, t_1\,t_2, T_2)$.  By the inductive hypothesis,
  we have that $\FV(t_1)\sub\dom(\Gamma)$ and $\FV(t_2)\sub\dom(\Gamma)$.
  Thus $\FV(t_1\,t_2)=\FV(t_1)\cup\FV(t_2)\sub\dom(\Gamma)\cup
  \dom(\Gamma) = \dom(\Gamma)$.
\end{description}

\section*{Exercise 6}
\renewcommand{\thesection}{6}
\theoremnumber{1}

Suppose $v$ is a closed value.
\begin{itemize}
\item Suppose $\emptyset\vdash v\typrel\Unit$.
  By inversion on the typing relation, there are four cases to
  consider.
  \begin{description}
  \item[\quad(T-Unit)] Suppose $v=\unit$ and $\Unit=\Unit$.  Then
    $v=\unit$.

  \item[\quad(T-Var)] Suppose there is a variable $x$ such that $v=x$
    and $(x, \Unit)\in\Gamma$.  But this contradicts the fact that
    $v$ is closed.  Thus $v=\unit$.

  \item[\quad(T-Abs)] Suppose there are a variable $x$, a term $t$ and types
    $T_1$ and $T_2$ such that $v=\lambda x.\,t$, $\Unit=T_1\fun T_2$ and
    $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$.  But 
    $\Unit=T_1\fun T_2$ is a contradiction, and thus $v=\unit$.

  \item[\quad(T-App)] Suppose there are terms $t_1$ and $t_2$ and
    types $T_1$ and $T_2$ such that $v=t_1\,t_2$, $\Unit=T_2$,
    $\Gamma\vdash t_1\typrel T_1\fun T_2$ and $\Gamma\vdash t_2\typrel
    T_1$.  Thus $t_1\,t_2=v$ is a value---contradiction.  Hence
    $v=\unit$.
\end{description}

\item Suppose $\emptyset\vdash v\typrel S_1\fun S_2$ for some types
  $S_1$ and $S_2$.  By inversion on the typing relation, there are
  four cases to consider.
  \begin{description}
  \item[\quad(T-Unit)] Suppose $v=\unit$ and $S_1\fun S_2=\Unit$.
    But $S_1\fun S_2=\Unit$ is a contradiction, and thus
    $v$ is an abstraction.

  \item[\quad(T-Var)] Suppose there is a variable $x$ such that $v=x$
    and $(x, S_1\fun S_2)\in\Gamma$.  But this contradicts the fact that
    $v$ is closed.  Thus $v$ is an abstraction.

  \item[\quad(T-Abs)] Suppose there are a variable $x$, a term $t$ and types
    $T_1$ and $T_2$ such that $v=\lambda x.\,t$, $S_1\fun S_2=T_1\fun T_2$ and
    $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$.  Thus $v$ is an abstraction.

  \item[\quad(T-App)] Suppose there are terms $t_1$ and $t_2$ and
    types $T_1$ and $T_2$ such that $v=t_1\,t_2$, $S_1\fun S_2=T_2$,
    $\Gamma\vdash t_1\typrel T_1\fun T_2$ and $\Gamma\vdash t_2\typrel
    T_1$.  Thus $t_1\,t_2=v$ is a value---contradiction.  Hence
    $v$ is an abstraction.
\end{description}
\end{itemize}

\section*{Exercise 7}
\renewcommand{\thesection}{7}
\theoremnumber{1}

We prove the proposition.  Let $P=\setof{(t,t')}{t\eqtxt{and}t'\eqtxt{are
closed terms, and, for all closed terms}t'', \eqtxt{if} t\fun t'',
\eqtxt{then} t'=t''}$.  We use induction on the evaluation relation to
show that, for all closed terms $t$ and $t'$, if $t\fun t'$,
then $P(t,t')$.  (Then, suppose $t$, $t'$ and $t''$ are closed terms,
$t\fun t'$ and $t\fun t''$.  Thus $P(t,t')$, so that, if $t\fun t''$,
then $t'=t''$.  But $t\fun t''$, and thus $t'=t''$.)
\begin{description}
\item[\quad(E-App1)] Suppose $t_1$, $t'_1$ and $t_2$ are closed terms,
  and $t_1\fun t'_1$, and assume the inductive hypothesis,
  $P(t_1,t'_1)$.  We must show that $P(t_1\,t_2, t'_1\,t_2)$.  Suppose
  $u$ is a closed term and $t_1\,t_2\fun u$.  Because $t_1\fun t'_1$,
  we have that $t_1$ is not a normal form, and thus that $t_1$ is not
  a value.  Thus, by inversion of the evaluation relation, only
  (E-App1) applies, and we have that there is a closed term $t''_1$
  such that $u=t''_1\,t_2$ and $t_1\fun t''_1$.  By the inductive
  hypothesis, it follows that $t'_1=t''_1$.  Thus $t'_1\,t_2 =
  t''_1\,t_2 = u$.

\item[\quad(E-App2)] Suppose $v_1$ is a closed value, $t_2$ and $t'_2$
  are closed terms, and $t_2\fun t'_2$, and assume the inductive
  hypothesis, $P(t_2,t'_2)$.  We must show that $P(v_1\,t_2,
  v_1\,t'_2)$.  Suppose $u$ is a closed term and $v_1\,t_2\fun u$.
  Because $t_2\fun t'_2$, we have that $t_2$ is not a normal form, and
  thus that $t_2$ is not a value.  Furthermore, $v_1$ is a value, and
  thus is a normal form.  Hence, by inversion of the evaluation
  relation, only (E-App2) applies, and we have that there is a closed
  term $t''_2$ such that $u=v_1\,t''_2$ and $t_2\fun t''_2$.  By the
  inductive hypothesis, it follows that $t'_2=t''_2$.  Thus $v_1\,t'_2
  = v_1t''_2 = u$.

\item[\quad(E-AppAbs)] Suppose $x$ is a variable, $t$ is a term such
  that $\FV(t)\sub\{x\}$, and $v$ is a value.  We must show that
  $P((\lambda x.\,t)v, [x\mapsto v]t)$.  Suppose $u$ is a closed term
  such that $(\lambda x.\,t)v\fun u$.  Because $\lambda x.\,t$ is a
  value, it is a normal form.  And, because $v$ is a value, it is a
  normal form.  Thus, by inversion of the evaluation relation, only
  (E-AppAbs) applies, and we have that $u=[x\mapsto v]t$, so that
  $[x\mapsto v]t = u$.
\end{description}

\section*{Exercise 8}
\renewcommand{\thesection}{8}
\theoremnumber{1}

We disprove the proposition.  Let $x$ be a variable.  We will show
that $\emptyset\vdash\lambda x.\,x\typrel\Unit\fun\Unit$ and
$\emptyset\vdash \lambda
x.\,x\typrel(\Unit\fun\Unit)\fun(\Unit\fun\Unit)$.
\begin{itemize}
\item Because $(x,\Unit)\in\{(x,\Unit)\}$, (T-Var) tells us that
  $\{(x,\Unit)\}\vdash x\typrel\Unit$.  But $\{(x,\Unit)\} =
  \emptyset[x\mapsto\Unit]$, and thus $\emptyset[x\mapsto\Unit]\vdash
  x\typrel\Unit$.  Thus $\emptyset\vdash\lambda
  x.\,x\typrel\Unit\fun\Unit$, by (T-Abs).

\item Because $(x,\Unit\fun\Unit)\in\{(x,\Unit\fun\Unit)\}$, (T-Var)
  tells us that $\{(x,\Unit\fun\Unit)\}\vdash x\typrel\Unit\fun\Unit$.
  But $\{(x,\Unit\fun\Unit)\} = \emptyset[x\mapsto\Unit\fun\Unit]$,
  and thus $\emptyset[x\mapsto\Unit\fun\Unit]\vdash
  x\typrel\Unit\fun\Unit$.  Thus $\emptyset\vdash\lambda
  x.\,x\typrel(\Unit\fun\Unit)\fun(\Unit\fun\Unit)$, by (T-Abs).
\end{itemize}

\section*{Exercise 9}
\renewcommand{\thesection}{9}
\theoremnumber{1}

We prove the theorem.  Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is
    a context}, t\eqtxtl{is a term}, T \eqtxt{is a type and, if}
  \Gamma=\emptyset, \eqtxt{then} t \eqtxtl{is a closed term that is not
    stuck}}$.  We use induction on the typing relation to prove that,
for all contexts $\Gamma$, terms $t$ and types $T$, if $\Gamma\vdash
t\typrel T$, then $P(\Gamma,t,T)$.  (Then, suppose $t$ is a closed
term, and $t$ is well-typed.  Thus $\emptyset\vdash t\typrel T$ for
some type $T$.  By the result of the induction, we have that
$P(\emptyset,t,T)$.  Thus, if $\emptyset = \emptyset$, then
$t$ is a closed term that is not stuck.  But $\emptyset=\emptyset$,
and thus $t$ is not stuck.)
\begin{description}
\item[\quad(T-Unit)] Suppose $\Gamma$ is a context.  We must show that
  $P(\Gamma,\unit,\Unit)$, and this follows because $\unit$ is a
  closed value, and so is a closed term that is not stuck.

\item[\quad(T-Var)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $T$ is a type and $(x,T)\in\Gamma$.  We must show that
  $P(\Gamma,x,T)$.  Suppose $\Gamma=\emptyset$.  We must show that $x$
  is a closed term that is not stuck.  But
  $(x,T)\in\Gamma=\emptyset$---contradiction.  Thus $x$ is
  a closed term that is not stuck.

\item[\quad(T-Abs)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $t$ is a term, $T_1$ and $T_2$ are types, and $\Gamma[x\mapsto
  T_1]\vdash t\typrel T_2$, and assume the inductive hypothesis,
  $P(\Gamma[x\mapsto T_1], t, T_2)$.  We must show that
  $P(\Gamma,\lambda x.\,t,T_1\fun T_2)$.  Suppose $\Gamma=\emptyset$.
  We must show that $\lambda x.\,t$ is a closed term that is not
  stuck.  Because $\emptyset[x\mapsto T_1]\vdash t\typrel T_2$, the
  Free Variables Lemma (Exercise~5) tells us that
  $\FV(t)\sub\dom(\emptyset[x\mapsto T_1]) = \dom(\emptyset)\cup \{x\}
  = \emptyset\cup\{x\}=\{x\}$.  Thus $\FV(\lambda x.\,t) =
  \FV(t)\setminus \{x\} \sub \{x\}\setminus\{x\} = \emptyset$, showing
  that $\lambda x.\,t$ is closed.  And $\lambda x.\,t$ is a value,
  showing that $\lambda x.\,t$ is not stuck.  Thus $\lambda x.\,t$ is a
  closed term that is not stuck.

\item[\quad(T-App)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$ are
  terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1\fun
  T_2$ and $\Gamma\vdash t_2\typrel T_1$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1\fun T_2)$ and $P(\Gamma,t_2,T_1)$.  We
  must show that $P(\Gamma, t_1\,t_2, T_2)$.  Suppose
  $\Gamma=\emptyset$.  We must show that $t_1\,t_2$ is a closed term
  that is not stuck.  By $P(\Gamma,t_1,T_1\fun T_2)$, we have that
  $t_1$ is a closed term that is not stuck.  And, by
  $P(\Gamma,t_2,T_1)$, $t_2$ is a closed term that is not stuck.  Thus
  $t_1\,t_2$ is closed.  Because $t_1$ is a closed term that is not
  stuck, there are two cases to consider.
  \begin{itemize}
  \item Suppose $t_1$ is not a normal form.  Thus there is a closed
    term $t'_1$ such that $t_1\fun t'_1$.  Hence $t_1\,t_2\fun
    t'_1\,t_2$, by (E-App1), showing that $t_1\,t_2$ is not a normal
    form, and so is not stuck.  Thus $t_1\,t_2$ is a closed term
    that is not stuck.

  \item Suppose $t_1$ is a closed value.  Because $t_2$ is a closed
    term that is not stuck, there are two subcases to consider.
    \begin{itemize}
    \item Suppose $t_2$ is not a normal form.  Thus there is a closed
      term $t'_2$ such that $t_2\fun t'_2$.  Hence $t_1\,t_2\fun
      t_1\,t'_2$, by (E-App2), because $t_1$ is a value, showing that
      $t_1\,t_2$ is not a normal form, and so is not stuck.
      Thus $t_1\,t_2$ is a closed term that is not stuck.

    \item Suppose $t_2$ is closed value.  Because $t_1$ is a closed
      value and $\emptyset\vdash t_1\typrel T_1\fun T_2$, the
      Canonical Forms Lemma (Exercise~6) tells us that $t_1$ is an
      abstraction.  Thus $t_1=\lambda x.\,u$ for a variable $x$ and a
      term $u$.  Because $t_2$ is a value, we have that
      $t_1\,t_2=(\lambda x.\,u)t_2\fun [x\mapsto t_2]u$, by
      (E-AppAbs), showing that $t_1\,t_2$ is not a normal form, and so
      is not stuck.  Thus $t_1\,t_2$ is a closed term that is not stuck.
    \end{itemize}
  \end{itemize}
\end{description}

\section*{Exercise 10}
\renewcommand{\thesection}{10}
\theoremnumber{1}

We prove the theorem.  First, we prove two supporting lemmas.

\begin{lemma}[Weakening]
\label{Weakening}
For all contexts $\Gamma$ and $\Gamma'$, terms $t$ and types $T$, if
$\Gamma\vdash t\typrel T$ and $\Gamma\sub\Gamma'$, then $\Gamma'\vdash
t\typrel T$.
\end{lemma}

\begin{proof}
Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is a context}, t\eqtxtl{is
    a term}, T \eqtxt{is a type and, for all contexts}\Gamma',
    \eqtxt{if} \Gamma\sub\Gamma', \eqtxt{then} \Gamma'\vdash
    t\typrel T}$.  We
use induction on the typing relation to prove that, for all contexts
$\Gamma$, terms $t$ and types $T$, if $\Gamma\vdash t\typrel T$, then
$P(\Gamma,t,T)$.
(Then, suppose $\Gamma$ and $\Gamma'$ are contexts, $t$ is a term,
$T$ is a type, $\Gamma\vdash t\typrel T$ and $\Gamma\sub\Gamma'$.
By the result of the induction, $P(\Gamma,t,T)$.  Thus,
since $\Gamma\sub\Gamma'$, we have that $\Gamma'\vdash t\typrel T$.)
\begin{description}
\item[\quad(T-Unit)] Suppose $\Gamma$ is a context.  We must show that
  $P(\Gamma,\unit,\Unit)$.  Suppose $\Gamma'$ is a context and
  $\Gamma\sub\Gamma'$.  We must show that
  $\Gamma'\vdash\unit\typrel\Unit$, and this follows by (T-Unit).

\item[\quad(T-Var)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $T$ is a type and $(x,T)\in\Gamma$.  We must show that
  $P(\Gamma,x,T)$.  Suppose $\Gamma'$ is a context and $\Gamma\sub
  \Gamma'$.  We must show that $\Gamma'\vdash x\typrel T$, and this
  follows by (T-Var), since $(x,T)\in\Gamma\sub\Gamma'$.

\item[\quad(T-Abs)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $t$ is a term, $T_1$ and $T_2$ are types, and $\Gamma[x\mapsto
  T_1]\vdash t\typrel T_2$, and assume the inductive hypothesis,
  $P(\Gamma[x\mapsto T_1], t, T_2)$.  We must show that
  $P(\Gamma,\lambda x.\,t,T_1\fun T_2)$.  Suppose $\Gamma'$ is a
  context and $\Gamma\sub\Gamma'$.  We must show that
  $\Gamma'\vdash\lambda x.\,t\typrel T_1\fun T_2$.  Because of
  (T-Abs), it will suffice to show that $\Gamma'[x\mapsto T_1]\vdash
  t\typrel T_2$.  Thus, by the inductive hypothesis, it will suffice to
  show that $\Gamma[x\mapsto T_1]\sub\Gamma'[x\mapsto T_1]$.  Suppose
  $(y,T)\in\Gamma[x\mapsto T_1]$.  We must show that
  $(y,T)\in\Gamma'[x\mapsto T_1]$.  There are two cases to consider.
  \begin{itemize}
  \item Suppose $y=x$.  Because $(y,T)\in\Gamma[x\mapsto T_1]$,
    we have that $\Gamma[x\mapsto T_1](y) = T$.  But
    $y=x$, and thus $\Gamma[x\mapsto T_1](y) = T_1$, so that
    $T=T_1$.  Thus $(y,T)=(x,T_1)\in\Gamma'[x\mapsto T_1]$.

  \item Suppose $y\neq x$.  Since $\Gamma[x\mapsto T_1](y)=T$ and
    $y\neq x$, it follows that $T=\Gamma[x\mapsto T_1](y) =
    \Gamma(y)$, so that $(y,T)\in\Gamma\sub\Gamma'$.  Thus
    $\Gamma'[x\mapsto T_1](y) = \Gamma'(y)=T$, showing that
    $(y,T)\in\Gamma'[x\mapsto T_1]$.
  \end{itemize}

\item[\quad(T-App)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$ are
  terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1\fun
  T_2$ and $\Gamma\vdash t_2\typrel T_1$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1\fun T_2)$ and $P(\Gamma,t_2,T_1)$.  We
  must show that $P(\Gamma, t_1\,t_2, T_2)$.  Suppose $\Gamma'$ is
  a context and $\Gamma\sub\Gamma'$.  We must show that
  $\Gamma'\vdash t_1\,t_2\typrel T_2$.  By the inductive hypothesis,
  we have that $\Gamma'\vdash t_1\typrel T_1\fun T_2$ and
  $\Gamma'\vdash t_2\typrel T_1$.  Thus, by (T-App), it follows
  that $\Gamma'\vdash t_1\,t_2\typrel T_2$.
\end{description}
\end{proof}

\begin{lemma}[Typing of Substitutions]
\label{Substitution}
For all contexts $\Gamma$, variables $y$, terms $t$, values $v$ and
types $T,T'$, if $\Gamma[y\mapsto T']\vdash t\typrel T$ and
$\emptyset\vdash v\typrel T'$, then $\Gamma\vdash[y\mapsto v]t\typrel
T$.
\end{lemma}

\begin{proof}
It suffices to show that, for all variables $y$, values $v$ and types $T'$,
if $\emptyset\vdash v\typrel T'$, then, for all
contexts $\Gamma$, terms $t$ and types $T$, if $\Gamma[y\mapsto T']\vdash
t\typrel T$, then $\Gamma\vdash[y\mapsto v]t\typrel T$.

Suppose $y$ is a variable, $v$ is a value, $T'$ is a type, and
$\emptyset\vdash v\typrel T'$.  We must show that, for all 
contexts $\Gamma$, terms $t$ and types $T$, if $\Gamma[y\mapsto T']\vdash
t\typrel T$, then $\Gamma\vdash[y\mapsto v]t\typrel T$.

Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is a context}, t\eqtxtl{is
    a term}, T \eqtxt{is a type and, for all contexts}\Gamma',
    \eqtxt{if} \Gamma=\Gamma'[y\mapsto T'], \eqtxt{then} \Gamma'\vdash
    [y\mapsto v]t\typrel T}$.  We
use induction on the typing relation to prove that, for all contexts
$\Gamma$, terms $t$ and types $T$, if $\Gamma\vdash t\typrel T$, then
$P(\Gamma,t,T)$.
(Then, suppose $\Gamma$ is a context, $t$ is a term, $T$ is a type and
$\Gamma[y\mapsto T']\vdash t\typrel T$.  We must show that
$\Gamma\vdash[y\mapsto v]t\typrel T$.  By the result of the induction,
we have that $P(\Gamma[y\mapsto T'],t,T)$, so that for all contexts $\Gamma'$,
if $\Gamma[y\mapsto T']=\Gamma'[y\mapsto T']$, then $\Gamma'\vdash
[y\mapsto v]t\typrel T$.  Thus $\Gamma\vdash[y\mapsto v]t\typrel T$.)

\begin{description}
\item[\quad(T-Unit)] Suppose $\Gamma$ is a context.  We must show that
  $P(\Gamma,\unit,\Unit)$.  Suppose $\Gamma'$ is a context and
  $\Gamma=\Gamma'[y\mapsto T']$.  We must show that
  $\Gamma'\vdash[y\mapsto v]\unit\typrel\Unit$.  Since $[y\mapsto
  v]\unit = \unit$, it will suffice to show $\Gamma'\vdash
  \unit\typrel\Unit$, which follows by (T-Unit).

\item[\quad(T-Var)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $T$ is a type and $(x,T)\in\Gamma$.  We must show that
  $P(\Gamma,x,T)$.  Suppose $\Gamma'$ is a context and
  $\Gamma=\Gamma'[y\mapsto T']$.  We must show that
  $\Gamma'\vdash[y\mapsto v]x\typrel T$.  There are two cases
  to consider.
  \begin{itemize}
  \item Suppose $y=x$.  Thus $[y\mapsto v]x = v$, so we must show that
    $\Gamma'\vdash v\typrel T$.  Because $\emptyset\vdash v\typrel
    T'$, the Weakening Lemma (Lemma~\ref{Weakening}), tells us that
    $\Gamma'\vdash v\typrel T'$.  Because $(x,T)\in\Gamma=
    \Gamma'[y\mapsto T']$ and $x=y$, we have that $T =
    \Gamma'[y\mapsto T'](x) = T'$.  Thus $\Gamma'\vdash v\typrel T$,
    as required.

  \item Suppose $y\neq x$.  Thus $[y\mapsto v]x = x$, so we must show
    that $\Gamma'\vdash x\typrel T$.  Because $(x,T)\in\Gamma =
    \Gamma'[y\mapsto T']$ and $x\neq y$, we have that
    $T=\Gamma'[y\mapsto T'](x)= \Gamma'(x)$.  Thus $(x,T)\in\Gamma'$,
    so that $\Gamma'\vdash x\typrel T$ follows by (T-Var).
  \end{itemize}

\item[\quad(T-Abs)] Suppose $\Gamma$ is a context, $x$ is a variable,
  $t$ is a term, $T_1$ and $T_2$ are types, and $\Gamma[x\mapsto
  T_1]\vdash t\typrel T_2$, and assume the inductive hypothesis,
  $P(\Gamma[x\mapsto T_1], t, T_2)$.  We must show that
  $P(\Gamma,\lambda x.\,t,T_1\fun T_2)$.  Suppose $\Gamma'$ is a
  context and $\Gamma=\Gamma'[y\mapsto T']$.  We must show that
  $\Gamma'\vdash[y\mapsto v]\lambda x.\,t\typrel T_1\fun T_2$.
  There are two cases to consider.
  \begin{itemize}
  \item Suppose $x=y$.  Then $[y\mapsto v]\lambda x.\,t = \lambda
    x.\,t$, so it will suffice to show that $\Gamma'\vdash\lambda
    x.\,t\typrel T_1\fun T_2$.  By (T-Abs), it will suffice to show
    that $\Gamma'[x\mapsto T_1]\vdash t\typrel T_2$.  Because
    $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$ and
    $\Gamma=\Gamma'[y\mapsto T']$, we have that $\Gamma'[y\mapsto
    T'][x\mapsto T_1]\vdash t\typrel T_2$.  Thus it will suffice to
    show that $\Gamma'[x\mapsto T_1] = \Gamma'[y\mapsto T'][x\mapsto
    T_1]$.  Because $x=y$, both of these contexts have domain
    $\dom(\Gamma')\cup\{x\}$, both yield $T_1$ when called with
    $x$, and both yield $\Gamma'(z)$ when called with
    some $z\in\dom(\Gamma')\setminus\{x\}$.  Thus, they are equal.

  \item Suppose $x\neq y$.  Then $[y\mapsto v]\lambda x.\,t = \lambda
    x.\,[y\mapsto v] t$, so it will suffice to show that
    $\Gamma'\vdash\lambda x.\,[y\mapsto v]t \typrel T_1\fun T_2$.  By
    (T-Abs), it will suffice to show that $\Gamma'[x\mapsto T_1]\vdash
    [y\mapsto v]t\typrel T_2$.  By the inductive hypothesis, we have
    that, for all contexts $\Gamma'$, if $\Gamma[x\mapsto
    T_1]=\Gamma'[y\mapsto T']$, then $\Gamma'\vdash [y\mapsto
    v]t\typrel T_2$.  Thus, if $\Gamma[x\mapsto T_1]=\Gamma'[x\mapsto
    T_1][y\mapsto T']$, then $\Gamma'[x\mapsto T_1]\vdash[y\mapsto
    v]t\typrel T_2$.  Thus, it will suffice to show that
    $\Gamma[x\mapsto T_1]=\Gamma'[x\mapsto T_1][y\mapsto T']$.
    Because $\Gamma=\Gamma'[y\mapsto T']$, it will suffice to show
    that $\Gamma'[y\mapsto T'][x\mapsto T_1]=\Gamma'[x\mapsto
    T_1][y\mapsto T']$.  The contexts both have domain $\dom(\Gamma')
    \cup \{x\} \cup \{y\}$, both yield $T_1$ when called with $x$,
    both yield $T'$ when called with $y$, and both yield $\Gamma'(z)$
    when called with a $z\in\dom(\Gamma')\setminus\{x,y\}$.  Thus,
    they are equal.
  \end{itemize}

\item[\quad(T-App)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$ are
  terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1\fun
  T_2$ and $\Gamma\vdash t_2\typrel T_1$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1\fun T_2)$ and $P(\Gamma,t_2,T_1)$.  We
  must show that $P(\Gamma, t_1\,t_2, T_2)$.  Suppose $\Gamma'$ is a
  context and $\Gamma=\Gamma'[y\mapsto T']$.  We must show that
  $\Gamma'\vdash[y\mapsto v](t_1\,t_2)\typrel T_2$.  Since $[y\mapsto
  v](t_1\,t_2) = [y\mapsto v]t_1\,[y\mapsto v]t_2$, we must show that
  $\Gamma'\vdash[y\mapsto v]t_1\,[y\mapsto v]t_2\typrel T_2$.  By
  $P(\Gamma,t_1,T_1\fun T_2)$, we have that $\Gamma'\vdash[y\mapsto
  v]t_1\typrel T_1\fun T_2$.  And, by $P(\Gamma,t_2,T_1)$, we have
  that $\Gamma'\vdash[y\mapsto v]t_2\typrel T_1$.  Thus, by T-App, we
  have that $\Gamma'\vdash [y\mapsto v]t_1\,[y\mapsto v]t_2\typrel
  T_2$.
\end{description}
\end{proof}

Now, we prove the Preservation Theorem.  Let
$P=\setof{(t,t')}{t\eqtxt{and}t'\eqtxtl{are closed terms, and, for all}
    \eqtxtr{types} T,\eqtxt{if}\emptyset\vdash t\typrel T,\eqtxt{then}
  \emptyset\vdash t'\typrel T}$.
We use induction on the evaluation relation to show that, for all closed
terms $t$ and $t'$, if $t\fun t'$, then $P(t,t')$.  (Then, suppose
$t$ and $t'$ are closed terms, $T$ is a type,
$\emptyset\vdash t\typrel T$ and $t\fun t'$.
By the result of the induction, we have that $P(t,t')$.  But
$\emptyset\vdash t\typrel T$, and thus $\emptyset\vdash t'\typrel T$.)
\begin{description}
\item[\quad(E-App1)] Suppose $t_1$, $t'_1$ and $t_2$ are closed terms,
  and $t_1\fun t'_1$, and assume the inductive hypothesis,
  $P(t_1,t'_1)$.  We must show that $P(t_1\,t_2, t'_1\,t_2)$.
  Suppose $T$ is a type and $\emptyset\vdash t_1\,t_2\typrel T$.
  We must show that $\emptyset\vdash t'_1\,t_2\typrel T$.
  By inversion of the typing relation, we have that there is a type
  $T'$ such that $\emptyset\vdash t_1\typrel T'\fun T$ and
  $\emptyset\vdash t_2\typrel T'$.  By the inductive hypothesis,
  it follows that $\emptyset\vdash t'_1\typrel T'\fun T$.  Thus by
  (T-App), we have that $\emptyset\vdash t'_1\,t_2\typrel T$.

\item[\quad(E-App2)] Suppose $v_1$ is a closed value, $t_2$ and $t'_2$
  are closed terms, and $t_2\fun t'_2$, and assume the inductive
  hypothesis, $P(t_2,t'_2)$.  We must show that $P(v_1\,t_2,
  v_1\,t'_2)$.
  Suppose $T$ is a type and $\emptyset\vdash v_1\,t_2\typrel T$.
  We must show that $\emptyset\vdash v_1\,t'_2\typrel T$.
  By inversion of the typing relation, we have that there is a type
  $T'$ such that $\emptyset\vdash v_1\typrel T'\fun T$ and
  $\emptyset\vdash t_2\typrel T'$.  By the inductive hypothesis,
  it follows that $\emptyset\vdash t'_2\typrel T'$.  Thus by
  (T-App), we have that $\emptyset\vdash v_1\,t'_2\typrel T$.

\item[\quad(E-AppAbs)] Suppose $x$ is a variable, $t$ is a term such
  that $\FV(t)\sub\{x\}$, and $v$ is a value.  We must show that
  $P((\lambda x.\,t)v, [x\mapsto v]t)$.  Suppose $T$ is a type and
  $\emptyset\vdash(\lambda x.\,t)v\typrel T$.  We must show that
  $\emptyset\vdash[x\mapsto v]t\typrel T$.  By the inversion of the
  typing relation, we have that there is a type $T'$ such that
  $\emptyset\vdash\lambda x.\,t\typrel T'\fun T$ and $\emptyset\vdash
  v\typrel T'$.  Because $\emptyset\vdash\lambda x.\,t\typrel T'\fun
  T$, by another inversion of the typing relation, we have that
  $\emptyset[x\mapsto T']\vdash t\typrel T$.  Since
  $\emptyset[x\mapsto T']\vdash t\typrel T$ and $\emptyset\vdash
  v\typrel T'$, Lemma~\ref{Substitution} tells us that
  $\emptyset\vdash[x\mapsto v]t\typrel T$.
\end{description}
\end{document}

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