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\begin{document}
\begin{center}\large\bf
CIS 705 --- Programming Languages --- Spring 2009
\end{center}
\begin{center}\Large\bf
Assignment 6
\end{center}
\begin{center}\large\bf
Model Answers
\end{center}

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

Let $P=\setof{t}{t\eqtxt{is a term and, for all substitutions} \sigma,
  \eqtxt{if} \FV(t)\cap\dom(\sigma)=\emptyset, \eqtxt{then} t\,\sigma
  = t}$.
We use structural induction to show that, for all terms $t$,
$P(t)$.
\begin{description}
\item[\quad(Unit Constant)] We must show $P(\unit)$.  Suppose $\sigma$
  is a substitution and $\FV(\unit)\cap\dom(\sigma)=\emptyset$.  We
  have that $\unit\,\sigma = \unit$.

\item[\quad(Pairing)] Suppose $t_1$ and $t_2$ are terms, and assume
  the inductive hypothesis: $P(t_1)$ and $P(t_2)$.  We must show that
  $P((t_1,t_2))$.  Suppose $\sigma$ is a substitution and
  $\FV((t_1,t_2))\cap\dom(\sigma)=\emptyset$.  Because $\FV((t_1,t_2))
  = \FV(t_1)\cup\FV(t_2)$, it follows that
  $\FV(t_1)\cap\dom(\sigma)=\emptyset$ and
  $\FV(t_2)\cap\dom(\sigma)=\emptyset$.  Thus, by the inductive
  hypothesis, we have that $t_1\,\sigma = t_1$ and $t_2\,\sigma =
  t_2$, so that $(t_1,t_2)\,\sigma = (t_1\,\sigma, t_2\,\sigma) =
  (t_1,t_2)$.

\item[\quad(First Projection)] Similar to pairing, but slightly
  simpler.

\item[\quad(Second Projection)] Similar to pairing, but slightly
  simpler.

\item[\quad(Variable)] Suppose $x$ is a variable.  We must show
  $P(x)$.  Suppose $\sigma$ is a substitution and
  $\FV(x)\cap\dom(\sigma)=\emptyset$.  Since $\FV(x) = \{x\}$, this
  means that $x\not\in\dom(\sigma)$.  Thus $x\,\sigma = x$.

\item[\quad(Abstraction)] Suppose $x$ is a variable, $T$ is a type,
  and $t$ is a term, and assume the inductive hypothesis: $P(t)$.  We must
  show $P(\lambda x\typrel T.\,t)$.  Suppose $\sigma$ is a substitution and
  $\FV(\lambda x\typrel T.\,t)\cap\dom(\sigma)=\emptyset$.  We must
  show that $(\lambda x\typrel T.\,t)\,\sigma = \lambda x\typrel
  T.\,t$.  Because $(\lambda x\typrel T.\,t)\,\sigma = \lambda
  x\typrel T.\,t\,(\sigma/\{x\})$, it will suffice to show that
  $t\,(\sigma/\{x\}) = t$.  Thus, by the inductive hypothesis, it will
  suffice to show that $\FV(t)\cap\dom(\sigma/\{x\})=\emptyset$.
  Since $\dom(\sigma/\{x\}) = \dom(\sigma)\setminus\{x\}$, we must
  show that $\FV(t)\cap(\dom(\sigma)\setminus\{x\}) = \emptyset$.
  Suppose, toward a contradiction, that $z\in\FV(t)$ and
  $z\in\dom(\sigma)\setminus\{x\}$.  Thus $z\in\dom(\sigma)$ and
  $z\neq x$.  Since $z\in\FV(t)$ and $z\neq x$, we have that $z\in
  \FV(t)\setminus\{x\} = \FV(\lambda x\typrel T.\,t)$.  Hence
  $z\in\FV(\lambda x\typrel T.\,t)$ and $z\in\dom(\sigma)$,
  contradicting our assumption that $\FV(\lambda x\typrel
  T.\,t)\cap\dom(\sigma)=\emptyset$.  Thus
  $\FV(t)\cap(\dom(\sigma)\setminus\{x\}) = \emptyset$, as required.

\item[\quad(Application)] Similar to pairing.
\end{description}

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

\begin{lemma}
\label{MinusLem1} For all sets $A$, $B$ and $C$,
$(A\setminus C)\cup(B\setminus C) = (A\cup B)\setminus C$.
\end{lemma}

\begin{proof}
Suppose $A$, $B$ and $C$ are sets.  We must show that
$(A\setminus C)\cup(B\setminus C) \sub (A\cup B)\setminus C \sub
(A\setminus C)\cup(B\setminus C)$.
\begin{itemize}
\item ($(A\setminus C)\cup(B\setminus C) \sub (A\cup B)\setminus C$)\quad
  Suppose $x\in(A\setminus C)\cup(B\setminus C)$.  There are two
  cases to consider.
  \begin{itemize}
  \item Suppose $x\in A\setminus C$, so that $x\in A$ and $x\not\in C$.
    Thus $x\in A\cup B$ and $x\not\in C$, so that
    $x\in(A\cup B)\setminus C$.

  \item Suppose $x\in B\setminus C$, so that $x\in B$ and $x\not\in C$.
    Thus $x\in A\cup B$ and $x\not\in C$, so that
    $x\in(A\cup B)\setminus C$.
  \end{itemize}

\item ($(A\cup B)\setminus C\sub(A\setminus C)\cup(B\setminus C)$)\quad
  Suppose $x\in(A\cup B)\setminus C$.  Thus $x\in A\cup B$ and
  $x\not\in C$.  Thus there are two cases to consider.
  \begin{itemize}
  \item Suppose $x\in A$.  Thus $x\in A$ and $x\not\in C$, so that
    $x\in(A\setminus C)\sub(A\setminus C)\cup(B\setminus C)$.
    
  \item Suppose $x\in B$.  Thus $x\in B$ and $x\not\in C$, so that
    $x\in(B\setminus C)\sub(A\setminus C)\cup(B\setminus C)$.
  \end{itemize}
\end{itemize}
\end{proof}

Let $P=\setof{t}{t\eqtxt{is a term and, for all substitutions}
  \sigma, \FV(t\,\sigma) = FV(t)\setminus\dom(\sigma)}$.
We use structural induction to show that, for all terms $t$, $P(t)$.
\begin{description}
\item[\quad(Unit Constant)] We must show $P(\unit)$.  Suppose $\sigma$
  is a substitution.  We have that $\FV(\unit\,\sigma) = \FV(\unit) =
  \emptyset = \emptyset\setminus\dom(\sigma) =
  \FV(\unit)\setminus\dom(\sigma)$.

\item[\quad(Pairing)] Suppose $t_1$ and $t_2$ are terms, and assume
  the inductive hypothesis: $P(t_1)$ and $P(t_2)$.  We must show that
  $P((t_1,t_2))$.  Suppose $\sigma$ is a substitution.  We must show
  that $\FV((t_1,t_2)\,\sigma) = \FV((t_1,t_2))\setminus\dom(\sigma)$.
  By the inductive hypothesis and Lemma~\ref{MinusLem1}, we have that
  \begin{align*}
    \FV((t_1,t_2)\,\sigma) &=
    \FV((t_1\,\sigma, t_2\,\sigma)) =
    \FV(t_1\,\sigma)\cup\FV(t_2\,\sigma) \\
    &= (\FV(t_1)\setminus\dom(\sigma)) \cup
    (\FV(t_2)\setminus\dom(\sigma)) \\
    &= (\FV(t_1)\cup\FV(t_2))\setminus\dom(\sigma) \\
    &= \FV((t_1,t_2))\setminus\dom(\sigma).
  \end{align*}

\item[\quad(First Projection)] Similar to pairing, but simpler.

\item[\quad(Second Projection)] Similar to pairing, but simpler.

\item[\quad(Variable)] Suppose $x$ is a variable.  We must show that
  $P(x)$.  Suppose $\sigma$ is a substitution.  We must show that
  $\FV(x\,\sigma) = \FV(x)\setminus\dom(\sigma)$.  There are two cases
  to consider.
  \begin{itemize}
  \item Suppose $x\in\dom(\sigma)$.  Then
    $\FV(x\,\sigma) = \FV(\sigma(x)) = \emptyset =
    \{x\}\setminus\dom(\sigma) = \FV(x)\setminus\dom(\sigma)$.

  \item Suppose $x\not\in\dom(\sigma)$.  Then $\FV(x\,\sigma) = \FV(x)
    = \{x\} = \{x\}\setminus\dom(\sigma) =
    \FV(x)\setminus\dom(\sigma)$.
  \end{itemize}

\item[\quad(Abstraction)] Suppose $x$ is a variable, $T$ is a type,
  and $t$ is a term, and assume the inductive hypothesis: $P(t)$.  We
  must show that $P(\lambda x\typrel T.\,t)$.  Suppose $\sigma$ is a
  substitution.  We must show that $\FV((\lambda x\typrel
  T.\,t)\,\sigma) = \FV(\lambda x\typrel T.\,t)\setminus\dom(\sigma)$.
  By the inductive hypothesis, we have that
  \begin{align*}
    \FV((\lambda x\typrel T.\,t)\,\sigma)
    &= \FV(\lambda x\typrel T.\,t\,(\sigma/\{x\})) \\
    &= \FV(t\,(\sigma/\{x\})) \setminus \{x\} \\
    &= (\FV(t)\setminus\dom(\sigma/\{x\})) \setminus \{x\} \\
    &= (\FV(t)\setminus(\dom(\sigma)\setminus\{x\})) \setminus \{x\}.
  \end{align*}
  And $\FV(\lambda x\typrel T.\,t)\setminus\dom(\sigma)
    = (\FV(t)\setminus\{x\})\setminus\dom(\sigma)$,
  so it will suffice to show that
  \begin{displaymath}
    (\FV(t)\setminus(\dom(\sigma)\setminus\{x\})) \setminus \{x\} =
    (\FV(t)\setminus\{x\})\setminus\dom(\sigma) .
  \end{displaymath}
  There are two inclusions to show.
  \begin{itemize}
  \item Suppose $w\in(\FV(t)\setminus(\dom(\sigma)\setminus\{x\}))
    \setminus \{x\}$.  Thus $w\in
    \FV(t)\setminus(\dom(\sigma)\setminus\{x\})$ and $w\neq x$, so
    that $w\in\FV(t)$ and $w\not\in\dom(\sigma)\setminus\{x\}$.  Hence
    $w\not\in\dom(\sigma)$.  Because $w\in\FV(t)$ and $w\neq x$, we
    have that $w\in\FV(t)\setminus\{x\}$.  Thus
    $w\in(\FV(t)\setminus\{x\})\setminus\dom(\sigma)$.

  \item Suppose $w\in(\FV(t)\setminus\{x\})\setminus\dom(\sigma)$.
    Thus $w\in\FV(t)\setminus\{x\}$ and $w\not\in\dom(\sigma)$, so
    that $w\in\FV(t)$ and $w\neq x$.  Because $w\not\in\dom(\sigma)$,
    we have that $w\not\in\dom(\sigma)\setminus\{x\}$.  Since
    $w\in\FV(t)$ and $w\not\in\dom(\sigma)\setminus\{x\}$, it follows
    that $w\in\FV(t)\setminus(\dom(\sigma)\setminus\{x\})$.  And
    $w\neq x$, and thus
    $w\in(\FV(t)\setminus(\dom(\sigma)\setminus\{x\})) \setminus
    \{x\}$.
  \end{itemize}

\item[\quad(Application)] Similar to pairing.
\end{description}

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

\begin{lemma}
\label{MinusLem2}
For all sets $A$, $B$ and $C$,
$A\setminus(B\setminus C) = (A\setminus B)\cup(A\cap C)$.
\end{lemma}

\begin{proof}
Suppose $A$, $B$ and $C$ are sets.  It will suffice to show that
$A\setminus(B\setminus C) \sub (A\setminus B)\cup(A\cap C) \sub
A\setminus(B\setminus C)$.
\begin{itemize}
\item ($A\setminus(B\setminus C) \sub (A\setminus B)\cup(A\cap
  C)$)\quad Suppose $x\in A\setminus(B\setminus C)$, so that $x\in A$
  and $x\not\in B\setminus C$.  There are two subcases to consider.
  \begin{itemize}
  \item Suppose $x\in C$.  Thus $x\in A$ and $x\in C$, so that $x\in
    A\cap C\sub (A\setminus B)\cup(A\cap C)$.

  \item Suppose $x\not\in C$.  Because $x\not\in B\setminus C$, it
    follows that $x\not\in B$.  Thus $x\in A$ and $x\not\in B$,
    so that $x\in A\setminus B\sub(A\setminus B)\cup(A\cap C)$.
  \end{itemize}

\item ($(A\setminus B)\cup(A\cap C)\sub A\setminus(B\setminus C)$)\quad
  Suppose $x\in (A\setminus B)\cup(A\cap C)$.  There are two cases to
  consider.
  \begin{itemize}
  \item Suppose $x\in A\setminus B$.  Thus $x\in A$ and $x\not\in B$,
    so that $x\not\in B\setminus C$.  Hence $x\in
    A\setminus(B\setminus C)$.

  \item Suppose $x\in A\cap C$.  Thus $x\in A$ and $x\in C$, so that
    $x\not\in B\setminus C$.  Hence $x\in A\setminus(B\setminus
    C)$.
  \end{itemize}
\end{itemize}
\end{proof}

Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is a context}, t\eqtxtl{is
    a term}, T \eqtxt{is a type and, for all substitutions} \sigma,
  \eqtxt{if, for all} x\in\dom(\Gamma)\cap\dom(\sigma),\,
  \emptyset\vdash\sigma(x)\typrel\Gamma(x), \eqtxt{then}
  \Gamma/\dom(\sigma) \vdash t\,\sigma \typrel T}$.
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 $P(\Gamma,t,T)$.
\begin{description}
\item[\quad(T-Unit)] Suppose $\Gamma$ is a context.  We must show that
  $P(\Gamma,\unit,\Unit)$.  Suppose $\sigma$ is a substitution, and
  for all $x\in\dom(\Gamma)\cap\dom(\sigma)$,
  $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  We must show that
  $\Gamma/\dom(\sigma)\vdash \unit\,\sigma\typrel \Unit$.
  And this follows by (T-Unit), since $\unit\,\sigma=\unit$.

\item[\quad(T-Pair)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$
  are terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1$
  and $\Gamma\vdash t_2\typrel T_2$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1)$ and $P(\Gamma,t_2,T_2)$.  We must
  show that $P(\Gamma, (t_1,t_2), T_1\times T_2)$.  Suppose $\sigma$
  is a substitution, and for all $x\in\dom(\Gamma)\cap\dom(\sigma)$,
  $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  We must show that
  $\Gamma/\dom(\sigma)\vdash(t_1,t_2)\,\sigma\typrel T_1\times T_2$.
  Because of $P(\Gamma,t_1,T_1)$, we have that
  $\Gamma/\dom(\sigma)\vdash t_1\,\sigma\typrel T_1$, and because of
  $P(\Gamma,t_2,T_2)$, we have that $\Gamma/\dom(\sigma)\vdash
  t_2\,\sigma\typrel T_2$.  Thus $\Gamma/\dom(\sigma)\vdash
  (t_1\,\sigma, t_2\,\sigma)\typrel T_1\times T_2$, by (T-Pair).  But
  $(t_1,t_2)\,\sigma = (t_1\,\sigma, t_2\,\sigma)$, and so
  $\Gamma/\dom(\sigma)\vdash(t_1,t_2)\,\sigma\typrel T_1\times T_2$.

\item[\quad(T-Fst)] Similar to (T-Pair), but slightly simpler.

\item[\quad(T-Snd)] Similar to (T-Pair), but slightly simpler.

\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 $\sigma$ is a substitution, and for all
  $x\in\dom(\Gamma)\cap\dom(\sigma)$,
  $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  We must show that
  $\Gamma/\dom(\sigma)\vdash x\,\sigma\typrel T$.
  There are two cases to consider.
  \begin{itemize}
  \item Suppose $x\in\dom(\sigma)$.  Since $(x,T)\in\Gamma$, we have that
    $x\in\dom(\Gamma)\cap\dom(\sigma)$.  Thus by our assumption we have
    that $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  But
    $\sigma(x) = x\,\sigma$ and $\Gamma(x) = T$, so that
    $\emptyset\vdash x\,\sigma\typrel T$.  Thus the weakening lemma
    tells us that $\Gamma/\dom(\sigma)\vdash x\,\sigma\typrel T$.

  \item Suppose $x\not\in\dom(\sigma)$.  Then $x\,\sigma = x$, so we
    must show that $\Gamma/\dom(\sigma)\vdash x\typrel T$.  By
    (T-Var), it will suffice to show that
    $(x,T)\in\Gamma/\dom(\sigma)$.  Since $\dom(\Gamma/\dom(\sigma)) =
    \dom(\Gamma)\setminus\dom(\sigma)$, $x\in\dom(\Gamma)$ and
    $x\not\in\dom(\sigma)$, we have that $x\in
    \dom(\Gamma/\dom(\sigma))$.  Thus $(\Gamma/\dom(\sigma))(x) =
    \Gamma(x) = T$, since $(x,T)\in\Gamma$, showing that
    $(x,T)\in\Gamma/\dom(\sigma)$.
  \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
  $P(\Gamma,\lambda x\typrel T_1.\,t,T_1\fun T_2)$.  Suppose $\sigma$
  is a substitution, and, (\dag) for all
  $x\in\dom(\Gamma)\cap\dom(\sigma)$,
  $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  We must show that
  $\Gamma/\dom(\sigma)\vdash (\lambda x\typrel T_1.\,t)\,\sigma\typrel
  T_1\fun T_2$.  Since $(\lambda x\typrel T_1.\,t)\,\sigma = \lambda
  x\typrel T_1.\,t\,(\sigma/\{x\})$, we must show that
  $\Gamma/\dom(\sigma)\vdash \lambda x\typrel
  T_1.\,t\,(\sigma/\{x\})\typrel T_1\fun T_2$.  And, by T-Abs, it will
  suffice to show that $(\Gamma/\dom(\sigma))[x\mapsto T_1]\vdash
  t\,(\sigma/\{x\})\typrel T_2$.

  Next, we show that, (\ddag) for all $y\in\dom(\Gamma[x\mapsto
  T_1])\cap \dom(\sigma/\{x\})$,
  $\emptyset\vdash(\sigma/\{x\})(y)\typrel \Gamma[x\mapsto T_1](y)$.
  Suppose $y\in\dom(\Gamma[x\mapsto T_1])\cap \dom(\sigma/\{x\})$.
  Thus $y\in\dom(\Gamma[x\mapsto T_1]) = \dom(\Gamma)\cup\{x\}$ and
  $y\in\dom(\sigma/\{x\}) = \dom(\sigma)\setminus\{x\}$, so that
  $y\in\dom(\sigma)$ and $y\neq x$.  Thus
  $y\in\dom(\Gamma)\cap\dom(\sigma)$, so that (\dag) tells us that
  $\emptyset\vdash\sigma(y)\typrel\Gamma(y)$.  We have that $\sigma(y)
  = (\sigma/\{x\})(y)$ and (because ($y\neq x$)
  $\Gamma(y) = \Gamma[x\mapsto T_1](y)$.  Thus
  $\emptyset\vdash(\sigma/\{x\})(y)\typrel\Gamma[x\mapsto T_1](y)$.
  This completes the proof of (\ddag).

  Because of (\ddag) and the inductive hypothesis, $P(\Gamma[x\mapsto
  T_1], t, T_2)$, we have that
  \begin{displaymath}
    \Gamma[x\mapsto T_1]/\dom(\sigma/\{x\})\vdash t\,(\sigma/\{x\})\typrel T_2.
  \end{displaymath}
  Thus it will suffice to show that $(\Gamma/\dom(\sigma))[x\mapsto
  T_1] = \Gamma[x\mapsto T_1]/\dom(\sigma/\{x\})$.  By
  Lemma~\ref{MinusLem2}, we have that
  \begin{align*}
    \dom((\Gamma/\dom(\sigma))[x\mapsto T_1])
    &= \dom(\Gamma/\dom(\sigma)) \cup \{x\} \\
    &= (\dom(\Gamma)\setminus\dom(\sigma)) \cup \{x\} \\
    &= ((\dom(\Gamma)\cup\{x\})\setminus\dom(\sigma)) \cup \{x\} \\
    &= ((\dom(\Gamma)\cup\{x\})\setminus\dom(\sigma)) \cup
       ((\dom(\Gamma)\cup\{x\})\cap\{x\}) \\
    &= (\dom(\Gamma)\cup\{x\})\setminus(\dom(\sigma)\setminus\{x\}) \\
    &= \dom(\Gamma[x\mapsto T_1])\setminus\dom(\sigma/\{x\}) \\
    &= \dom(\Gamma[x\mapsto T_1]/(\sigma/\{x\})).
  \end{align*}
  Finally, suppose $y\in\dom((\Gamma/\dom(\sigma))[x\mapsto T_1]) =
  \dom(\Gamma[x\mapsto T_1]/(\sigma/\{x\}))$.
  We must show that $(\Gamma/\dom(\sigma))[x\mapsto T_1](y) =
  (\Gamma[x\mapsto T_1]/(\sigma/\{x\}))(y)$.  There are two
  cases to consider.
  \begin{itemize}
  \item Suppose $y=x$.  Then $(\Gamma/\dom(\sigma))[x\mapsto T_1](y) =
    T_1 = \Gamma[x\mapsto T_1](y) = (\Gamma[x\mapsto
    T_1]/(\sigma/\{x\}))(y)$.

  \item Suppose $y\neq x$.  Then $(\Gamma/\dom(\sigma))[x\mapsto
    T_1](y) = (\Gamma/\dom(\sigma))(y) = \Gamma(y) = \Gamma[x\mapsto
    T_1](y) = (\Gamma[x\mapsto T_1]/(\sigma/\{x\}))(y)$.
  \end{itemize}

\item[\quad(T-App)] Similar to (T-Pair).
\end{description}

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

Let $P=\setof{T}{T\eqtxt{is a type and, for all closed terms}
  t\eqtxt{and}t', \eqtxt{if} \emptyset\vdash t\typrel T \eqtxt{and}
  t\fun t', \eqtxt{then} R_T(t) \eqtxt{iff} R_T(t')}$.
We use structural induction to show that, for all types $T$,
$P(T)$.
\begin{description}
\item[\quad(Unit Type)] We must show $P(\Unit)$.  Suppose $t$ and $t'$
  are closed terms, $\emptyset\vdash t\typrel\Unit$ and $t\fun t'$.
  We must show $R_\Unit(t)$ iff $R_\Unit(t')$.  By preservation,
  we know that $\emptyset\vdash t'\typrel\Unit$.
  There are two directions to show.
  \begin{itemize}
  \item Suppose $R_\Unit(t)$.  Thus $t$ converges, so that
    $t\fun^*v$ for some closed value $v$.  By the determinacy of
    the evaluation relation, it follows that $t'\fun^*v$, showing the
    $t'$ converges.  And $\emptyset\vdash t'\typrel\Unit$, so
    that $R_\Unit(t')$.

  \item Suppose $R_\Unit(t')$.  Thus $t'$ converges, so that
    $t'\fun^*v$ for some closed value $v$.  Thus $t\fun^*v$, showing
    the $t$ converges.  And $\emptyset\vdash t\typrel\Unit$, so
    that $R_\Unit(t)$.
  \end{itemize}

\item[\quad(Product Type)] Suppose $T_1$ and $T_2$ are types, and
  assume the inductive hypothesis, $P(T_1)$ and $P(T_2)$.  We must
  show that $P(T_1\times T_2)$.  Suppose $t$ and $t'$ are closed
  terms, $\emptyset\vdash t\typrel T_1\times T_2$ and $t\fun t'$.  We
  must show $R_{T_1\times T_2}(t)$ iff $R_{T_1\times T_2}(t')$.  By
  preservation, we know that $\emptyset\vdash t'\typrel T_1\times
  T_2$.  There are two directions to show.
  \begin{itemize}
  \item Suppose $R_{T_1\times T_2}(t)$.  Thus $t$ converges,
    $R_{T_1}(\fst\,t)$ and $R_{T_2}(\snd\,t)$, so that $t\fun^*v$ for
    some closed value $v$, $\emptyset\vdash\fst\,t\typrel T_1$ and
    $\emptyset\vdash\snd\,t\typrel T_2$.  By the determinacy of the
    evaluation relation, it follows that $t'\fun^*v$, showing the $t'$
    converges.  Because $t\fun t'$, we have that $\fst\,t\fun\fst\,t'$.
    And $R_{T_1}(\fst\,t)$, so that $P(T_1)$ tells us that
    $R_{T_1}(\fst\,t')$.  Because $t\fun t'$, we have that
    $\snd\,t\fun\snd\,t'$.  And $R_{T_2}(\snd\,t)$, so that $P(T_2)$
    tells us that $R_{T_2}(\snd\,t')$.  Since $\emptyset\vdash
    t'\typrel T_1\times T_2$, $t'$ converges, $R_{T_1}(\fst\,t')$ and
    $R_{T_2}(\snd\,t')$, we can conclude that $R_{T_1\times T_2}(t')$.

  \item Suppose $R_{T_1\times T_2}(t')$.  Thus $t'$ converges,
    $R_{T_1}(\fst\,t')$ and $R_{T_2}(\snd\,t')$, so that $t'\fun^*v$
    for some closed value $v$.  Thus $t\fun^*v$, showing the $t$
    converges.  Since $\emptyset\vdash t\typrel T_1\times T_2$, we
    have that $\emptyset\vdash\fst\,t\typrel T_1$ and
    $\emptyset\vdash\snd\,t\typrel T_2$.  Because $t\fun t'$, we have
    that $\fst\,t\fun\fst\, t'$.  And $R_{T_1}(\fst\,t')$, so that
    $P(T_1)$ tells us that $R_{T_1}(\fst\,t)$.  Because $t\fun t'$, we
    have that $\snd\,t\fun\snd\,t'$.  And $R_{T_2}(\snd\,t')$, so that
    $P(T_2)$ tells us that $R_{T_2}(\snd\,t)$.  Since $\emptyset\vdash
    t\typrel T_1\times T_2$, $t$ converges, $R_{T_1}(\fst\,t)$ and
    $R_{T_2}(\snd\,t)$, we can conclude that $R_{T_1\times T_2}(t)$.
  \end{itemize}

\item[\quad(Function Type)] Suppose $T_1$ and $T_2$ are types, and
  assume the inductive hypothesis, $P(T_1)$ and $P(T_2)$.  We must
  show that $P(T_1\fun T_2)$.  Suppose $t$ and $t'$ are closed
  terms, $\emptyset\vdash t\typrel T_1\fun T_2$ and $t\fun t'$.  We
  must show $R_{T_1\fun T_2}(t)$ iff $R_{T_1\fun T_2}(t')$.  By
  preservation, we know that $\emptyset\vdash t'\typrel T_1\fun
  T_2$.  There are two directions to show.
  \begin{itemize}
  \item Suppose $R_{T_1\fun T_2}(t)$.  Thus $t$ converges, and, (\dag)
    for all terms $s$, if $R_{T_1}(s)$, then $R_{T_2}(t\,s)$.  Because
    $t$ converges, we have that $t\fun^*v$ for some closed value $v$.
    By the determinacy of the evaluation relation, it follows that
    $t'\fun^*v$, showing the $t'$ converges.  Because $\emptyset\vdash
    t'\typrel T_1\fun T_2$ and $t'$ converges, to show that
    $R_{T_1\fun T_2}(t')$, it remains to show that, for all terms $s$,
    if $R_{T_1}(s)$, then $R_{T_2}(t'\,s)$.  Suppose $s$ is a term and
    $R_{T_1}(s)$.  We must show that $R_{T_2}(t'\,s)$.  Since
    $R_{T_1}(s)$, we know that $\emptyset\vdash s\typrel T_1$, so that
    $\emptyset\vdash t\,s\typrel T_2$ and $s$ is closed.  And, from
    (\dag) and $R_{T_1}(s)$, we can conclude $R_{T_2}(t\,s)$.  Since
    $t\fun t'$, it follows that $t\,s\fun t'\,s$.  Thus $P(T_2)$ tells
    us that $R_{T_2}(t'\,s)$, as required.

  \item Suppose $R_{T_1\fun T_2}(t')$.  Thus $t'$ converges, and,
    (\dag) for all terms $s$, if $R_{T_1}(s)$, then $R_{T_2}(t'\,s)$.
    Because $t'$ converges, we have that $t'\fun^*v$ for some closed
    value $v$.  Thus $t\fun^*v$, showing the $t$ converges.  Because
    $\emptyset\vdash t\typrel T_1\fun T_2$ and $t$ converges, to
    show that $R_{T_1\fun T_2}(t)$, it remains to show that, for all
    terms $s$, if $R_{T_1}(s)$, then $R_{T_2}(t\,s)$.  Suppose $s$ is
    a term and $R_{T_1}(s)$.  We must show that $R_{T_2}(t\,s)$.
    Since $R_{T_1}(s)$, we know that $\emptyset\vdash s\typrel T_1$,
    so that $\emptyset\vdash t\,s\typrel T_2$ and $s$ is closed.
    And, from (\dag) and
    $R_{T_1}(s)$, we can conclude $R_{T_2}(t'\,s)$.  Since $t\fun t'$,
    it follows that $t\,s\fun t'\,s$.  Thus $P(T_2)$ tells us that
    $R_{T_2}(t\,s)$, as required.
  \end{itemize}
\end{description}

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

\begin{lemma}
\label{SlashLem1}
For all functions $f$ and sets $A$, $(f/A)/A = f/A$.
\end{lemma}

\begin{proof}
Suppose $f$ is a function and $A$ is a set.  Since $\dom((f/A)/A) =
\dom(f/A) \setminus A = (\dom(f)\setminus A) \setminus A = \dom(f)
\setminus A = \dom(f/A)$, we have that $(f/A)/A$ and $f/A$ have the
same domain, $\dom(f)\setminus A$.  And, for all $x$ in that common
domain, $((f/A)/A)(x) = (f/A)(x) = f(x) = (f/A)(x)$, completing the
proof that $(f/A)/A = f/A$.
\end{proof}

\begin{lemma}
\label{SlashLem2}
For all functions $f$ and sets $A$ and $B$, $(f/A)/B = (f/B)/A$.
\end{lemma}

\begin{proof}
Suppose $f$ is a function and $A$ and $B$ are sets.  Since
$\dom((f/A)/B) = \dom(f/A) \setminus B = (\dom(f)\setminus A)\setminus
B = (\dom(f)\setminus B)\setminus A = \dom(f/B)\setminus A =
\dom((f/B)/A)$, we have that $(f/A)/B$ and $(f/B)/A$ have the same
domain $(\dom(f)\setminus A)\setminus B = (\dom(f)\setminus
B)\setminus A$.  And, for all $x$ in that common domain, we have that
$((f/A)/B)(x) = (f/A)(x) = f(x) = (f/B)(x) = ((f/B)/A)(x)$, completing
the proof that $(f/A)/B = (f/B)/A$.
\end{proof}

\begin{lemma}
\label{SlashLem3}
For all functions $f$ and elements $x$ and $y$ of our universe,
$f[x\mapsto y] / \{x\} = f / \{x\}$.

\end{lemma}

\begin{proof}
Suppose $f$ is a function and $x$ are $y$ are elements of our
universe.  Since $\dom(f[x\mapsto y] / \{x\}) =
\dom(f[x\mapsto y]) \setminus \{x\} =
(\dom(f)\cup\{x\}) \setminus \{x\} =
\dom(f)\setminus \{x\} = \dom(f / \{x\})$,
we have that $f[x\mapsto y] / \{x\}$ and $f / \{x\}$ have the
same domain, $\dom(f)\setminus\{x\}$.  And, for all
$z\in\dom(f)\setminus\{x\}$, $(f[x\mapsto y] / \{x\})(z) =
f[x\mapsto y](z) = f(z) = (f/\{x\})(z)$, completing the
proof that $f[x\mapsto y] / \{x\} = f / \{x\}$.
\end{proof}

\begin{lemma}
\label{SlashLem4}
For all functions $f$ and elements $x$, $y$ and $z$ of our universe,
if $x\neq y$, then $(f/\{y\})[x\mapsto z] = f[x\mapsto z]/\{y\}$.
\end{lemma}

\begin{proof}
Suppose $f$ is a function, $x$, $y$ and $z$ are elements of our
universe, and $x\neq y$.  We have that
$\dom((f/\{y\})[x\mapsto z]) = \dom(f/\{y\})\cup\{x\} =
(\dom(f)\setminus\{y\})\cup\{x\} =
(\dom(f)\cup\{x\})\setminus\{y\} =
\dom(f[x\mapsto z])\setminus\{y\} =
\dom(f[x\mapsto z]/\{y\})$.  Suppose $w$ is in the common domain
of $(f/\{y\})[x\mapsto z]$ and $f[x\mapsto z]/\{y\}$.
We must show that $((f/\{y\})[x\mapsto z])(w) =
(f[x\mapsto z]/\{y\})(w)$.
There are two cases to consider.
\begin{itemize}
\item Suppose $w=x$.  Then $((f/\{y\})[x\mapsto z])(w) = z =
  f[x\mapsto z](w) = (f[x\mapsto z]/\{y\})(w)$.

\item Suppose $w\neq x$.  Then $((f/\{y\})[x\mapsto z])(w) =
  (f/\{y\})(w) = f(w) = f[x\mapsto z](w) = (f[x\mapsto z]/\{y\})(w)$.
\end{itemize}
\end{proof}

\begin{lemma}
\label{Ex5Lem}
For all terms $t$, closed values $v$, variables $x$ and
substitutions $\sigma$,
\begin{displaymath}
(t\,(\sigma/\{x\}))\,\{(x,v)\} = t\,\sigma[x\mapsto v] .
\end{displaymath}
\end{lemma}

\begin{proof}
Suppose $v$ is a closed value and $x$ is a variable.
Let $P=\setof{t}{t\eqtxtl{is a term, and, for all}\eqtxtr{substitutions}
  \sigma, (t\,(\sigma/\{x\}))\,\{(x,v)\} = t\,\sigma[x\mapsto v]}$.
We use structural induction to prove that, for all terms $t$,
$P(t)$.
\begin{description}
\item[\quad(Unit Constant)] We must show $P(\unit)$.  Suppose $\sigma$
  is a substitution.  We have that 
  $(\unit\,(\sigma/\{x\}))\,\{(x,v)\} =
  \unit\,\{(x,v)\} = \unit = \unit\,\sigma[x\mapsto v]$.

\item[\quad(Pairing)] Suppose $t_1$ and $t_2$ are terms, and assume
  the inductive hypothesis: $P(t_1)$ and $P(t_2)$.  We must show that
  $P((t_1,t_2))$.  Suppose $\sigma$ is a substitution.
  By the inductive hypothesis, we have that
  \begin{align*}
    ((t_1,t_2)\,(\sigma/\{x\}))\,\{(x,v)\}
    &= (t_1\,(\sigma/\{x\}), t_2\,(\sigma/\{x\}))\,\{(x,v)\} \\
    &= ((t_1\,(\sigma/\{x\}))\,\{(x,v)\},
    (t_2\,(\sigma/\{x\}))\,\{(x,v)\}) \\
    &= (t_1\,\sigma[x\mapsto v], t_2\,\sigma[x\mapsto v]) \\
    &= (t_1,t_2)\,\sigma[x\mapsto v] .
  \end{align*}

\item[\quad(First Projection)] Similar to pairing.

\item[\quad(Second Projection)] Similar to pairing.

\item[\quad(Variable)] Suppose $y$ is a variable.  We must show
  $P(y)$.  Suppose $\sigma$ is a substitution.
  There are two cases to consider.
  \begin{itemize}
  \item Suppose $y=x$.  Then $y\not\in\dom(\sigma)\setminus\{x\} =
    \dom(\sigma/\{x\})$, $y\in\dom\{(x,v)\}$ and
    $y\in\dom(\sigma)\cup\{x\} = \dom(\sigma[x\mapsto v])$, so that
    $(y\,(\sigma/\{x\}))\,\{(x,v)\} = y\,\{(x,v)\} = \{(x,v)\}(y) = v
    = \sigma[x\mapsto v](y) = y\,\sigma[x\mapsto v]$.

  \item Suppose $y\neq x$.  There are two sub-cases to consider
    \begin{itemize}
    \item Suppose $y\in\dom(\sigma)$.  Since
      $y\in\dom(\sigma)\setminus\{x\} = \dom(\sigma/\{x\})$, we have
      that $(y\,(\sigma/\{x\}))\,\{(x,v)\} =
      ((\sigma/\{x\})(y))\,\{(x,v)\} = (\sigma(y))\,\{(x,v)\}$.  But
      $\sigma(y)$ is closed, so that
      $\FV(\sigma(y))\cap\dom(\{(x,v)\}) = \emptyset$, and thus
      Exercise~1 tells us that $(\sigma(y))\,\{(x,v)\} = \sigma(y)$.
      Furthermore, because $y\in\dom(\sigma)\cup\{x\} =
      \dom(\sigma[x\mapsto v])$, we have that $\sigma(y) =
      \sigma[x\mapsto v](y) = y\,\sigma[x\mapsto v]$.  Hence
      $(y\,(\sigma/\{x\}))\,\{(x,v)\} = (\sigma(y))\,\{(x,v)\} =
      \sigma(y) = y\,\sigma[x\mapsto v]$.

    \item Suppose $y\not\in\dom(\sigma)$.  Because
      $y\not\in\dom(\sigma)\setminus\{x\}=\dom(\sigma/\{x\})$,
      $y\not\in\dom\{(x,v)\}$ and $y\not\in\dom(\sigma)\cup\{x\} =
      \dom(\sigma[x\mapsto v])$, we have that
      $(y\,(\sigma/\{x\}))\,\{(x,v)\} = y\,\{(x,v)\} = y =
      y\,\sigma[x\mapsto v]$.
    \end{itemize}
  \end{itemize}

\item[\quad(Abstraction)] Suppose $y$ is a variable, $T$ is a type,
  and $t$ is a term, and assume the inductive hypothesis: $P(t)$.  We
  must show $P(\lambda y\typrel T.\,t)$.  Suppose $\sigma$ is a
  substitution.  There are two cases to consider.
  \begin{itemize}
  \item Suppose $y=x$.  By Lemma~\ref{SlashLem1} and since
    $\{(y,v)\}/\{y\} = \emptyset$ ($\dom(\{(y,v)\}/\{y\}) =
    \dom(\{(y,v)\}\setminus\{y\} = \{y\}\setminus\{y\} = \emptyset$),
    $(t\,(\sigma/\{y\}))\,\emptyset = t\,(\sigma/\{y\})$
    (by Exercise~1, since $\FV(t\,(\sigma/\{y\}))\cap\dom(\emptyset) =
    \FV(t\,(\sigma/\{y\}))\cap\emptyset = \emptyset$)
    and $\sigma/\{y\} = \sigma/\{x\} = \sigma[x\mapsto v]/\{x\} =
    \sigma[x\mapsto v]/\{y\}$ (by Lemma~\ref{SlashLem3}), we have that
    \begin{align*}
      ((\lambda y\typrel T.\,t)(\sigma/\{x\}))\,\{(x,v)\}
      &= (\lambda y\typrel T.\,t\,((\sigma/\{x\})/\{y\}))\,\{(x,v)\} \\
      &= (\lambda y\typrel T.\,t\,((\sigma/\{y\})/\{y\}))\,\{(x,v)\} \\
      &= (\lambda y\typrel T.\,t\,(\sigma/\{y\}))\,\{(x,v)\} \\
      &= \lambda y\typrel T.\,(t\,(\sigma/\{y\}))\,(\{(x,v)\}/\{y\}) \\
      &= \lambda y\typrel T.\,(t\,(\sigma/\{y\}))\,(\{(y,v)\}/\{y\}) \\
      &= \lambda y\typrel T.\,(t\,(\sigma/\{y\}))\,\emptyset \\
      &= \lambda y\typrel T.\,t\,(\sigma/\{y\}) \\
      &= \lambda y\typrel T.\,t\,(\sigma[x\mapsto v]/\{y\}) \\
      &= (\lambda y\typrel T.t)\,\sigma[x\mapsto v].
    \end{align*}

  \item Suppose $y\neq x$.  By Lemmas~\ref{SlashLem2} and
    \ref{SlashLem4}, the inductive hypothesis, and since
    $(\{(x,v)\}/{y}) = \{(x,v)\}$, we have that
    \begin{align*}
      ((\lambda y\typrel T.\,t)(\sigma/\{x\}))\,\{(x,v)\}
      &= (\lambda y\typrel T.\,t\,((\sigma/\{x\})/\{y\}))\,\{(x,v)\} \\
      &= \lambda y\typrel T.\,(t\,((\sigma/\{x\})/\{y\}))\,(\{(x,v)\}/{y}) \\
      &= \lambda y\typrel T.\,(t\,((\sigma/\{x\})/\{y\}))\,\{(x,v)\} \\
      &= \lambda y\typrel T.\,(t\,((\sigma/\{y\})/\{x\}))\,\{(x,v)\} \\
      &= \lambda y\typrel T.\,t\,((\sigma/\{y\})[x\mapsto v]) \\
      &= \lambda y\typrel T.\,t\,(\sigma[x\mapsto v]/\{y\}) \\
      &= (\lambda y\typrel T.t)\,\sigma[x\mapsto v].
    \end{align*}
  \end{itemize}

\item[\quad(Application)] Similar to pairing.
\end{description}
\end{proof}

Let $P = \setof{(\Gamma,t,T)}{\Gamma\eqtxtl{is a context}, t\eqtxtl{is
    a term}, T \eqtxt{is a type and, for all substitutions} \sigma,
  \eqtxt{if} \dom(\Gamma)\sub\dom(\sigma) \eqtxt{and, for all}
  x\in\dom(\Gamma), R_{\Gamma(x)}(\sigma(x)), \eqtxt{then}
  R_T(t\,\sigma)}$.  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
  $P(\Gamma,\unit,\Unit)$.  Suppose $\sigma$ is a substitution,
  $\dom(\Gamma)\sub \dom(\sigma)$, and for all $x\in\dom(\Gamma)$,
  $R_{\Gamma(x)}(\sigma(x))$.  We must show that
  $R_\Unit(\unit\,\sigma)$.  Since $\unit\,\sigma = \unit$, it will
  suffice to show that $R_\Unit(\unit)$, and this follows since
  $\emptyset\vdash\unit\typrel\Unit$ (by (T-Unit)) and $\unit$ is a
  value and thus converges.

\item[\quad(T-Pair)] Suppose $\Gamma$ is a context, $t_1$ and $t_2$
  are terms, $T_1$ and $T_2$ are types, $\Gamma\vdash t_1\typrel T_1$
  and $\Gamma\vdash t_2\typrel T_2$, and assume the inductive
  hypothesis, $P(\Gamma,t_1,T_1)$ and $P(\Gamma,t_2,T_2)$.  We must
  show that $P(\Gamma, (t_1,t_2), T_1\times T_2)$.  Suppose $\sigma$
  is a substitution, $\dom(\Gamma)\sub \dom(\sigma)$, and for all
  $x\in\dom(\Gamma)$, $R_{\Gamma(x)}(\sigma(x))$.  We must show that
  $R_{T_1\times T_2}((t_1,t_2)\,\sigma)$.  Since $(t_1,t_2)\,\sigma =
  (t_1\,\sigma, t_2\,\sigma)$, it will suffice to show $R_{T_1\times
    T_2}((t_1\,\sigma, t_2\,\sigma))$.  By the inductive hypothesis,
  we have that $R_{T_1}(t_1\,\sigma)$ and $R_{T_2}(t_2\,\sigma)$.
  Since $R_{T_1}(t_1\,\sigma)$, we have that $\emptyset\vdash
  t_1\,\sigma\typrel T_1$ and $t_1\,\sigma$ converges, so that
  $t_1\,\sigma\fun^*v_1$, for some closed value $v_1$.  By Exercise~4,
  it follows that $R_{T_1}(v_1)$.  And, since $R_{T_2}(t_2\,\sigma)$,
  we have that $\emptyset\vdash t_2\,\sigma\typrel T_2$ and
  $t_2\,\sigma$ converges, so that $t_2\,\sigma\fun^*v_2$, for some
  closed value $v_2$.  By Exercise~4, it follows that $R_{T_2}(v_2)$.
  Since $\emptyset\vdash t_1\,\sigma\typrel T_1$ and $\emptyset\vdash
  t_2\,\sigma\typrel T_2$, (T-Pair) gives us
  $\emptyset\vdash(t_1\,\sigma, t_2\,\sigma)\typrel T_1\times T_2$.
  Since $t_1\,\sigma\fun^*v_1$ and $t_2\,\sigma\fun^*v_2$, we have
  that $(t_1\,\sigma, t_2\,\sigma)\fun^*(v_1, t_2\,\sigma)\fun^*
  (v_1,v_2)$.  Thus, by Exercise~4, it will suffice to show that
  $R_{T_1\times T_2}((v_1,v_2))$.  By the preservation theorem, we
  have that $\emptyset\vdash(v_1,v_2)\typrel T_1\times T_2$.  Since
  $(v_1,v_2)$ is a value, it converges.  It remains to show that
  $R_{T_1}(\fst\,(v_1,v_2))$ and $R_{T_2}(\snd\,(v_1,v_2))$.  Since
  $\emptyset\vdash(v_1,v_2)\typrel T_1\times T_2$, we have that
  $\emptyset\vdash\fst\,(v_1,v_2)\typrel T_1$ and
  $\emptyset\vdash\snd\,(v_1,v_2)\typrel T_2$.  Furthermore,
  $\fst\,(v_1,v_2)\fun v_1$ and $\snd\,(v_1,v_2)\fun v_2$.  Thus,
  since $R_{T_1}(v_1)$ and $R_{T_2}(v_2)$, Exercise~4 tells us that
  $R_{T_1}(\fst\,(v_1,v_2))$ and $R_{T_2}(\snd\,(v_1,v_2))$.

\item[\quad(T-Fst)] Suppose $\Gamma$ is a context, $t$ is a term,
  $T_1$ and $T_2$ are types, $\Gamma\vdash t\typrel T_1\times T_2$,
  and assume the inductive hypothesis, $P(\Gamma,t,T_1\times T_2)$.
  We must show that $P(\Gamma, \fst\,t, T_1)$.  Suppose $\sigma$ is a
  substitution, $\dom(\Gamma)\sub \dom(\sigma)$, and for all
  $x\in\dom(\Gamma)$, $R_{\Gamma(x)}(\sigma(x))$.  We must show that
  $R_{T_1}((\fst\,t)\,\sigma)$.  Since $(\fst\,t)\,\sigma =
  \fst(t\,\sigma)$, it will suffice to show that
  $R_{T_1}(\fst(t\,\sigma))$.  By the inductive hypothesis, we have
  that $R_{T_1\times T_2}(t\,\sigma)$.  Thus $\emptyset\vdash t\,\sigma
  \typrel T_1\times T_2$ and $t\,\sigma$ converges, so that
  $\emptyset\vdash\fst\,(t\,\sigma)\typrel T_1$ and $t\,\sigma\fun^*v$
  for some closed value $v$.  By Exercise~4, we have that
  $R_{T_1\times T_2}(v)$, so that, by the definition of $R_{T_1\times
    T_2}$, $R_{T_1}(\fst\,v)$.  Since $t\,\sigma\fun^*v$, we have that
  $\fst\,(t\,\sigma)\fun^* \fst\,v$.  Thus $R_{T_1}(\fst(t\,\sigma))$,
  by Exercise~4.

\item[\quad(T-Snd)] Similar to (T-Fst).

\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 $\sigma$ is a substitution,
  $\dom(\Gamma)\sub \dom(\sigma)$, and for all $x\in\dom(\Gamma)$,
  $R_{\Gamma(x)}(\sigma(x))$.  We must show that $R_T(x\,\sigma)$.
  Since $x\in\dom(\Gamma)$, we have that $R_{\Gamma(x)}(\sigma(x))$.
  Since $(x,T)\in\Gamma$, we have that $\Gamma(x)=T$.  And, because
  $x\in\dom(\Gamma)$, we have that $x\in\dom(\sigma)$, so that
  $\sigma(x) = x\,\sigma$.  Thus $R_T(x\,\sigma))$.

\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
  $P(\Gamma,\lambda x\typrel T_1.\,t,T_1\fun T_2)$.  Suppose $\sigma$ is a
  substitution, $\dom(\Gamma)\sub \dom(\sigma)$, and for all
  $x\in\dom(\Gamma)$, $R_{\Gamma(x)}(\sigma(x))$.  We must show that
  $R_{T_1\fun T_2}((\lambda x\typrel T_1.\,t)\,\sigma)$.  Since
  $(\lambda x\typrel T_1.\,t)\,\sigma = \lambda x\typrel T_1.\,
  t\,(\sigma/\{x\})$, it will suffice to show $R_{T_1\fun T_2}(\lambda
  x\typrel T_1.\, t\,(\sigma/\{x\}))$.  

  Because $\Gamma[x\mapsto T_1]\vdash t\typrel T_2$, (T-Abs) gives us
  $\Gamma\vdash\lambda x\typrel T_1.\,t\typrel T_1\fun T_2$.  Since
  $\dom(\Gamma)\sub \dom(\sigma)$, and for all $x\in\dom(\Gamma)$,
  $R_{\Gamma(x)}(\sigma(x))$, we have that, for all
  $x\in\dom(\Gamma)\cap\dom(\sigma)$,
  $\emptyset\vdash\sigma(x)\typrel\Gamma(x)$.  Thus Exercise~3 tells
  us that $\Gamma/\dom(\sigma)\vdash (\lambda x\typrel
  T_1.\,t)\,\sigma \typrel T_1\fun T_2$.  But $\Gamma/\dom(\sigma) =
  \emptyset$ and $(\lambda x\typrel T_1.\,t)\,\sigma = \lambda x\typrel
  T_1.\, t\,(\sigma/\{x\})$, and thus $\emptyset\vdash\lambda x\typrel
  T_1.\, t\,(\sigma/\{x\}) \typrel T_1\fun T_2$.  Furthermore, $\lambda
  x\typrel T_1.\, t\,(\sigma/\{x\})$ is a value, and thus converges.  So
  to show $R_{T_1\fun T_2}(\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))$, it
  remains to show that, for all terms $s$, if $R_{T_1}(s)$, then
  $R_{T_2}((\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))\,s)$.

  Suppose $s$ is a term and $R_{T_1}(s)$.  We must show
  $R_{T_2}((\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))\,s)$.  Since
  $R_{T_1}(s)$, we have that $\emptyset\vdash s\typrel T_1$ and $s$
  converges, so that $s\fun^*v$ for some closed value $v$.  By
  Exercise~4, it follows that $R_{T_1}(v)$.  Since
  $\emptyset\vdash\lambda x\typrel T_1.\, t\,(\sigma/\{x\}) \typrel
  T_1\fun T_2$ and $\emptyset\vdash s\typrel T_1$, we have that
  $\emptyset\vdash(\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))\,s \typrel
  T_2$.  By Lemma~\ref{Ex5Lem}, we have that
  \begin{align*}
    (\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))\,s
    &\fun^*
    (\lambda x\typrel T_1.\, t\,(\sigma/\{x\}))\,v \\
    &\fun (t\,(\sigma/\{x\}))\,\{(x,v)\} \\
    &= t\,\sigma[x\mapsto v] .
  \end{align*}
  Thus, by Exercise~4, it will suffice to show that
  $R_{T_2}(t\,\sigma[x\mapsto v])$.  By the inductive hypothesis,
  $P(\Gamma[x\mapsto T_1], t, T_2)$, it will suffice to show that
  $\dom(\Gamma[x\mapsto T_1])\sub \dom(\sigma[x\mapsto v])$ and, for all
  $y\in\dom(\Gamma[x\mapsto T_1])$, $R_{\Gamma[x\mapsto
    T_1](y)}(\sigma[x\mapsto v](y))$.  Since
  $\dom(\Gamma)\sub\dom(\sigma)$, we have that $\dom(\Gamma[x\mapsto
  T_1]) = \dom(\Gamma)\cup\{x\} \sub \dom(\sigma)\cup\{x\} =
  \dom(\sigma[x\mapsto v])$.  Suppose $y\in \dom(\Gamma[x\mapsto
  T_1]) = \dom(\Gamma)\cup\{x\}$.  We must show that $R_{\Gamma[x\mapsto
    T_1](y)}(\sigma[x\mapsto v](y))$.  There are two cases to consider.
  \begin{itemize}
  \item Suppose $y=x$.  Then $R_{T_1}(v)$, $T_1 = \Gamma[x\mapsto
    T_1](y)$ and $v = \sigma[x\mapsto v](y)$, so that
    $R_{\Gamma[x\mapsto T_1](y)}(\sigma[x\mapsto v](y))$.

  \item Suppose $y\neq x$, so that $y\in\dom(\Gamma)$.  Thus
    $R_{\Gamma(y)}(\sigma(y))$, $\Gamma(y) = \Gamma[x\mapsto T_1](y)$
    and $\sigma(y) = \sigma[x\mapsto v](y)$, so that
    $R_{\Gamma[x\mapsto T_1](y)}(\sigma[x\mapsto v](y))$.
  \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 $\sigma$ is a
  substitution, $\dom(\Gamma)\sub \dom(\sigma)$, and for all
  $x\in\dom(\Gamma)$, $R_{\Gamma(x)}(\sigma(x))$.  We must show that
  $R_{T_2}((t_1\,t_2)\,\sigma)$.  Since $(t_1\,t_2)\,\sigma =
  (t_1\,\sigma)(t_2\,\sigma)$, it will suffice to show
  $R_{T_2}((t_1\,\sigma)(t_2\,\sigma))$.  By the inductive hypothesis,
  we have that $R_{T_1\fun T_2}(t_1\,\sigma)$ and
  $R_{T_1}(t_2\,\sigma)$.  Thus, by the definition of $R_{T_1\fun
    T_2}(t_1\,\sigma)$, we have $R_{T_2}((t_1\,\sigma)(t_2\,\sigma))$.

\end{description}

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

Suppose $t$ is a closed, well-typed term.  Thus $\emptyset\vdash
t\typrel T$ for some type $T$.  Since $\emptyset\vdash t\typrel T$,
$\dom(\emptyset)\sub\dom(\emptyset)$, and, for all
$x\in\dom(\emptyset)$, $R_{\emptyset(x)}(\emptyset(x))$, Exercise~5
tells us that $R_T(t\,\emptyset)$.  Since $\FV(t)\cap\dom(\emptyset) =
\emptyset\cap\emptyset = \emptyset$, Exercise~1 tells us that
$t\,\emptyset = t$.  Thus $R_T(t)$, so that $t$ converges.
\end{document}

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