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\lecture{26}                  %% Lecture number
\title{Strong Normalization}   %% Title of lecture
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\begin{document}
\maketitle

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\section{Introduction}

In Lecture 24, we proved that each term in the simply typed $\lambda$-calculus
would never get stuck.  Today, we want to show that it will actually terminate.
This property is known as _strong normalization_.

Formally, we want to prove that if $\judge{}e\tau$, then $e\Downarrow$.
We will prove this by induction, but we will need a fairly sophisticated induction
hypothesis that takes both the typing and the reduction order into account.
We cannot just do induction on the subterm relation.  For example,
even if $e_1$ and $e_2$ terminate, we cannot conclude that $e_1~e_2$ does:
consider $e_1=e_2=\lam x{xx}$.

\section{Church vs.~Curry}

We will prove this theorem in the pure simply-typed $\lambda$-calculus in Curry style.
This differs from Church style in that the binding occurrence of a variable in a $\lambda$-abstraction is not annotated with its type.

Let $\alpha,\beta,\ldots$ denote type variables, $x,y,\ldots$ term variables, $\sigma,\tau,\ldots$ types, and $d,e,\ldots$ terms.  In the Curry-style simply typed $\lambda$-calculus, terms and types are defined by
\(
e\ \ ::=\ \ x \bnf e\,d \bnf \lam xe &\qquad& \tau\ \ ::=\ \ \alpha \bnf \sigma\to\tau
\)
and the typing rules are
\(
\judge{\Gamma,\,x\ty\tau}x\tau \qquad
\frac{\Gjudge e{\sigma\to\tau}\quad\Gjudge d\sigma}{\Gjudge{(e\,d)}\tau} \qquad
\frac{\judge{\Gamma,\,x\ty\sigma}e\tau}{\Gjudge{(\lam xe)}{\sigma\to\tau}}
\)
Note that in Church style, a closed term can have at most one type, but in Curry style, if it has any type at all, then it has infinitely many.  For example, $\judge{}{\lam xx}{((\alpha\to\beta)\to\gamma)\to((\alpha\to\beta)\to\gamma)}$.  In general, if $\judge{}e\tau$, then also $\judge{}e{\tau'}$, where $\tau'$ is any substitution instance of $\tau$.

A term $e$ is _typable_ if there exists a type environment $\Gamma$ and a type $\tau$ such that $\Gjudge e\tau$.  One can show by induction that if $\Gjudge e\tau$, then $\FV e\subseteq\dom\Gamma$.

\section{Strong Normalization}

By the Church--Rosser theorem, normal forms are unique up to $\alpha$-equivalence, so any two reduction strategies starting from the same term that terminate must yield the same result up to $\alpha$-equivalence.  However, there may be some strategies that terminate and some that do not.

A term is _strongly normalizing_ (SN) if all $\beta$-reduction sequences starting from that term converge to a normal form; equivalently, if there is no infinite $\beta$-reduction sequence starting from that term.  Our main theorem is

\begin{theorem}
\label{thm:SN}
All typable terms are strongly normalizing.
\end{theorem}

\subsection{Ultra-Strong Normalization}

We say that a term $e$ is \emph{ultra-strongly normalizing with respect to $\Gamma$ and $\sigma$} and write $\judgeUSN\Gamma e\sigma$ if
\begin{enumerate}
\renewcommand\labelenumi{(\roman{enumi})}
\item
$\judge\Gamma e\sigma$
\item
for all $n\geq 0$, if $\sigma$ is of the form $\sigma_1\to\sigma_2\to\cdots\to\sigma_n\to\tau$ and $\judgeUSN\Gamma{e_i}{\sigma_i}$, $1\leq i\leq n$, then $e~e_1~e_2~\cdots~e_n$ is SN.
\end{enumerate}

A term $e$ is \emph{ultra-strongly normalizing} (USN) if it is ultra-strongly normalizing with respect to some $\Gamma$ and $\sigma$.

The definition of the relation $\forceUSN$ may seem circular, but it is not: $\judgeUSN\Gamma e\sigma$ is defined in terms of $\judgeUSN\Gamma{e_i}{\sigma_i}$, where the $\sigma_i$ are strict subexpressions of $\sigma$, so it is well-defined by structural induction on types.

Almost all the work we need to do is contained in the following lemma:

\begin{lemma}
\label{lem:SN}
Let $\seq x1n$ be distinct variables.  If
\begin{enumerate}
\renewcommand\labelenumi{\emph{(\roman{enumi})}}
\item
$\judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}e\tau$,
\item
$\judgeUSN\Gamma{d_i}{\sigma_i}$, $1\leq i\leq n$, and
\item
$x_j\notin\FV{d_i}$ for $j\gt i$,
\end{enumerate}
then $\judgeUSN\Gamma{\subst{\subst e{d_1}{x_1}\cdots}{d_n}{x_n}}\tau$.
\end{lemma}

\begin{proof}
Suppose the three premises (i)--(iii) hold.  The proof is by induction on the structure of $e$.

\paragraph{Case 1} Variable $x$.

\paragraph{Case 1A} $x=x_i$ for some $i$.  We have $\tau=\sigma_i$ by assumption (i) and $\subst{\subst x{d_1}{x_1}\cdots}{d_n}{x_n}=d_i$ by assumption (iii).  The desired conclusion is therefore $\judgeUSN\Gamma{d_i}{\sigma_i}$, which follows from assumption (ii).

\paragraph{Case 1B} $x\notin\{\seq x1n\}$.  We have $\judge\Gamma x\tau$ by assumption (i), and $\subst{\subst x{d_1}{x_1}\cdots}{d_n}{x_n}=x$.  The desired conclusion is therefore $\judgeUSN\Gamma x\tau$.  We already have $\judge\Gamma x\tau$, so we need only show that $x~e_1~\cdots~e_m$ is SN for all appropriately typed USN terms $e_i$.  But in any infinite $\beta$-reduction sequence starting from $x~e_1~\cdots~e_m$, every reduction must be inside one of the $e_i$, since there are no other $\beta$-redexes; therefore some $e_i$ must contain an infinite subsequence.  But this is impossible, since the $e_i$ are USN.

\paragraph{Case 2} Application $e_1~e_2$.  For some type $\sigma$,
\begin{eqnarray}
\lefteqn{\judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}{(e_1~e_2)}\tau}\nonumber\\
&\Rightarrow& \judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}{e_1}{\sigma\to\tau}\ \wedge\ \judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}{e_2}\sigma\nonumber\\
&\Rightarrow& \judgeUSN\Gamma{\subst{\subst{e_1}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma\to\tau}\ \wedge\ \judgeUSN\Gamma{\subst{\subst{e_2}{d_1}{x_1}\cdots}{d_n}{x_n}}\sigma\label{eqn:SN0}
\end{eqnarray}
by the induction hypthesis.  By clause (i) in the definition of USN, this implies
\(
\lefteqn{\judge\Gamma{\subst{\subst{e_1}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma\to\tau}\ \wedge\ \judge\Gamma{\subst{\subst{e_2}{d_1}{x_1}\cdots}{d_n}{x_n}}\sigma}\\
&\Rightarrow& \judge\Gamma{\subst{\subst{(e_1~e_2)}{d_1}{x_1}\cdots}{d_n}{x_n}}\tau\hspace{5cm}.
\)
This establishes clause (i) in the definition of USN for $e_1~e_2$.  For clause (ii), we must show that if $\tau=\tau_3\to\cdots\to\tau_m$ and if $\judgeUSN\Gamma{e_i}{\tau_i}$ for $3\leq i\leq m$, then
\begin{eqnarray}
\lefteqn{\subst{\subst{(e_1~e_2)}{d_1}{x_1}\cdots}{d_n}{x_n}~e_3~\cdots~e_m}\nonumber\\
&=&
(\subst{\subst{e_1}{d_1}{x_1}\cdots}{d_n}{x_n})~(\subst{\subst{e_2}{d_1}{x_1}\cdots}{d_n}{x_n})~e_3~\cdots~e_m\label{eqn:SN1}
\end{eqnarray}
is SN.  But by (\ref{eqn:SN0}),
\(
&& \judgeUSN\Gamma{\subst{\subst{e_1}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma\to\tau_3\to\cdots\to\tau_m}\\
&& \judgeUSN\Gamma{\subst{\subst{e_2}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma}\\
&& \judgeUSN\Gamma{e_i}{\tau_i},\quad 3\leq i\leq m,
\)
thus (\ref{eqn:SN1}) is SN.  This proves that $\judgeUSN\Gamma{\subst{\subst{(e_1~e_2)}{d_1}{x_1}\cdots}{d_n}{x_n}}\tau$.

\paragraph{Case 3} Abstraction $\lam xe$.  We can assume without loss of generality that $\lam xe$ has been $\alpha$-converted so that $x\notin\FV{d_i}$ and $x\neq x_i$ for any $i$, $1\leq i\leq n$.  Instead of $x$, let us call this bound variable $x_{n+1}$.  Then for some $\sigma_{n+1}$, we have
\begin{enumerate}
\renewcommand\labelenumi{(\roman{enumi})}
\item
$\judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}{(\lam{x_{n+1}}e)}{\sigma_{n+1}\to\tau}$,
\item
$\judgeUSN\Gamma{d_i}{\sigma_i}$, $1\leq i\leq n$, and
\item
$x_j\notin\FV{d_i}$ for $j\gt i$ (including $j=n+1$),
\end{enumerate}
and we wish to show $\judgeUSN\Gamma{\subst{\subst{(\lam{x_{n+1}}e)}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma_{n+1}\to\tau}$.

Starting from assumption (i), we have
\(
\lefteqn{\judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}{(\lam{x_{n+1}}e)}{\sigma_{n+1}\to\tau}}\\
&\Rightarrow& \judge{\Gamma,\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1,\,x_{n+1}:\sigma_{n+1}}e\tau\\
&\Rightarrow& \judge{\Gamma,\,x_{n+1}:\sigma_{n+1},\,x_n:\sigma_n,\,\ldots,\,x_1:\sigma_1}e\tau.
\)
If $d_{n+1}$ is any term such that $\judgeUSN\Gamma{d_{n+1}}{\sigma_{n+1}}$, then by the induction hypothesis we have both
\begin{eqnarray}
&& \judgeUSN{\Gamma,\,x_{n+1}:\sigma_{n+1}}{\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n}}\tau\label{eqn:SN2}\\
&& \judgeUSN\Gamma{\subst{\subst{e}{d_1}{x_1}\cdots}{d_{n+1}}{x_{n+1}}}\tau.\label{eqn:SN3}
\end{eqnarray}

For clause (i) in the definition of USN, starting from (\ref{eqn:SN2}), we have
\reasoning{\judge\Gamma{\subst{\subst{(\lam{x_{n+1}}e)}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma_{n+1}\to\tau}}
\(
\lefteqn{\judge{\Gamma,\,x_{n+1}:\sigma_{n+1}}{\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n}}\tau}\\
&\Rightarrow& \judge\Gamma{\lam{x_{n+1}}{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})}}{\sigma_{n+1}\to\tau}\\
&\Rightarrow& \reason{\judge\Gamma{\subst{\subst{(\lam{x_{n+1}}e)}{d_1}{x_1}\cdots}{d_n}{x_n}}{\sigma_{n+1}\to\tau}}{since $x_{n+1}\notin\FV{d_i}$.}
\)

For clause (ii), we wish to show that if in addition to the assumptions (i)--(iii) above, $\tau=\sigma_{n+2}\to\cdots\to\sigma_m\to\rho$ and $\judgeUSN{\Gamma}{d_i}{\sigma_i}$, $n+1\leq i\leq m$, then
\(
\lefteqn{\subst{\subst{(\lam{x_{n+1}}e)}{d_1}{x_1}\cdots}{d_n}{x_n}~d_{n+1}~\cdots~d_m}\\
&=& (\lam{x_{n+1}}{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})})~d_{n+1}~\cdots~d_m
\)
is SN.  Consider any infinite reduction sequence starting from this term.  We know that $\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n}$ is SN by (\ref{eqn:SN2}), and we know that the $d_i$ are SN by assumption, $n+1\leq i\leq m$.  Therefore, eventually a head reduction must be performed:
\(
\lefteqn{(\lam{x_{n+1}}{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})})~d_{n+1}~\cdots~d_m}\\
&\stackrel*\to& (\lam{x_{n+1}}{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})'})~d_{n+1}'~\cdots~d_m'\\
&\to& \subst{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})'}{d_{n+1}'}{x_{n+1}}~d_{n+2}'~\cdots~d_m'.
\)
But we could have done the head reduction initially:
\(
\lefteqn{(\lam{x_{n+1}}{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})})~d_{n+1}~\cdots~d_m}\\
&\to& \subst{\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n}}{d_{n+1}}{x_{n+1}}~d_{n+2}~\cdots~d_m\\
&\stackrel*\to& \subst{(\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n})'}{d_{n+1}'}{x_{n+1}}~d_{n+2}'~\cdots~d_m',
\)
leading to an infinite reduction sequence from $\subst{\subst{\subst{e}{d_1}{x_1}\cdots}{d_n}{x_n}}{d_{n+1}}{x_{n+1}}~d_{n+2}~\cdots~d_m$.  But this contradicts (\ref{eqn:SN3}).
\end{proof}

\begin{proofof}{Theorem \ref{thm:SN}}
Any typable term is USN: take $n=0$ in Lemma \ref{lem:SN}.  Any term that is USN is SN: take $n=0$ in the definition of USN.
\end{proofof}

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