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% 13/11/2024 11AM

\begin{fclemma}[]
  \label{lemma:2.1.6}
  % Lemma 2.1.6
  Assuming:
  - $M \tob N$
  Then:
  $M^{\tstar} \tobev N^{\tstar}$.
\end{fclemma}

\begin{proof}
  Induction over the structure of $M$. We'll leave
  the easier cases as exercises, and focus on when
  $M$ is a \gls{redex}, or when $M = P_1 P_2$,
  where $P_1$ is not a \gls{lab} and $N = Q_1 P_2$
  with $P_1 \tob Q_1$.

  Suppose that $M = (\lambda x . P_1) P_2$ is a
  \gls{redex}. Then there are three possibilities
  for $N$.
  \begin{enumerate}[(1)]
    \item $N = P_1\subst[x \defeq P_2]$: here
      $M^{\tstar} \tobev N^{\tstar}$ by the previous
      lemma.
    \item $N = (\lambda x . Q_1) P_2$, where
      $P_1 \tob Q_1$: here $N^{\tstar} =
      Q_1^{\tstar}
      \subst[x \defeq P_2^{\tstar}]$. By the
      induction hypothesis, $P_1^{\tstar} \tobev
      Q_1^{\tstar}$, so
      \[
        M^{\tstar} = P_1^{\tstar} \subst[x \defeq
        P_2^{\tstar}] \tobev Q_1^{\tstar} \subst[x
        \defeq P_2^{\tstar}] = N
      .\]
    \item $N = (\lambda x . Q_1) Q_2$, where $P_2
      \tob Q_2$: is similar.
  \end{enumerate}
  Now suppose $M = P_1 P_2$, where $P_1$ is not a
  \gls{lab}, and $N = Q_1 P_2$ with $P_1 \tob
  Q_1$. Here $M^{\tstar} = P_1^{\tstar}
  P_2^{\tstar}$. If
  $Q_1$ is not a \gls{lab}, the result is clear.
  So let $Q_1 = \lambda y . R$. Applying the
  induction hypothesis to $P_1 \tob \lambda y .
  R$, we get $P_1^{\tstar} \tobev \lambda y .
  R^{\tstar}$. Thus
  \[
    M^{\tstar} = P_1 ^{\tstar} P_2^{\tstar} \tobev (\lambda
    y . R^{\tstar}) P_2^{\tstar} \tob R^{\tstar} \subst[y
    \defeq P_2^{\tstar}] = N^{\tstar}
    .
    \qedhere
  \]
\end{proof}

\begin{fccoro}[]
  % Corollary 2.1.7
  If $M \tobev N$, then $M^{\tstar} \tob
  N^{\tstar}$.
\end{fccoro}

\begin{proof}
  Induction over the length of the chain $M \tobev
  N$, using \cref{lemma:2.1.6}.
\end{proof}

Applying this multiple times, $M \tobev N$ implies
$M^{\tstar[n]} \tobev N^{\tstar[n]}$ for all $n <
\omega$.

\begin{fcthm}[]
  \label{thm:2.1.8}
  % Theorem 2.1.8
  Assuming:
  - $M$ $\beta$-\gls{red} to $N$ in $n$ steps
  Then:
  $N \tobev M^{\tstar[n]}$.
\end{fcthm}

\begin{proof}
  By induction over $n$. The base case is clear,
  as $n = 0$ implies $M = N$.

  For $n > 0$, there is a \gls{uterm} $R$ with $M \tob R
  \to_{(n - 1)\beta} N$. By induction hypothesis,
  $N \tobev R^{\tstar[n - 1]}$. Since $M \tob R$, we
  have $R \tobev M^{\tstar}$ by \cref{lemma:2.1.4}.
  Thus we get $R^{\tstar[n - 1]} \tobev
  M^{\tstar[n]}$
  by the previous observation. Putting it all
  together:
  \[
    N \tobev R^{\tstar[n - 1]} \tobev
    M^{\tstar[n]}
    .
    \qedhere
  \]
\end{proof}

\begin{fcthm}[Church, Rosser, 1936]
  % Theorem 2.1.9
  Assuming:
  - $M, N_1, N_2$ are \glspl{ulterm} such that $M
  \tobev N_1, N_2$
  Then:
  there is a \gls{ulterm} $N$ such that $N_1, N_2
  \tobev N$.
\end{fcthm}

\begin{proof}
  Say $M \to_{r\beta} N_1$, $M \to_{s\beta} N_2$.
  Without loss of generality, say $r \le s$. By
  \cref{thm:2.1.8}, we have that $N_1 \tobev
  M^{\tstar[r]}$ and $N_2 \tobev M^{\tstar[s]}$. But
  $M^{\tstar[r]} \tobev M^{\tstar[s]}$ by successive
  applications of \cref{lemma:2.1.4} (as $r \le
  s$). So take $N = M^{\tstar[s]}$.
\end{proof}

Reminder of the picture to think of:
\begin{picmath}
  \begin{tikzcd}
    & M \ar[ld, "s\beta", swap] \ar[rd, "r\beta"] \\
    N_2 \ar[rdd, two heads, "\beta", swap]
    &
    & N_1 \ar[d, two heads, "\beta"] \\
    &
    & M^{r*} \ar[ld, two heads, "\beta"] \\
    & M^{s*}
  \end{tikzcd}
\end{picmath}

This has some important consequences:
\begin{itemize}
  \item If $M \equivb N$, then they $\tobev$ to
    the same \gls{uterm};
  \item If the \gls{bnf} of a \gls{uterm} exists,
    it is unique;
  \item We can use this to show that two
    \glspl{uterm} are not $\beta$-equivalent.
\end{itemize}

\begin{example*}
  $\lambda x . x$ and $\lambda x . \lambda y . x$
  are different \glspl{uterm} in \gls{bnf}, so
  they can't be $\beta$-equivalent.
\end{example*}

\begin{fcdefn}[Church numeral]
  \glsnoundefn{cn}{Church numeral}{Church
  numerals}%
  \glssymboldefn{cn}%
  % Definition 2.1.10
  Let $n$ be a natural number. Its corresponding
  \emph{Church numeral} $c_n$ is the \gls{ulterm}
  $c_n \defeq \lambda s . \lambda z . s^n(z)$,
  where $s^n(z)$ denotes
  \[
    \ub{s(s(\ldots(s}_{\text{$n$ times}}z)\ldots)
  .\]
\end{fcdefn}

\begin{example}
  % Example 2.1.11
  $\cn_0 = \lambda s . \lambda z . z$ is the
  `function' that takes $s$ to the identity map.

  $\cn_1 = \lambda s . \lambda z . \lambda s(z)$ is
  the `function' that takes $s$ to itself.

  $\cn_2 = \lambda s . \lambda z . ss(z)$ takes a
  function $s$ to its $2$-fold composite $z
  \mapsto s(s(z))$.
\end{example}

\begin{fcdefn}[lambda-definability]
  % Definition 2.1.2
  \glsadjdefn{ldef}{$\lambda$-definable}{function}%
  \glsnoundefn{ldefine}{$\lambda$-define}{$\lambda$-defines}%
  A partial function $f : \Nbb^k \to \Nbb$ is
  \emph{$\lambda$-definable} if there is a
  \gls{ulterm} $F$ such that $F \cn_{n_1} \ldots
  \cn_{n_k} \equivb \cn_{f(n_1, \ldots, n_k)}$.
\end{fcdefn}

\begin{fcprop}[Rosser]
  % Proposition 2.1.3
  Define the following \gls{ulterm}:
  \begin{citems}
    \item $A_+ \defeq \lambda x . \lambda y .
      \lambda s . \lambda z . xs(ys(z))$,
    \item $A_* \defeq \lambda x . \lambda y .
      \lambda s . x(ys)$,
    \item $A_e \defeq \lambda x . \lambda y . yx$.
  \end{citems}
  Then for all $n, m \in \Nbb$:
  \begin{citems}
    \item $A_+ \cn_n \cn_m \equivb \cn_{n + m}$;
    \item $A_* \cn_n \cn_m \equivb \cn_{nm}$;
    \item $A_e \cn_n \cn_m \equivb \cn_{n^m}$ if $m >
      0$.
  \end{citems}
\end{fcprop}