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% 04/12/2024 11AM

\begin{fclemma}[]
  \label{lemma:2.2.24}
  % Lemma 2.2.24
  Assuming:
  - $M \models \PA$ non-standard
  Then:
  there is a non-\gls{rec} set $S$ which is
  canonically coded in $\mathcal{M}$.
\end{fclemma}

\begin{proof}
  Say $A, B \subseteq \Nbb$ are \gls{renum} and
  \gls{ri}. By \cref{coro:2.2.11}, there are
  $\sigone$-formulae $\exists u . a(u, x)$ and
  $\exists u . b(u, x)$ defining $A$ and $B$
  respectively (so $a$ and $b$ are
  $\delz$-formulae).

  Fix $n \in \Nbb$. As the sets are disjoint, we
  have:
  \[
    \Nbb \models \forall v < n . \forall w < n .
    \forall x < n . \neg (a(v, x) \wedge b(w, x))
  .\]
  As this sentence is $\delz$, it follows, for any
  non-standard $\mathcal{M} \models \PA$ and
  $\ul{n} \in \mathcal{M}$ that:
  \[
    \mathcal{M} \models \forall v < \ul{n} .
    \forall w < \ul{n} . \forall x < \ul{n} .
    \neg(a(v, x) \wedge b(w, x))
  .\]
  By \nameref{lemma:2.2.19}, there is some
  non-standard $c \in \mathcal{M}$ such that
  \[
    \mathcal{M} \models \forall v < c . \forall w <
    c . \forall x < x . \neg(a(v, x) \wedge b(w,
    x))
    \tag{$*$}
    \label{lec24_eq}
  .\]

  Now define $X \defeq \{n \in \Nbb : \exists v <
  c . a(v, \ul{n})\}$. Note that:
  \begin{itemize}
    \item $A \subseteq X$: let $n \in A$, so that
      $\Nbb \models a(m, n)$ for some $m \in
      \Nbb$ (a $A$ is defined by $\exists u . a(u,
      x)$). Then $\mathcal{M} \models a(\ul{m},
      \ul{n})$, as $a$ is $\delz$. Hence
      $\mathcal{M} \models \exists v < c . a(v,
      \ul{n})$ as any standard $\ul{m}$ is below
      $c$ as it is non-standard. But then $n \in
      X$.
    \item $B \cap X = \emptyset$: if $n \in B$,
      then $\Nbb \models b(m, n)$ for some $m$, so
      arguing as before we get $\mathcal{M}
      \models \exists w < c . b(w, \ul{n})$. By
      \eqref{lec24_eq}, we can deduce $\mathcal{M}
      \models \neg \exists v < c . a(v, \ul{n})$.
      So $n \notin X$.
  \end{itemize}
  As $A$ and $B$ are \gls{ri}, $X$ can't be
  \gls{rec}. This shows that $\mathcal{M}$ must
  encode a non-\gls{rec} set, which implies that
  it must \glsref[cc]{canonically} encode a non-\gls{rec} set
  by \cref{prop:2.2.21}.
\end{proof}

\begin{fcthm}[Tennenbaum]
  \label{thm:2.2.25}
  % Theorem 2.2.25
  Assuming:
  - $\mathcal{M} = (M, \oplus, \otimes,
  \preccurlyeq, n_0, n_1)$ a countable
  non-standard model of $\PA$
  Then:
  $\oplus$ is not \gls{rec}.
\end{fcthm}

\begin{proof}
  As $\mathcal{M}$ is countable, we may as well
  assume that $M = \Nbb$, $n_0 = 0$, $n_1 = 1$.

  By \cref{lemma:2.2.24}, there is some $c \in M$
  that \glsref[cc]{canonically codes} a
  non-\gls{rec} subset $X = \{n : M \models
  \exists y . (\pi_{\ul{n}} \times y = c)\}
  \subseteq \Nbb$.

  As $\PA$ proves that
  \[
    \pi_{\ul{n}} \times x = \ub{x + \cdots + x}_{\text{$\pi_n$
    times}}
  ,\]
  we have that
  \[
    \pi_{\ul{n}} \times y = \ub{y + \cdots + y}_{\text{$\pi_n$
    times}}
  \]
  for all $y \in M$. So $n \in X$ if and only if
  there is $d \in M$ such that
  \[
    c = \ub{d \oplus \cdots \oplus d}_{\text{$\pi_n$ times}}
  .\]
  Suppose $\oplus$ is \gls{rec}. Then we can can
  through $\Nbb$ (which is $M$) and look for some
  $d \in M$ that realises the disjunction of:
  \[
    \begin{cases}
      c = \ub{x \oplus \cdots \oplus
      x}_{\text{$\pi_n$ $x$'s}} \\
      c = \ub{x \oplus \cdots \oplus
      x}_{\text{$\pi_n$ $x$'s}} \oplus 1 \\
      \cdots
      c = \ub{x \oplus \cdots \oplus
      x}_{\text{$\pi_n$ $x$'s}} \oplus \ub{1
      \oplus \cdots \oplus 1}_{\text{$\pi_n - 1$
      ones}}
    \end{cases}
  \]
  As $\oplus$ is \gls{rec}, we can decide whether
  the disjunction holds of a given $d$. Moreover,
  the search for such $d$ always terminates:
  \begin{itemize}
    \item Euclidean division is provable in $\PA$:
      for any $u, v \in M$ with $v \neq 0$, there
      are unique $q, r \in M$ such that $r
      \preccurlyeq v$ and $u = (v \otimes q)
      \oplus r$.
    \item \[
        \PA \syn \forall x . (x < \pi_1
        \leftrightarrow (x = 0 \wedge x = 1 \wedge
        \cdots \wedge x = (1 + \cdots + 1))
      ;\]
  \end{itemize}
  Combining these, we get that division of $c$ by
  $\pi_{\ul{n}}$ in $M$ leaves a unique quotient
  $d \in M$, and remainder $r \preccurlyeq
  \pi_{\ul{n}}$, which is either $0$ or $1$ or $1
  \oplus 1$ or \ldots or $1 \oplus 1 \oplus \cdots
  \oplus 1$ ($\pi_n - 1$ times); i.e. one of the
  disjunctions from before.

  Now we see that $X$ is \gls{rec}: if our
  search provides $d$ such that
  \[
    \mathcal{M} \models c = \ub{d \oplus \cdots \oplus
    d}_{\text{$\pi_n$ times}}
  ,\]
  then $n \in X$, and if the search gives $d$
  satisfying one of the other disjunctions, then
  $n \notin X$.

  This contradicts the choice of $X$, so $\oplus$
  can't be \gls{rec}.
\end{proof}