%! TEX root = LC.tex
% vim: tw=50
% 02/12/2024 11AM

\textbf{Strategy:}
\begin{enumerate}[(1)]
  \item Given a countable model $\mathcal{M}$ of
    $\PA$, we note that we encode subsets of
    $\Nbb$ as elements of $\mathcal{M}$;
  \item If $\mathcal{M}$ is non-standard, then
    there is an element that codes a non-\gls{rec}
    set;
  \item If $\mathcal{M}$ also has \gls{rec}
    $\oplus$, then there is a membership
    \glsref[dec]{decision}
    procedure for any subset that it codes.
\end{enumerate}
\newcommand\pr{\operatorname{pr}}%
Note that there is a $\Sigma_1$-formula $\pr(x,
y)$ that captures $y$ being the $x$-th prime, and
$\PA \syn \forall x . \exists ! y . \pr(x, y)$.
So if $\Nbb$ thinks that $k$ is the $n$-th prime,
then any model of $\PA$ thinks so too. Write
$\pi_n$ for the $n$-th prime.

\begin{fclemma}[Overspill]
  \label{lemma:2.2.19}
  % Lemma 2.2.19
  Assuming:
  - $\mathcal{M}$ a non-standard model of $\PA$
  - $\varphi(x)$ an $L_{\PA}$-formula
  - $\mathcal{M} \models \varphi(n)$ for all
  standard natural numbers $n$
  Then:
  there is a nonstandard natural number $e$ such
  that $\mathcal{M} \models \varphi(e)$.
\end{fclemma}

\begin{proof}
  Say $\mathcal{M} \models \varphi(n)$ for all
  standard $n$, but only them.
  Then $\mathcal{M} \models \varphi(0)$ and
  $\mathcal{M} \models \forall n . (\varphi(n) \to
  \varphi(n + 1))$ holds (if $\varphi(n)$ holds,
  then $n$ and hence $n + 1$ are standard).

  By $I\varphi$ (induction), we conclude that
  $\mathcal{M} \models \forall n . \varphi(n)$.
  But $\mathcal{M}$ is non-standard, so there is
  non-standard $e \in \mathcal{M}$ with
  $\varphi(e)$, contradiction.
\end{proof}

Fix some $m \in \Nbb$, and a property $\varphi(x)$
of the natural numbers.
\begin{itemize}
  \item There is a number $c$ such that $\forall k
    < m . (\varphi(k) \leftrightarrow \pi_k \mid
    c)$, namely the product of all primes $\pi_k$
    with $k < m$ and $\varphi(k)$.
  \item We perceive $c$ as a code for the numbers
    with the property $\varphi$ below $m$, which
    we can decode by prime factorisation.
\end{itemize}

\begin{fcdefn}[Canonically coded]
  \glsadjdefn{cc}{canonically coded}{subset}%
  \label{defn:2.2.20}%
  % Definition 2.2.20
  A subset $S \subseteq \Nbb$ is \emph{canonically
  coded} in a model $\mathcal{M}$ of $\PA$ if
  there is $c \in \mathcal{M}$ such that
  \[
    S = \{n \in \Nbb : \exists y . (\pi_{\ul{n}}
    \times y = c)\}
  \]
  where $\ul{n}$ denotes the standard number $n$
  in the model.
\end{fcdefn}

We could use other formulas to code subsets. Th
subsets of $\Nbb$ coded in $\mathcal{M}$ are those
$S \subseteq \Nbb$ for which there is a
$\PA$-formula $\varphi(x, y)$ and $c \in
\mathcal{M}$ such that $S = \{n \in \Nbb :
\mathcal{M} \models \varphi(\ul{n}, c)\}$.

As it turns out, coding via $\sigone$-formulae
gives nothing new:

\begin{fcprop}[]
  \label{prop:2.2.21}
  % Proposition 2.2.21
  Assuming:
  - $C(u, x)$ be a $\delz$-formula
  - $\mathcal{M}$ a non-standard model of $\PA$
  Then:
  given any $\tilde{b} \in \mathcal{M}$, there is
  $c \in \mathcal{M}$ such that, for any $n \in
  \Nbb$:
  \[
    \mathcal{M} \models \exists k < \tilde{b} .
    C(k, n) \leftrightarrow \exists y .
    (\pi_{\ul{n}} \times y) = c
  .\]
\end{fcprop}

\begin{proof}[Proof (sketch*)]
  The following formula holds in $\Nbb$ for any
  $n$:
  \[
    \forall b . \exists a . \forall u < n .
    (\exists k < b . C(k, u) \leftrightarrow
    \exists y . (\pi_u \times y) = a)
  .\]
  This is by the reasoning we gave when
  introducing codes, which works due to the bound
  on $k$ and $u$. This can be proved in $\PA$*.

  Thus
  \[
    \mathcal{M} \models \forall b . \exists a .
    \forall u < \ul{n} . (\exists k < b . C(k, u)
    \leftrightarrow \exists y . (\pi_u \times y =
    a))
  \]
  for any $n \in \Nbb$. So by \cref{lemma:2.2.19}
  there is a non-standard $w \in \mathcal{M}$ such
  that
  \[
    \mathcal{M} \models \forall b . \forall a .
    \forall u < w . (\exists k < b . C(k, u)
    \leftrightarrow \exists y . (\pi_u \times y =
    a))
  .\]
  So for any $\tilde{b} \in \mathcal{M}$, there
  must be $c \in \mathcal{M}$ such that
  \[
    \mathcal{M} \models \forall u < w . (\exists k
    < \tilde{b} . C(k, u) \leftrightarrow \exists
    y . (\pi_u \times y = c))
  .\]
  Now $w$ is non-standard, so $\mathcal{M} \models
  \ul{n} < w$ for all $n \in \Nbb$. So for any
  $\tilde{b} \in M$ there is $c \in \mathcal{M}$
  with
  \[
    \mathcal{M} \models \exists k < \tilde{b} .
    C(k, n) \leftrightarrow \exists y .
    (\pi_{\ul{n}} \times y = c)
  \]
  for all $n \in \Nbb$.
\end{proof}

\begin{fcdefn}[Recursively inseparable]
  \glsadjdefn{ri}{recursively inseparable}{sets}%
  \label{defn:2.2.22}
  % Definition 2.2.22
  We say that subsets $A, B \subset \Nbb$ are
  \emph{recursively inseparable} if they are
  disjoint and there is no \gls{rec} $C \subseteq
  \Nbb$ with $B \cap C = \emptyset$ and $A
  \subseteq C$.
\end{fcdefn}

\begin{fcprop}[]
  There are \gls{renum} subsets $A, B \subseteq
  \Nbb$ that are \gls{ri}.
\end{fcprop}

\begin{proof}
  Fix an effective enumeration $\{\varphi_n : n <
  \omega\}$ of the \glspl{prf}. Define $A = \{n
  \in \Nbb : \varphi_n(n) = 0\}$ and $B = \{n \in
  \Nbb : \varphi_n(n) = 1\}$, which are clearly
  disjoint and are clearly \gls{renum}.

  Suppose there is a \gls{rec} $C$ with $A
  \subseteq C$ and $B \cap C = \emptyset$, and
  write $\chi_C$ for its (\gls{tr}) characteristic
  function. There must be $u \in \Nbb$ such that
  $\chi_C = \varphi_u$, as $\chi_C$ is \gls{tr}.

  Since $\chi_C(u)\downarrow$ and is either $0$ or
  $1$, we have either $u \in A$ or $u \in B$.

  If $u \in A$, then $\chi_C(u) = \varphi_u(u) =
  0$, so $u \notin C$, contradicting $A \subseteq
  C$; so $u \in B$. But then $\chi_C(u) =
  \varphi_u(u) = 1$, so $u \in C$, contradicting
  $B \cap C = \emptyset$. Thus $A$ and $B$ are
  \gls{ri}.
\end{proof}
