%! TEX root = LC.tex
% vim: tw=50
% 27/11/2024 11AM

  If $f(\ol{x})$ is $\mu y . g(\ol{x}, y) = 0$ and
  the graph of $g$ is definable by a
  $\sigone$-formula, then the graph of $f$ is
  definable by
  \[
    \exists u . ((u)_y = 0 \wedge \forall i < y .
    ((u)_i \neq 0 \wedge \ub{\forall j \le y . \exists
    v (v = g(\ol{x}, j) \wedge v = (u)_j)}_{(*)}))
  \]
  by using \nameref{lemma:2.2.9} to code
  $g(\ol{x}, 0), g(\ol{x}, 1), \ldots, g(\ol{x},
  f(\ol{x}))$.

  Again, this is equal to a $\sigone$-formula if
  the graph of $g$ is given by $\exists \ol{w}
  \varphi(\ol{x}, y, z, \ol{w})$ with $\varphi \in
  \delz$, then $(*)$ is equal in $\Nbb$ to
  \[
    \exists s . \forall j \le y . \exists v,
    \ol{w} \le s . (v = (u)_j \wedge
    \varphi(\ol{x}, j, v, \ol{w}))
    .
    \qedhere
  \]
\end{enumerate}
\end{proof}

\begin{fccoro}[]
  \label{coro:2.2.11}
  % Corollary 2.2.11
  \begin{iffc}
    \lhs
    A subset $A \subseteq \Nbb^k$ is \gls{renum}
    \rhs
    there is a $\sigone$-formula $\psi(x_1,
    \ldots, x_k)$ such that, given $\ol{x} \in
    \Nbb^k$, we have $\ol{x} \in A$ if and only if
    $\Nbb \models \psi(x)$.
  \end{iffc}
\end{fccoro}

\begin{iffproof}
  \rightimpl
  If $A$ is \gls{renum}, then there is a \gls{rec}
  $f$ such that $A = \dom(f)$. Given $\ol{x} \in
  \Nbb^k$, we thus have $x \in A$ if and only if
  $\Nbb \models \exists v . v = f(\ol{x})$. But
  $\exists v . v = f(\ol{x})$ is equal to a
  $\sigone$-formula by \cref{thm:2.2.10}.

  \leftimpl
  Conversely, if $A$ is defined in $\Nbb$ by a
  $\sigone$-formula $\psi$, define $f(\ol{x}) = 0$
  if $\Nbb \models \psi(\ol{x})$, and $f(\ol{x})
  \uparrow$ otherwise. The graph of $f$ is given
  by $y = 0 \wedge \psi(\ol{x})$, which is
  $\sigone$, and so $f$ is \gls{rec} by
  \cref{thm:2.2.10}. But $A = \dom(f)$, so $A$ is
  \gls{renum}.
\end{iffproof}

Any model of \PAm{} includes a copy of $\Nbb$ inside
of it: consider the \emph{standard natural
numbers}
\[
  \ul{n} = \ub{SSS \ldots S}_{n}0
.\]

In fact, $\Nbb$ embeds in any model \PAm{} as an
initial segment: essentially because
\[
  \PAm \syn \forall x . (x \le \ul{k} \to x =
  \ul{0} \wedge x = \ul{1} \wedge \cdots \wedge x
  = \ul{k})
.\]
In Example Sheet 4, you will see that $\Nbb$ is a
$\delz$-elementary substructure of any model of
\PAm: every $\delz$-sentence $\varphi(\ul{n})$
true in $\Nbb$ is also true in the model.

\begin{fcdefn}[Representation of a total function]
  \glsadjdefn{reped}{represented}{function}%
  \glsadjdefn{reple}{representable}{function}%
  \glsnoundefn{repon}{representation}{representations}%
  \glsnoundefn{rep}{represent}{represents}%
  Let $f : \Nbb^k \to \Nbb$ be total and $T$ be
  any $L_{\PA}$-theory extending \PAm. We say that
  $f$ is \emph{represented in $T$} if there is an
  $L_{\PAm}$ formula $\theta(x_1, \ldots, x_k, y)$
  such that, for all $\ol{n} \in \Nbb^k$:
  \begin{cenum}[(a)]
    \item $T \syn \exists! y . \theta(\ol{n}, y)$
    \item If $k = f(\ol{n})$, then $T \syn
      \theta(\ol{n}, \ul{k})$
  \end{cenum}
\end{fcdefn}

\begin{fclemma}
  \label{lemma:2.2.13}
  % Lemma 2.2.13
  Every total recursive function $f : \Nbb^k \to
  \Nbb$ is $\sigone$-\gls{reped} in \PAm.
\end{fclemma}

\begin{proof}
  The graph of $f$ is given by a $\sigone$-formula
  by \cref{thm:2.2.10}, say $\exists \ol{z} .
  \varphi(\ol{x}, y, \ol{z})$ where $\varphi \in
  \delz$. Without loss of generality, we may
  assume that $\ol{z}$ is a single variable (for
  example, rewrite $\exists z . \exists \ol{w} < z
  . \varphi(\ol{x}, y, \ol{w})$).

  Let $\psi(\ol{x}, y, z)$ be the $\delz$-formula
  \[
    \varphi(\ol{x}, y, z) \wedge \forall u, v \le y
    + z . (u + v < y + z \to \neg \varphi(\ol{x},
    u, v))
  .\]
  Then the $\sigone$-formula $\theta(\ol{x}, y)
  \defeq \exists z . \psi(\ol{x}, y, z)$
  \glspl{rep} $f$ in \PAm.

  We show $\PAm \syn \theta(\ol{n}, k)$ first,
  where $k = f(\ol{n})$. Note that $k$ is the
  unique element of $\Nbb$ such that $\Nbb \models
  \exists z . \varphi(\ol{n}, k, z)$, as $f$ is a
  function.

  Take $l$ to be the first natural number such
  that $\Nbb \models \varphi(\ol{n},
  k, l)$. Then $\Nbb \models \psi(\ol{n}, k, l)$
  too, whence $\Nbb \models \exists z .
  \psi(\ol{n}, k, z)$. But any $\sigone$-sentence
  true in $\Nbb$ is true in any model of \PAm
  (c.f. Example Sheet 4), so $\PAm \syn \exists z
  . \psi(\ol{n}, k, z)$, i.e. $\PAm \syn
  \theta(\ol{n}, k)$.

  To see that $\PAm \syn \exists ! y .
  \theta(\ol{n}, y)$, let $l$ be the first number
  such that $\Nbb \models \varphi(\ol{n}, k, l)$,
  where $k = f(\ol{n})$. Suppose $a, b \in
  \mathcal{M} \models \PAm$, with $\mathcal{M}
  \models \psi(\ol{n}, a, b)$. We will show that
  $a = k$. Completeness settles the claim. Again,
  $\varphi(\ol{n}, k, l)$ is a $\delz$-sentence
  true in $\Nbb$, thus true in $\mathcal{M}$.

  Using the fact that $<$ is a linear ordering in
  $\mathcal{M}$, we have $a, b \le k + l \in
  \Nbb$, so $a, b \in \Nbb$ (as $\Nbb$ is an
  initial segment of $\mathcal{M}$). Now
  $\mathcal{M} \models \psi(\ol{n}, a, b) \in
  \delz$, hence $\Nbb \models \psi(\ol{x}, a, b)$
  and thus $\Nbb \models \exists z .
  \varphi(\ol{n}, a, z)$. Thus $a = k$ as needed.
\end{proof}

\begin{fccoro}
  \label{coro:2.2.14}
  % Corollary 2.2.14
  Every \gls{rec} set $A \subseteq \Nbb^k$ is
  $\sigone$-\gls{reple} in \PAm.
\end{fccoro}

\begin{proof}
  The characteristic function $\chi_A$ of $A$ is
  \gls{tr}, so $\chi_A(\ol{x}) = y$ is \gls{reped}
  by some $\sigone$-formula $\theta(\ol{x}, y)$ in
  \PAm. But then $\theta(\ol{x}, 1)$ \glspl{rep}
  $A$ in \PAm.
\end{proof}
