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\begin{fclemma}[Diagonalisation Lemma]
  \label{lemma:2.2.15}
  % Lemma 2.2.15
  Assuming:
  - $T$ an $L_{\PA}$-theory
  - in $T$,
  every \gls{tr} function is $\sigone$-\gls{reped}
  - $\theta(x)$ an $L_{\PA}$-formula with one
  free variable $x$
  Then:
  there is an
  $L_{\PA}$-sentence $G$ such that
  \[
    T \syn G \leftrightarrow \theta(\Gn{G})
  .\]
  Moreover, if $\theta$ is a $\pione$-formula,
  then we can take $G$ to be a $\pione$-sentence.
\end{fclemma}

\begin{proof}
  Define a \gls{tr} function $\diag$ this way: on
  input $n \in \Nbb$, check if $n =
  \Gn{\sigma(x)}$ is the \gls{Gn} of some
  $L_{\PA}$-formula $\sigma(x)$. If so, return
  $\Gn{\forall y . (y = \ul{n} \to \sigma(y))}$,
  else return $0$.

  As $\diag$ is \gls{tr}, it is
  $\sigone$-\gls{reped} in $T$ by some $\delta(x,
  y)$. Consider the formula
  \[
    \psi(x) \defeq \forall z . (\delta(x, z) \to
    \theta(z))
  .\]
  Let $n = \Gn{\psi(x)}$ and $G \defeq \forall y .
  (y = \ul{n} \to \psi(y))$. This makes $G$ the
  sentence whose \gls{Gn} is
  $\diag(\Gn{\psi(x)})$. It is obvious that $T
  \syn G \leftrightarrow \psi(\ul{n})$, so we know
  that
  \[
    T \syn G \leftrightarrow \forall z .
    (\delta(\ul{n}, z) \to \theta(z))
    \label{lec22_alpha}
    \tag{$\alpha$}
  .\]
  Now $\delta(x, y)$ \glspl{rep} $\diag$ in $T$,
  and $\diag(n) = \Gn{G}$ by construction, hence
  \[
    T \syn \forall z . (\delta(\ul{n}, z)
    \leftrightarrow z = \Gn{G})
    \label{lec22_beta}
    \tag{$\beta$}
  .\]
  Combining \eqref{lec22_alpha} and
  \eqref{lec22_beta}, we get $T \syn G
  \leftrightarrow \theta(\Gn{G})$ as needed.

  Finally, note that if $\theta \in \pione$, then
  both $\psi$ and $G$ are equal to a
  $\pione$-formula.
\end{proof}

\begin{fcthm}[Crude Incompleteness]
  \label{thm:2.2.16}
  % Theorem 2.2.16
  Assuming:
  - $T$ be a \gls{rec} set of (\glspl{Gn} of)
  $L_{\PA}$-sentences
  - $T$ is consistent (never
  includes both $\varphi$ and $\neg \varphi$)
  - $T$ contains all the $\sigone$ and $\pione$
  sentences provable in \PAm
  Then:
  there is a $\pione$-sentence $\tau$ such that $\tau
  \notin T$ and $\neg \tau \notin T$.
\end{fcthm}

\begin{proof}
  Let $\theta(x)$ be a $\sigone$-formula that
  \glspl{rep} $T$ in \PAm, so that
  \[
    x \in T \iff \PAm \syn \theta(x) \qquad
    \text{and} \qquad x \notin T \iff \PAm \syn
    \neg \theta(x)
  .\]
  This exists since $T$ is \gls{rec}. By the
  \nameref{lemma:2.2.15}, there is a
  $\pione$-sentence $\tau$ such that $\PAm \syn
  \tau \leftrightarrow \neg \theta(\Gn{\tau})$.

  If $\Gn{\tau} \in T$, then $\PAm \syn
  \theta(\Gn{\tau})$, and thus $\PAm \syn \neg
  \tau$. But then $\Gn{\neg\tau} \in T$ (as $\neg
  \tau \in \sigone$ and $\PAm$ proves it).

  If $\Gn{\neg\tau} \in T$, then $\tau \notin T$,
  so $\PAm \syn \neg \theta(\Gn{\tau})$, and thus
  $\PAm \syn \tau$. As $\tau \in \pione$ and
  $\PAm \syn \tau$, we have $\Gn{\tau} \in T$.

  Since $T$ is consistent, we can't have either of
  $\Gn{\tau}$ or $\Gn{\neg\tau}$ in $T$.
\end{proof}

\begin{fccoro}[Godel-Rosser Theorem]
  \label{coro:2.2.17}
  % Corollary 2.2.17
  Let $T$ be a consistent $L_{\PA}$-theory
  extending $\PAm$ and admitting a \gls{renum}
  axiomatisation. Then $T$ is $\pione$-incomplete:
  there is a $\pione$-sentence $\tau$ such that
  $T \not\syn \tau$ and $T \not\syn \neg \tau$.
\end{fccoro}

\begin{proof}
  By \nameref{thm:2.2.4}'s Theorem, we may assume that $T$ is
  \gls{rec}. Suppose that $T$ is
  $\pione$-complete, and consider the set $S$ of
  (\glspl{Gn} of) all the $\sigone$ and $\pione$
  sentences in $L_{\PA}$ that $T$ proves.

  The set $S$ is \gls{rec}: we can effectively
  decide if a given sentence is $\sigone$ or
  $\pione$, then check if $\Gn{\sigma} \in S$ by
  systematically searching through all proofs
  using the axioms in $T$, until we either find a
  proof of $\sigma$ or a proof of $\neg \sigma$.
  Since $T$ is $\pione$-complete, there is always
  such a proof, and we'll find it in finite
  time.

  But then $S$ satisfies the hypotheses of
  \cref{thm:2.2.16}, so there is a
  $\pione$-sentence $\tau$ with $\Gn{\tau} \notin
  S$ and $\Gn{\neg \tau} \notin S$, contradicting
  $\pione$-completeness of $T$.
\end{proof}

\begin{fcdefn}[Recursive structure]
  \glsadjdefn{recm}{recursive}{model}%
  \label{defn:2.2.18}
  % Definition 2.2.18
  A (countable) $L_{\PA}$-structure $\mathcal{M}$
  is \emph{recursive} if there are \gls{tr}
  functions $\oplus : \Nbb^2 \to \Nbb$, $\otimes
  : \Nbb^2 \to \Nbb$, a binary recursive relation
  $\preccurlyeq \subseteq \Nbb^2$, and natural
  numbers $n_0, n_1 \in \Nbb$ such that
  $\mathcal{M} \cong (\Nbb, \oplus, \otimes,
  \preccurlyeq, n_0, n_1)$ as
  $L_{\PA}$-structures.
\end{fcdefn}

We will show that the usual $\Nbb$ is the only
\gls{recm} model of $\PA$ (up to $\cong$).