Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic
Theorem
Let $T$ be the set of theorems of some consistent theory in the language of arithmetic which contains minimal arithmetic.
The set of Gödel numbers of the theorems of $T$ is not definable in $T$.
Proof
The proof is by contradiction.
Let $\Theta$ be the set of Gödel numbers of the theorems of $T$.
Suppose it is defined in $T$ by the formula $\map \theta y$.
Since $T$ contains $Q$, we may apply the Diagonal Lemma to $\neg \map \theta y$.
This gives us a sentence $G$ such that
- $T \vdash G \leftrightarrow \neg \map \theta {\hat G}$
where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times).
Suppose $G$ is not a theorem of $T$.
- Then the Gödel number of $G$ is not in $\Theta$.
- Since $\theta$ defines $\Theta$ in $T$, this means that:
- $T \vdash \neg \map \theta {\hat G}$
- But, by choice of $G$ (specifically, the bi-implication above), this gives us:
- $T\vdash G$
- which contradicts $G$ not being a theorem of $T$
Thus, $G$ is a theorem of $T$.
- But, then the Gödel number of $G$ is in $\Theta$, and
- since $\theta$ defines $\Theta$ in $T$, this means that
- $T \vdash \map \theta {\hat G}$
- But, then this gives us
- $T \vdash \neg G$
- which contradicts $G$ being a theorem of $T$, since $T$ is consistent.
Since assuming $\Theta$ was definable in $T$ necessarily leads to a contradiction, we conclude that it is impossible.
$\blacksquare$