Does $\Sigma_1 \cup \Pi_1$ generate the complete first order theory of arithmetic? If a set $T$ of sentences in the language of arithmetic


*

*is deductively closed under the usual inference rules of first order logic, and

*includes all true $\Sigma_1$ sentences and all true $\Pi_1$ sentences, and

*includes no false sentences;


is $T$ necessarily the complete first-order theory of $\mathbb{N}$?
 A: The consistency of $T$ is a true arithmetical sentence yet is not provable in $T$.  This observation depends on two well-known facts:  First, one can define, in the language of arithmetic, the notion of a "true $\Sigma_1$ or $\Pi_1$ sentence".  (In fact, although there is no arithmetical truth definition for the entire language of arithmetic, there are truth definitions for any bounded part of the arithmetical hierarchy.)  So one can define what $T$ is and then formalize "$T$ is consistent" in the language of arithmetic.  Second, Gödel's second incompleteness theorem (about the unprovability of consistency), though often stated only for recursively axiomatized theories, can be extended to more general arithmetically definable theories like $T$.
A: The set $T$ is computable in $0''$. The theory of $\mathbb{N}$ contains the entire arithmetical hierarchy. Therefore $T$ can't be the theory of $\mathbb{N}$.
In response to the comment I'll show how to construct a sentence that is independent of $T$. There might be a better way, but this is the easiest that I can see.
$T$ is computably enumerable in $0'$, so by Post's theorem there is a $\Sigma_2$ formula $\phi(x)$ such that for any sentence $\psi$, $\mathbb{N} \models \phi(\ulcorner \psi \urcorner)$ if and only if $\psi \in T$. Then, I think the proof of Gödel's incompleteness theorem still applies using $\phi$ as a provability predicate, so we can construct a sentence independent of $T$ by diagonalization.
