Complexity of true $\Pi_1$-sentences

Question

The set of Gödel numbers of $\Pi_1$-sentences $\varphi(x)$ such that $\varphi(0)$ is true is not recursive. Is it recursively enumerable?

Answer 1

Expanding on Andreas Blass' comment above:

No, it is not r.e., and in fact it is $\Pi^0_1$-complete. To see this, suppose I have a $\Pi^0_1$ formula $\psi(x)\equiv\forall y\theta(x,y)$ with $\theta$ using only bounded quantifiers. For each $n$ consider the new formula $$\psi_n(x)\equiv \forall y\theta(\underline{n},y).$$ We have that $\psi_n$ is in your set iff $\psi(n)$ holds, so the map $n\mapsto\psi_n$ (which is easily checked to be computable) gives a reduction of the set defined by $\psi$ to your set.

In general, the set of true $\Sigma^0_n$ (resp. $\Pi^0_n$) sentences in arithmetic is $\Sigma^0_n$ (resp. $\Pi^0_n$) complete, and shifting from sentences to formulas evaluated at a specific input (your "$\varphi(0)$"s) isn't a substantive change since we can always pull a "dummy variable" trick a la the above.