Does the property $T\vdash Pvbl_T(\ulcorner \sigma \urcorner) \implies T\vdash \sigma$ apply to set theories?

Question

I know from other posts that $PA\vdash Pvbl_{PA}(\ulcorner \sigma \urcorner ) \implies PA\vdash \sigma$ and this applies to other extensions/restrictions of PA as well. Does it also apply to set theories where Godel numbering can be carried out but not every element of a model's universe corresponds to a Godel number? Specifically, I'm wondering if it is possible for $Pvbl_{ZF}$ to get "fooled" by a set in the universe claiming to be a proof which is not a genuine Godel number. I'm thinking it might be avoidable because hereditary finite sets are definable but I don't know if that helps. Thanks.

Answer 1

This is ultimately an arithmetic soundness assumption on the theory. For instance, it fails for $\sf PA+\lnot Con(PA),$ or any other theory that is consistent but proves its own inconsistency.

There’s no real satisfactory way to prove $\sf ZF$ is ($\Sigma^0_1$-) sound any more than there’s a way to prove $\sf ZF$ is consistent (it’s a stronger assumption). So it’s possible in that sense (e.g. I’m sure if you search around you’ll find people acknowledging the possibility that $\sf ZF$ is consistent but proves its own inconsistency), but generally we assume such theories are sound and this kind of thing doesn’t happen.

Note the the difference between $\sf PA$ and $\sf ZF$ here is only one of degree: $\sf PA$ is weaker and more transparent, so we’re just more comfortable assuming outright that it is sound. Also, $\sf ZF$ can prove $\sf PA$ is sound and not that $\sf ZF$ is sound, but it seems weird to trust $\sf ZF$ if you’re skeptical about $\sf PA$.