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.
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@spaceisdarkgreen https://math.stackexchange.com/a/1622337/164699 – Ari Aug 15 '23 at 17:11
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@spaceisdarkgreen Also, https://math.stackexchange.com/a/2818117/164699 the last part of the answer – Ari Aug 15 '23 at 17:12
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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 $\sf PA$.
spaceisdarkgreen
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@spacedarkgreen With respect to PA isn't the reason also that we have the standard model so we know any proofs correspond to actual Godel numbers. I'm just wondering if the same situation can be transferred to set theory if the Pvbl predicate has a clause limiting proofs to finite sets. – Ari Aug 15 '23 at 18:06
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Having a standard model (in the set-theoretic sense) is a stronger assumption than the $\Sigma^0_1$-soundness mentioned by spaceisdarkgreen. – Andreas Blass Aug 15 '23 at 18:18
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@Ari For set theory the question becomes “does the theory have a model whose natural numbers are isomorphic to the standard model of arithmetic?” If not, it can be wrong about what “finite” is. – spaceisdarkgreen Aug 15 '23 at 18:49
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@Ari And per my answer, assuming these model theoretical things is a similar commitment to assuming the soundness of a theory. Even for PA “the standard model of arithmetic is a thing and PA is a model of it” means the same thing as “PA is sound”. – spaceisdarkgreen Aug 15 '23 at 18:57
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2@PW_246 it proves the inconsistency of PA and is stronger than PA. – spaceisdarkgreen Aug 15 '23 at 23:52
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Just to be clear, you’re saying $T \vdash Prv(\bot)$ as opposed to $T \vdash \bot$, right? – PW_246 Aug 16 '23 at 01:42
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2@PW_246 Not sure exactly what you're asking, but $\sf PA+\lnot Con(PA)$ is an example of a theory $T$ such that $T\vdash \operatorname{Prv}_T(\bot)$, but (assuming $\sf PA$ is consistent), $T\not\vdash \bot.$ – spaceisdarkgreen Aug 16 '23 at 01:55
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@spaceisdarkgreen You commented "For set theory the question becomes “does the theory have a model whose natural numbers are isomorphic to the standard model of arithmetic?” " Can it be any other way? I mean if we have just the basic axioms of pairing and finite unions don't we automatically have the Von Neuman numerals? – Ari Aug 17 '23 at 14:08
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@Ari yes, but the Von Neumann naturals constructed in a model need not be isomorphic to the natural numbers. This can be proved by a similar compactness argument to the one that shows there are nonstandard models of arithmetic in the first place. – spaceisdarkgreen Aug 17 '23 at 14:18