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The question arise from a statement that I have heard a lot "ZFC cannot prove the existence of a model of itself by Godel's 2nd incompleteness theorem", in particular referred to the existence of inaccessible cardinals, and I cannot understand why that should be obvious.

I know Godel's 2nd incompleteness theorem can be translated to any theory T that uniformly interprets PA. This way you get a formula $Con(T)$ that is equivalent to say "T is consistent" in any standard model (i.e. a model in which the natural numbers are the usual ones). The formula has a form like $\forall n \neg proof(n,\bot)$ (where $\bot$ any contradiction). (As described for example here: In Godel's first incompleteness theorem, what is the appropriate notion of interpretation function?).

But in non-standard model $Con(T)$ may fail since there are number grater then any standard natural number, and for those numbers $proof(n,\bot)$ may hold or not without having any meaning.


So now I get that if M is a model of ZFC, and $I_M$ be a formula in ZFC for "Exist M", and if we have $ZFC \vdash I_M \to Con(ZFC)$, then assuming $ZFC\vdash I_M$ would lead to $ZFC\vdash Con(ZFC)$ which is a contradiction by 2nd incompleteness theorem.

My question is: but how can we prove $ZFC \vdash I_M \to Con(ZFC)$? And why that should be obvious?



From a formal point of view, I have no idea how to derive that.

But even from an informal point of view I have some problem in understanding:

Con(ZFC) has the meaning "ZFC is consistent" only in a model where there are no strange natural numbers. So even if ZFC proves it has a model M, to get Con(ZFC) I think we should not only be able to prove that the model is standard, but also that ZFC itself recognize the model as standard, and I think this is something impossible since "being standard" (so not having any natural number different from the usual ones) it's not a first order statement.

If M is a model of $ZFC \cup \{\neg Con(ZFC)\}$ would it be true anyway $ZFC \vdash I_M \to Con(ZFC)$?

Or even if M is actually a model in which $Con(ZFC)$ is true, we just stated that ZFC proves it has a model in which $Con(T)$ holds, not that every model satisfy it. And being just satisfied in a single model does not mean it is a theorem.

I would be happy if I can understand even just one of the 2 possible cases, so if that is true with $M$ a general model (so may have non-standard $\omega$) or if it is true and how to do it when $M$ is concrete easy standard model (like $M=V_k$ for $k$ inaccessible and $I_M$ is "exist $k$ inaccessible")

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Your argument that a non-standard model would make the $\operatorname{Con}$ "meaningless" is a bit of misleading to yourself.

In a model of $\sf ZFC$ with a non-standard $\omega$ (e.g. a model of $\lnot\operatorname{Con}\sf(ZFC)$) the way the model would interpret $\sf ZFC$ internally will also be different, because $\sf ZFC$ has axiom schemata: the non-standard integers will contribute some new non-standard axioms. So you should also say that $\sf ZFC$ is "meaningless" in that model.

But in either case, as I said, you'll be lying to yourself. The arithmetic predicate for $\sf ZFC$ (or rather "a Gödel number of an axiom in $\sf ZFC$") or the arithmetic predicate $\operatorname{Con}$, are descriptions. They are still valid in non-standard models, and they just produce "some extra noise". These alleged inconsistencies which might happen are because in the internal version of logic and the theory in discussion, the model will produce more axioms, more inference rules, and even more axioms to the theory. And these can—and sometimes do—create inconsistencies.

But it is only misleading yourself to say "Oh, but if it's not the standard model, then it's a bunch of formal nonsense".

The way to prove what you want to prove is to show that:

  1. $\sf ZFC$ internalizes logic in a very strong way: it has a completeness theorem. So if a theory is consistent, it has a model, and vice versa. Secondly, the arithmetic operator $\operatorname{Con}$ is compatible with this version internally. So a theory which can be Gödelized is consistent if and only if $\operatorname{Con}(T)$ is true in the $\omega$ of the model (which is the unique model for second-order arithmetic in that model).

  2. $\sf ZFC$ proves that if $\kappa$ is inaccessible, then $V_\kappa$ is a model of $\sf ZFC$, but not just of $\sf ZFC$, but in fact of the internal version of it.

Combine the two proofs, and you get that $\sf ZFC$ proves that if there is an inaccessible cardinal, then there is a model of $\sf ZFC$, and therefore $\sf ZFC$ is consistent, and therefore $\operatorname{Con}\sf(ZFC)$ holds.


Now, just a few words of terminology. An inner model is a technical term which refers—usually—to a proper class which is transitive, contains all the ordinals, and satisfies the axioms of $\sf ZF$ or $\sf ZFC$. Being a class model, the truth predicate for it is a meta-predicate, and what $\sf ZFC$ can do at best, is to prove for each axiom of $\sf ZF$ that it holds in the inner model. For example $L$ is an inner model of $V$.

Inner models are not enough to prove the consistency of $\sf ZFC$. They are only enough to establish the meta-theoretic equiconsistency results. For example if $\sf ZF$ is consistent, then by Gödel's construction of $L$ we know that $\sf ZFC$ is consistent. But this is a meta-theorem, it is taking place in the meta-theory, rather than inside the theory.

So it is not enough to conclude from the existence of an inner model that $\operatorname{Con}\sf(ZFC)$ holds or fails.

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  • $\begingroup$ It is perhaps worth pointing out though that you can use $L$ in a somewhat different way to prove equiconsistency without passing to the metatheory: if ZF has a model, then you can form $L$ inside that model to get a model of ZFC. $\endgroup$ – Eric Wofsey Mar 20 '18 at 1:15
  • $\begingroup$ Eric, that is a meta theorem, but here your meta theory is ZFC. $\endgroup$ – Asaf Karagila Mar 20 '18 at 8:26

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