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Let $I$ be the formula which states that there exists strongly inaccessible cardinals.

My question is regarding the proof of $ZFC\nvdash I$ appearing in Jech (part of theorem 12.12). He starts by proving (in $ZFC$) that if $\kappa$ is strongly inaccessible then $V_\kappa\models ZFC$. Why is it not done here? if $ZFC\vdash I$ then $ZFC$ proves its own consistency ($V_\kappa$ is a model) which contradicts the second incompleteness theorem.

Instead, he proceeds to claim that $V_\kappa\models \neg I$ (which requires some effort i think), and then says that if $ZFC\vdash I$ then any model for $ZFC$ is also a model for $I$, which contradicts $V_\kappa\models ZFC,\neg I$. Is this necessary? or can i stop after $V_\kappa \models ZFC$?

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    $\begingroup$ You forgot to indicate that $\kappa$ is the least strongly inaccessible in your argument. The proof that $V_\kappa\models\lnot I$ is much easier, both in outline and in details, than the proof of the second incompleteness theorem. $\endgroup$ – Andrés E. Caicedo Aug 24 '15 at 21:38
  • $\begingroup$ Thanks. I actually missed minimality. $\endgroup$ – Ariel Aug 25 '15 at 0:03
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Your argument is correct but the point is that you do not have to appeal to Godel's 2nd incompleteness theorem to prove this and Jech does that.

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  • $\begingroup$ hmmm, is there a special reason to wanting to avoid using Godel's 2nd incompleteness? $\endgroup$ – Ariel Aug 24 '15 at 17:49
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Actually I think there is a subtle obstacle for which you can't appeal to Godel's 2nd incompleteness theorem and so the proof of Jech is necessary:

When we talk about $V_k$ as a model of $ZFC$ and when we state $V_k\vDash ZFC$ we are referring to the element of a model of set theory, since $V_k$ to be a model must be a set, so must be an element of some model of set theory.

In practice the statement $ZFC \vdash I$ is not equivalent to say "$ZFC$ proves it has a model" (and so "$ZFC$ proves its own consistency") but it's equivalent instead to the following: "$ZFC$ proves that, if it has a model (and thus IF it is consistent), then it has also an inner model (a second model which is an element of the previous model)". The contradiction thus rely really only on what Jech stated.

To see this more clearly, just think about the other possibility: if $ZFC$ is not consistent we would have $ZFC\vdash I$ since a non consistent theory proves everything. So it's clear that "$ZFC\vdash I$" is not equivalent to "$ZFC$ has a model".

One final remark: remember that the 2nd incompleteness theorem is to be intended more as "there is no first order logic formula that truly express the concept of being consistent" and not "a theory can not prove its own consistency". In fact, if you read the proof of Godel's theorem the formula $Cons(T)$ is equivalent to the statement "T is consistent" only in the standard model of natural numbers. If you pick a non standard model the formula has no meaning at all (since it's not really defined for any non standard natural number which is an element of the non standard model) and that's the reason why it can be falsified.

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  • $\begingroup$ Of course you can appeal to the second incompleteness theorem. Since ZFC proves "If there exists an inaccessible cardinal, then ZFC is consistent", it follows that if ZFC is consistent and proves the existence of an inaccessible, it will prove its own consistency. This, as noted, is a contradiction to the incompleteness theorem. $\endgroup$ – Asaf Karagila Mar 14 '18 at 17:35
  • $\begingroup$ Moreover, the formula Con(T) is a first-order formula. It is just not a $\Sigma_1$ formula, but rather a $\Pi_1$ formula, and therefore cannot be proved just from the assumption that it is true in $\Bbb N$. And while it is true that in non-standard models the meaning of Con(T) is a bit "off", it is a useful construct nonetheless as it lets us talk about internal consistency proofs and so on. This is particularly useful when you check to see that a model of ZFC+"ZFC is inconsistent" must have non-standard integers. $\endgroup$ – Asaf Karagila Mar 14 '18 at 17:36
  • $\begingroup$ Sorry I apologize for my answer, it was written with the knowledge I have but it’s clear I’m missing something. Now I tried to search how you can do what you stated in the first comment, I found this: math.stackexchange.com/questions/464805/…. So I get you can prove that is not the case $ZFC \vdash Con(ZFC)$ for the formula Con(ZFC) builded in the link. But still I don't get, how can we prove $ZFC\vdash I\to Con(ZFC)$? If you any references where I can find the material I can study that. $\endgroup$ – Cla Mar 15 '18 at 15:51
  • $\begingroup$ The reason why I struggle to see the second implication is this: since $ZFC \cup \{\neg Con(ZFC)\}$ is consistent (relatively speaking), it has a model, call it M. This M is also a model for ZFC alone, so suppose can we want to show the same assertion as before for I’=“exist M”. What can we say about $ZFC \vdash I’$? (I mean, can we use here the same argument as above and thus prove $ZFC\vdash “exist M” \to Con(ZFC)$ so that we have a proof that ZFC doe not prove $I’$?) $\endgroup$ – Cla Mar 15 '18 at 15:52
  • $\begingroup$ Easy. ZFC proves the completeness theorem for first-order logic, so if ZFC proves that something has a model, it proves that that theory is consistent (internally!), since ZFC proves that if there is an inaccessible cardinal then there is a model of ZFC, it proves that implication. $\endgroup$ – Asaf Karagila Mar 15 '18 at 15:53

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