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I understand that Gödel's first incompleteness theorem applies to ZFC as a first-order theory. One can construct a sentence $G_F$ such that $ZFC \nvdash G_F$ and $ZFC \nvdash \neg G_F$.

However, Gödel's first incompleteness theorem itself is a theorem in ZFC. Additionally, in the meta-proof, we also show that $G_F$ is true under the standard model.

So, my question is: why can't the sentence (or its negation) be proven within the system (which is ZFC), but outside the system (which is still ZFC), it can be shown that the sentence is true?

I can think of two possible reasons: The first reason is that the meta-language and the object language are different. In other words, the meta-proof cannot be fully formalized (which seems incorrect to me). The second is that the meta-language has more 'information' than the object language, such as knowing how $G_F$ was constructed.

I would greatly appreciate any clarification.

Victor
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  • You can replace ZFC with PA everywhere in this question. – Asaf Karagila Oct 13 '24 at 19:43
  • @AsafKaragila I think that in the case of PA, the meta-language is still ZFC (or at least set theory), since concepts like recursive sets are used. Therefore, I don't see a contradiction in this case. – Victor Oct 13 '24 at 19:48
  • PA can be its own meta-theory. – Asaf Karagila Oct 13 '24 at 19:48
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    No, the meta-proof shows that if ZFC is consistent then G is true. – spaceisdarkgreen Oct 13 '24 at 19:49
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    I mean, the whole point of Gödel's work on incompleteness was to show that you can internalise first-order logic into some very basic arithmetic. He was very meticulous about this so that it doesn't get dismissed as "some theorem about logic, not about 'real' maths" or some such. So, as a meta-theory, we can go significantly weaker than PA, let alone ZFC. – Asaf Karagila Oct 13 '24 at 19:57
  • @AsafKaragila I see. Thanks for your patience! So, the meta-theory can be quite weak (I recall from some posts that it only needs to allow for the manipulation of finite strings). This makes me even more confused about why the internal and external perspectives can be so different. – Victor Oct 13 '24 at 20:04
  • @spaceisdarkgreen Wow, that helps! But in many proofs (formal or informal) I read from the math logic textbook, they use "proof by contradiction". So they always implicitly assume the consistency of the theory? – Victor Oct 13 '24 at 20:10
  • @Victor I'm not sure what exact usages of proof by contradiction you mean (in he theory, in the metatheory?) But either way, the answer is no... you are arguably implicitly assuming the consistency of your metatheory whenever you do anything, but that has nothing to do with using proof by contradiction, per se. – spaceisdarkgreen Oct 13 '24 at 21:14
  • @Victor "you are arguably implicitly assuming the consistency of your metatheory whenever you do anything" But let's be careful here... I only mean that in the sense that you want to "trust" your metatheory, so naturally you should believe it's consistent. But you can't have the assumption of the metatheory's own consistency formally baked into the metatheory itself or we run into issues with the incompleteness theorem, as you describe in the post (which leads to the 2nd incompleteness theorem). – spaceisdarkgreen Oct 13 '24 at 21:21
  • @spaceisdarkgreen Truly appreciate for your detail explanation! Would you mind pointing out in the proof I sketched which part the consistency is assumed? Is that "the existence of a model part"? – Victor Oct 13 '24 at 21:39
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    @Victor No, the argument for why $G$ is true is that it is equivalent to the statement that $G$ isn't provable in the system, and, given the incompleteness theorem, $G$ is indeed not provable. But what we're forgetting is that the incompleteness theorem actually only says if the system is consistent, then it can't prove $G$. And how could it not have that assumption? Of course if it were inconsistent, it would prove $G,$ since it would prove every statement, so in that case $G$ would be false. – spaceisdarkgreen Oct 13 '24 at 22:00

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