[cruegge]

That's a good intuition for first order PA, where the completeness theorem holds, but not the full story either. PA in second order logic only has a single model, but is still incomplete: there are statements that are true in the only model, but not provable.

[bedman12345]

Yeah but second order logic itself is incomplete. ZFC doesn't have a single model, too, btw. In the end unprovable sentences exist because there are multiple models satisfying your axioms.

[cruegge]

What I meant to say is that multiple models are not the only reason for something to be true but unprovable, the incompleteness theorem also holds in more general conditions.

Concerning multiple models of ZFC: I'm always confused by such statements about the foundations of set theory itself, they seem weirdly self-referential. ZFC certainly can't prove that it has multiple (or even any) models. Does such a statement need additional axioms, or is there a general theorem like "If a first order theory has any model, then it has multiple ones."?

[bedman12345]

You simply use "intuitive mathematics", in other words: no formalization. That's at least what I got when I read books on set theory.

Löwenheim-Skolem implies the existence of a countable model of ZFC. https://en.m.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skole...

"What I meant to say is that multiple models are not the only reason for something to be true but unprovable, the incompleteness theorem also holds in more general conditions." I can't give a clean rebuttal for this, but I believe this to be profoundly mistaken. It might be technically correct though, depending on how you'd formalize this statement. To formalize math you need a logic that has some properties: it should be decidable whether a proof is correct, you should be able to write it down, it should not be contradictory. If you take these together the only way a statement is unprovable, is if it is independent from the axioms, i.e. there exist multiple models.

[cruegge]

I see. Checking the formulation of the incompleteness theorem again, I noticed that I probably misunderstood something here: it indeed essentially requires proofs to be verifyable, which second order theories do not provide. So second order PA can (and in fact does) have a proof of every statement or its negation, without contradicting incompleteness, but provability is somewhat useless in this case. For theories that fit the conditions you listed, unprovability is indeed due to multiple models. Does that make sense?

Edit: however, consistent second order theories don't always have a model. In first order, if S is an undecideable statement in theory T, then both T+S and T+!S have models, both of which are models for T, so undecideable statements always come from multiple models. But that does not need to be the case in second order, so your claim "it is independent from the axioms, i.e. there exist multiple models" is not neccessarily true, i.e. there may be cases when decideability of some statement fails in a theory with a unique model. Or is there another argument for your claim?

Forgive me for spamming questions, I just try to understand how these things fit together. But maybe we should just stick to first order, since everything else is too weird anyway.

[cruegge]

Ok, the more I try to understand this, the more confused I get. Probably just scrap everything I said, I'll need to do some more reading.

[bo1024]

This was all a helpful/interesting discussion!