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I'm trying to understand the full extent of Gödel's Incompleteness Theorem with regards to what types of consistent and complete theories it rules out. Does it only prove that number theory/ZFC cannot be a subtheory of a complete consistent theory which is recursively axiomatizable or rather any complete consistent theory regardless of how it might be axiomatized? If it's only the former, are there any known consistent complete theories for which number theory is a subtheory? Thanks.

Bernard
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Ari
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    It is a subtheory of the theory of all true (in $\mathbb N$) arithmetical statements, which is complete and consistent (and, of course, not recursively axiomatizable). – spaceisdarkgreen Oct 25 '21 at 15:13
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    Moreover, any consistent theory has a complete, consistent extension (in general, many such extensions, if the theory isn’t already complete). This is a consequence of the completeness theorem, which says any consistent theory has a model. Then just pick a model consider the theory of all statements true in that model. – spaceisdarkgreen Oct 25 '21 at 15:24
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    @Peter You're wrong, in exactly the way that the OP is asking about: Godel only applies to sufficiently "simple" (= recursively axiomatizable) theories. (On a more minor note I'd also quibble with the idea that Presburger is the "most important" of the complete consistent r.e. theories; algebraically closed fields of characteristic zero, anyone?) – Noah Schweber Oct 25 '21 at 15:37
  • To the OP, what constitutes a "known" theory which is not recursively axiomatizable? For example, for any structure $\mathcal{A}$ the theory $$Th(\mathcal{A}):={\varphi:\mathcal{A}\models\varphi}$$ is trivially complete and consistent. So since $\mathcal{N}:=(\mathbb{N};+,\times)\models\mathsf{PA}$, we have $Th(\mathcal{N})$ is a complete consistent theory extending $\mathsf{PA}$. Does $Th(\mathcal{N})$ count as "known?" – Noah Schweber Oct 25 '21 at 15:39
  • Incidentally, while it's very plausible to me that this is a duplicate question, I couldn't find an exact dupe; that's why I've answered it. – Noah Schweber Oct 25 '21 at 15:51
  • I personally find it helpful to take a constructive, model-theoretic view of things. This way, the theorem becomes a statement about the inability of certain first-order theories to control their models - in fact, it is just one of many results along these lines (along with Lowenheim-Skolem and others). – user3716267 Oct 25 '21 at 15:56
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    (Reconisdering my above comment... I'm not sure it's best to view the existence of completions as a consequence of the completeness theorem, since showing this is often a step in proving the completeness theorem. Perhaps it's better to say that this is another way of looking at the completeness theorem, or an alternative statement of it.) – spaceisdarkgreen Oct 25 '21 at 16:15

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First, recall that for any structure $\mathcal{A}$ - including $\mathcal{N}:=(\mathbb{N};+,\times)$, the standard model of arithmetic - the theory of $\mathcal{A}$ $$Th(\mathcal{A}):=\{\varphi:\mathcal{A}\models\varphi\}$$ is trivially complete and consistent. Since $\mathcal{N}\models\mathsf{PA}$ this gives as a special case that $Th(\mathcal{N})$ - more commonly denoted "$\mathsf{TA}$" for "true arithmetic" - is a very natural complete consistent theory extending $\mathsf{PA}$.

So Godel's incompleteness theorem cannot possibly apply to arbitrary complexity theories, at least not without additional complicating hypotheses which would rule out things like $\mathsf{TA}$.

OK, now let's actually say a bit about Godel.

The proof of the first incompleteness theorem has a key technical limitation: the theory $T$ in question must exhibit a combination of weakness and strength. Roughly speaking, $T$ must be able to prove all "very basic" facts about its own provability predicate. This both requires $T$ to be reasonably strong, and requires $T$ to be not too complicated. For example, true arithmetic $\mathsf{TA}$ is not definable in $\mathcal{N}$, so it can't even talk about itself at all (let alone prove things about itself).

As such, GIT is in fact fairly limited. Here's one way to pose GIT which is pretty close to optimal:

No consistent complete r.e. theory can interpret Robinson arithmetic.

The notion of interpretation here is a particularly technical one, so this isn't necessarily a good formulation of the incompleteness theorem to start off focusing on. But I think it's a good thing to see, if only briefly, early on: the various ingredients (consistency, completeness, r.e.-ness, and "logical strength" in some capacity) are clearly displayed and there is no imprecise language being used.

That said, we can indeed push Godel beyond the r.e. theories, at least to a certain extent. Basically, we just need to strengthen the consistency hypothesis to rule out errors which are "low-level" compared to the theory itself. One result along these lines which I think is folklore is discussed at this old answer of mine (albeit focusing on theories in the language of arithmetic for simplicity - note that I just corrected that answer, since looking back on it the original version had an indexing error!).

Noah Schweber
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    Out of curiosity, why the downvote? – Noah Schweber Oct 25 '21 at 15:56
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    likely, someone feels you shouldn't have answered the question. people are petty. – user3716267 Oct 25 '21 at 15:58
  • @Noah Thanks for taking the time to answer. Re: your comment to my post itself, I guess I'm lacking a fundamental understanding of how a formal theory is dependent on any particular structure. In one place my textbook has "$\sigma\in T ;\text{iff};T\models \sigma$" which I take as meaning $T$ needs to imply $\sigma$ in all models, but then in other places it mentions specifically a structure in question. I tried looking at Wikipedia now and I'm also not quite getting it. Do you know of other sources? – Ari Oct 25 '21 at 16:47
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    @Ari First of all it's certainly not true that $\sigma\in T$ iff $T\models\sigma$ for an arbitrary theory $T$; it's only the case if $T$ is deductively closed. But beyond that I'm not sure what you mean by "a formal theory is dependent on any particular structure." The point is just that every structure $\mathcal{A}$ yields a corresponding theory $Th(\mathcal{A})$ which is guaranteed to have several nice properties (e.g. complete, consistent, deductively closed, ...). Note that the map $\mathcal{A}\mapsto Th(\mathcal{A})$ is not injective even up to isomorphism. – Noah Schweber Oct 25 '21 at 16:50
  • On the other hand, it is surjective in the context of complete consistent deductively closed theories: if $T$ is a complete consistent deductively closed theory then $T=Th(\mathcal{A})$ for some $\mathcal{A}$ (usually not unique even up to isomorphism though). Does this address your question? – Noah Schweber Oct 25 '21 at 16:51
  • @Noah Yes basically. My book also has that theories are closed under logical implication (as opposed to deductively closed). Is that right? But I thought logical implication "$\Gamma\models \sigma$" means that for every structure $\mathcal{A}$ if $\Gamma$ is satisfied then $\sigma$ is too. So I still don't get the connection between any specific structure and a theory. Also how is $Th(\mathcal{A})$ guaranteed to be deductively closed? Isn't a derivable contradiction possible? – Ari Oct 25 '21 at 17:27
  • @Ari "Closed under logical implication" is the same as "deductively closed." The logical implication "$\Gamma\models\sigma$" does indeed mean: "For every structure $\mathcal{A}$, if $\Gamma$ is satisfied in $\mathcal{A}$ then $\sigma$ is too." I don't see how that is in tension with anything I've said. Some theories come from specific structures, others don't. "How is $Th(\mathcal{A})$ guaranteed to be deductively closed?" See here. "Isn't a derivable contradiction possible?" No, why would it be? – Noah Schweber Oct 25 '21 at 17:34
  • @NoahSchweber I guess that's just another layer of my misunderstanding. Why is there still even any discussion on questioning a theory like ZF (or $Th(\mathcal{N})$) being consistent and instead having to phrase things in terms of $Cons(ZF)\implies Cons(ZFC)$? I know we have Godel's Second Incompleteness, but if we can find a model for them, why is that not good enough? – Ari Oct 25 '21 at 17:54
  • @Ari Can we find a model for something like $\mathsf{ZF}$? In what sense do we "have" the set-theoretic universe, and how do we "know" that it satisfies $\mathsf{ZF}$? (Meanwhile it has no computable model.) One use of the relative consistency phrasing is to make results more broadly acceptable - for example, I can accept "$\mathsf{PA}\vdash\mathsf{Con(ZF)\rightarrow Con(ZFC)}$" even if I'm dubious about the consistency of $\mathsf{ZF}$, and I can see that result as nontrivial even if I'm confident about the consistency of $\mathsf{ZFC}$. – Noah Schweber Oct 25 '21 at 17:57
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    @Ari To put a finer point on what Noah is saying, the 2nd incompleteness theorem implies that we cannot construct a model of ZF in ZF, (and Con(ZF)->Con(ZFC) means we can't construct one in ZFC either). It would be good enough since the existence of a model implies the theory is consistent, but GIT implies we can't have anything good enough (in ZF, anyway). (Meanwhile, we can work in a stronger theory than ZF where there are models for ZF, but why should anyone skeptical of the consistency of ZF trust something a stronger theory says?) – spaceisdarkgreen Oct 25 '21 at 19:48