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Gödel's incompleteness theorem applies to formal languages with countable alphabets. So it does not rule out the possibility that one might be able to prove 'everything' in a formal system with an uncountable alphabet OR expand the alphabet to account for new variables.

Two ideas follow from the above:

  • Hypothetically, a theorem can be "complete" if uncountable.
  • A theorem can "reach" completeness in a limited fashion before being considered "uncountable", hence a sort of "loophole".

Therefore, would it be possible to demonstrate a theorem that is as close as possible to being "complete" without it becoming uncountable?

Concerning the "loophole" the demarcation between countability of an alphabet and uncountability would render the theorem complete, if the demarcation can be ascertained a priori or a posteriori?

Shawn W
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    What does it mean for a theorem to be complete or uncountable? A theorem is a single sentence. Also, Godel applies to only certain formal systems - it's not really about countability at all. In particular, every (consistent) theory in a countable language has at least one consistent complete extension in that same language. This doesn't contradict Godel since Godel only applies to certain theories. – Noah Schweber May 17 '21 at 20:25
  • @NoahSchweber I mean that, provability in formal axiomatic theories. Being computable in other words? – Shawn W May 17 '21 at 20:28
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    The real key to Gödel is that the axioms are recursively enumerable, not countable. We can show there exist maximal consistent subsets of the countable set of all statements, and take those as axioms. Then we can show that maximality implies completeness. It’s just not useful for human or computer-read proofs, because there is no way to algorithmically prove each step is allowed. – Thomas Andrews May 17 '21 at 20:28
  • @ThomasAndrews It's also important that the axiom system be "rich enough" (e.g. interpret Robinson arithmetic) - there are lots of recursively enumerable complete axiom systems. – Noah Schweber May 17 '21 at 20:30
  • I know next to nothing (i.e. not "nothing", but "almost nothing") about this topic, but the references and searches in this answer might be of interest to you. – Dave L. Renfro May 17 '21 at 20:31
  • Yeah, I know that. I was objecting to the need for uncountability for completeness when uncomputability is all we need. @NoahSchweber – Thomas Andrews May 17 '21 at 20:31
  • @ThomasAndrews "We can show there exist maximal consistent subsets of the countable set of all statements, and take those as axioms. Then we can show that maximality implies completeness." Can you reference anything on this subject? Thanks. – Shawn W May 17 '21 at 20:32
  • @ShawnW See Lindenbaum's Lemma. – Noah Schweber May 17 '21 at 20:35
  • I’m just applying Zorn’s Lemma on countable sets to the collection of consistent subsets of all statements. If $$S_1\subseteq S_2\subseteq \cdots$$ is an increasing sequence of collections of consistent statements, then $\bigcup S_i$ is a consistent set of statements, because any proof of inconsistency would use only finitely man of the $S_i,$ because proofs are finite. @ShawnW – Thomas Andrews May 17 '21 at 20:40
  • @ShawnW "I mean that, provability in formal axiomatic theories. Being computable in other words?" Every theorem is provable in some system. Meanwhile, what does it mean for a theorem to be computable? – Noah Schweber May 17 '21 at 20:43
  • @NoahSchweber That they are recursively enumerable. – Shawn W May 17 '21 at 20:45
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    @ShawnW A theorem is a single sentence; it makes no sense to say that a theorem is recursively enumerable. (Separately, "recursively enumerable" is not the same thing as "computable" - "computable" is the same as "recursive".) – Noah Schweber May 17 '21 at 20:46
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    @ShawnW You seem to be mixing up theorems and theories, as well as computability/recursiveness and provability/completeness. Try to state your question clearly, without using terminology you're not yet fully familiar with. – Noah Schweber May 17 '21 at 20:47

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There is really no connection between Godel's incompleteness theorem and countability. Rather, Godel's theorem applies to every theory $T$ with the following three properties:

  • $T$ is consistent.

  • $T$ is "sufficiently rich" (I'm being deliberately vague about what this means since this isn't really the key point here, but if you want a precise condition it's enough for $T$ to interpret Robinson arithmetic, where "interpret" is a technical term I'm not going to go into here).

  • $T$ is recursively axiomatizable - this is basically a "simplicity" condition, saying that $T$ isn't too complicated.

Each of these hypotheses is crucial. In particular, suppose $T$ is any consistent theory whatsoever. Then we can find a complete consistent theory $T'\supseteq T$ in the same language; this is a quick application of Zorn's Lemma (consider the poset of consistent extensions of $T$ in the same language, ordered by inclusion), although Zorn isn't necessary when dealing with countable languages. This result is Lindenbaum's Lemma, and is in fact a key step in the proof of Godel's completeness theorem! Re: an apparent tension between Lindenbaum and incompleteness, see e.g. here (although note that the OP's comments are wrong).

(I'm skipping a step here; Zorn guarantees a maximal consistent extension of $T$, but we separately need to show that such an extension is complete. But this is easy: if $S$ isn't complete then there is some sentence $\varphi$ such that both $S\cup\{\varphi\}$ and $S\cup\{\neg\varphi\}$ are consistent, and each of these must be a proper extension of $S$.)

Noah Schweber
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  • I don’t think of the recursiveness of axioms as a simplicity condition, I think of it as a checkability condition. Checking a proof cannot be a manual process if the axioms are not recursive. There is no way for us to read a proof and say, “yes, this axiom referenced here is really an axiom.” – Thomas Andrews May 17 '21 at 20:48
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    @ThomasAndrews ... How is that not a simplicity condition? Computability literally is a measure of complexity (see "Turing reducibility"); saying that an axiom system is computably enumerable is saying that it's not too complicated. – Noah Schweber May 17 '21 at 20:49
  • And unless we are using some abstract definition of recursive, the usual meaning there does have something to do with countability. Specifically, it is a countable set of axioms, so if there are uncountably many symbols, there are some symbols which are never used in the axioms. Essentially, such a theory can only address countable sets of symbols. – Thomas Andrews May 17 '21 at 20:53
  • @ThomasAndrews Technically yes, but it's a total red herring for actually understanding GIT. In fact we can make this precise: if $T$ interprets $\mathsf{Q}$ then some countable fragment $T'$ of $T$ interprets $\mathsf{Q}$, and there is then a countable subtheory $S$ of $T$ in the language of $T'$ over which $T$ is conservative; this $S$ can then be the focus of "Godel-for-$T$." (Actually strictly speaking $S$ is just a subtheory of the deductive closure of $T$; if we want a genuine subtheory we may need to go to a slightly larger language, but this is a silly point.) – Noah Schweber May 17 '21 at 20:54
  • Technically, it is a simplicity issue, but it is a simplicity issue with a goal - that proofs be checkable. Calling it a simplicity issue gives the impression that simplicity is the goal, but being checkable is the goal. – Thomas Andrews May 17 '21 at 20:56
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    @ThomasAndrews Checkability is itself a kind of simplicity. This is a bizarre splitting of hairs. – Noah Schweber May 17 '21 at 20:58
  • Checkability is, again, a specific goal of simplicity. It is logically splitting hairs, but for communication purposes, it is important. I mean, as you say, if requiring something to be recursive is alway a simplicity issue, isn’t it also redundant or splitting hairs to mention simplicity? – Thomas Andrews May 17 '21 at 21:02
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    @ThomasAndrews I don't agree at all with your distinction. Re: "if requiring something to be recursive is alway a simplicity issue, isn’t it also splitting hairs to mention simplicity?," the point is that simplicity is a non-technical term which I'm using to try to motivate what's going on. I honestly have no idea what your point is, and I'm going to bow out at this point. – Noah Schweber May 17 '21 at 21:04
  • Drop your defenses. I was the first to +1 this. Re-read what I wrote, not the attack you seem to be interpreting me as making. – Thomas Andrews May 17 '21 at 21:25