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Is there any way in which a reasonably strong foundation of mathematics can get around the hypotheses of the incompleteness theorems?

Kyle Gannon
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pkjag
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    No, of course not. – Andrés E. Caicedo Jul 11 '14 at 21:24
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    No. If it is capable of arithmetic (essentially multiplication and addition), then it is subject to Gödel's theorems (in classical logic). – Kyle Gannon Jul 11 '14 at 21:24
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    One tiny loophole: one of the hypotheses is that your theory has to be consistent. Inconsistent theories (i.e., ones that prove everything) have nothing that they can't prove. – Steven Stadnicki Jul 11 '14 at 21:28
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    @StevenStadnicki: You still can't bypass Gödel's first theorem which states that any "sufficiently strong" recursive theory is not both $consistent$ and $complete$. You are no longer consistent in this case. – Kyle Gannon Jul 11 '14 at 21:30
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    Lead us not into contradiction // But deliver us from Gödel // Now and at the hour of our math. ${}\qquad{}$ – Michael Hardy Jul 11 '14 at 21:39
  • What about the consistency of the Peano axioms then? It seems to me the way this has been settled is by bypassing the problem... – Count Iblis Jul 11 '14 at 21:42
  • @CountIblis: I'm not exactly sure what you are arguing. Could you be more specific? – Kyle Gannon Jul 11 '14 at 21:45
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    @CountIblis: The proofs of consistency of $\sf PA$ are either from theories like $\sf ZF$ which are much stronger than $\sf PA$; or something like Gentzen's proof which uses a theory whose strength is incomparable with that of Peano (namely it doesn't prove everything that $\sf PA$ proves about arithmetic). – Asaf Karagila Jul 11 '14 at 21:50
  • @AsafKaragila What theory is incomparable with $\mathsf{PA}$? – Andrés E. Caicedo Jul 11 '14 at 21:51
  • Yes, we can prove that PA is consistent, but only from the standpoint of a stronger theory. Set theory too is proved consistent by stronger theories, like large cardinals. – Rene Schipperus Jul 11 '14 at 21:52
  • @Andres: I've corrected my comment. Thanks. I meant to say that Gentzen's proof from $\sf PRA$+induction doesn't prove everything that $\sf PA$ proves about arithmetic; but since it proves the consistency of $\sf PA$, it doesn't prove strictly less either. – Asaf Karagila Jul 11 '14 at 21:53
  • Gentzen uses the order type of $\epsilon_0$, a kind on consistency proof with minimal assumptions. – Rene Schipperus Jul 11 '14 at 21:54
  • @ReneSchipperus, so is there a limit where we can say that this theory is too weak or too strong to be proven (its consistency) by another or do we keep strengthening the theories ad infinitum?? – pkjag Jul 11 '14 at 21:56
  • @pkjag I think this may be one of the great philosophical challenges of modern mathematics. – Rene Schipperus Jul 11 '14 at 21:59
  • I apologize if this is naive, but I was under the impression that Godel's Theorem applied only to first-order logical systems capable of arithmetic. For that reason I thought it was a limitation not of the axioms but on first-order logic. Are second-order logics also vulnerable to Gödel's Theorems? – mweiss Jul 11 '14 at 22:21
  • @mweiss: This is a late, but in general the incompleteness theorems can be generalized to apply to any formal system that interprets arithmetic. See this post for details. It does not matter what is the underlying logic. And there is actually nothing special about arithmetic, except that it can be used to reason about programs. For a computability viewpoint, see this post, which shows that any system that can perform basic reasoning about finite strings (can interpret TC) is incomplete. – user21820 Oct 27 '17 at 12:50

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Essentially, the answer is no (in classical first-order logic). If your theory,T, is capable of "understanding" both addition and multiplication of $\mathbb{N}$, then one of the following fails: (1) $T$ is consistent, (2) $T$ is recursive, (3) $T$ is complete.

Giving up (1) is the absolute worst case for a logician/mathematician. The only time someone would think about giving up this property is if they thought that the theory of the natural numbers is inconsistent. Essentially, this position might be held by an ultrafinitist or someone who believe PA is inconsistent (see Edward Nelson).

Giving up (2) is useless for the logician/mathematician. There does exist a complete and consistent theory of arithmetic, namely $Th(\mathbb{N})$ in the language $L=\{+,\times,0,1\}$. However, the theorems of $Th(\mathbb{N})$ cannot be found recursively, i.e., there is no algorithm that tells us whether or not a sentence in the language of arithmetic is a theorem of $Th(\mathbb{N})$. Therefore, $Th(\mathbb{N})$ is boring to study.

Giving up (3) is the most reasonable for research purposes and for "real-world" study. There are still open problems in basic arithmetic (see Goldbach Conjecture) and using PA to try and solve this problems is still possible.

(As an asside: Note furthermore, that if GC is independent of PA, we can show, using ZFC, that $Th(\mathbb{N})\vdash GC$).

Kyle Gannon
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  • Your parenthetical remark at the end is better if you remove the "even". It may be false otherwise, if Goldbach's conjecture ends up being false. – Andrés E. Caicedo Jul 11 '14 at 23:03
  • @AndresCaicedo: Removed. Thanks – Kyle Gannon Jul 11 '14 at 23:08
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    There's no way to "escape" Gödel's Theorem. There might be a way to remove its sting. Gödel showed neither G nor ~G is provable in PA if PA is consistent. This stings because it means PA is not proof-theoretically complete. Suppose we modify PA and/or its underlying logic in such a way that some of its sentences are neither true nor false. Partial recursive functions give us statements that are neither, on most interpretations. Then Gödel's Theorem loses its sting because classical proof-theoretical completeness is no longer desirable. As Kyle says here, we give up completeness. – MikeC Jul 12 '14 at 17:22