Turing's proof that a Halting oracle is impossible and Gödel's proof that and omega-consistent first order theory of arithmetic must be incomplete are similar in that they use self-referential arguments. Is there an interesting relationship between them especially in light of the Curry-Howard Correspondence and especially the categorical version thereof i.e. the Curry-Howard-Lambek Correspondence.
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4Gödel's theorem is precisely that the set of theorems of arithmetic is a recursively enumerable set but not a recursive set. The halting theorem is precisely that the set of (turning machine, input) pairs that halt is a recursively enumerable set but not a recursive set. – MJD Mar 08 '15 at 18:40
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@MJD, could the Curry Howard Correspondence be useful in any way though? – 11Kilobytes Mar 08 '15 at 19:32
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1I don't know any way that the CH correspondence ties in, but I'm not at all an expert, and I think you should wait a while and see if you get any better answers than my little comment. I feel like I left out some important connections between Gödel's theorem and the halting problem, and mentioned only one similarity. – MJD Mar 08 '15 at 19:48
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1Related: Prove Gödel's incompleteness theorem using halting problem – MJD Mar 08 '15 at 19:50
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You can see Peter Smith, An Introduction to Gödel's Theorems (2nd ed - 2013) : Ch.43 Halting and incompleteness, page 328-on. – Mauro ALLEGRANZA Mar 09 '15 at 07:57
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Turing's proof that a Halting oracle is impossible and Gödel's proof that and omega-consistent first order theory of arithmetic must be incomplete are similar in that they use self-referential arguments. Is there an interesting relationship between them.
Well, Gödel's theorem is a simple consequence of Turing's proof.
Take a look at my Introduction to Gödel's Theorems, for example. §43.2 (in the numbering of the second edition) shows that the recursive unsolvability of the halting problem implies that the set of truths of the first-order language of arithmetic is not recursively enumerable. But the theorems in that language of a formalized theory $T$ are recursively enumerable. So there are truths that $T$ can't prove, and if $T$ is sound, can't disprove either. So it is incomplete.
§43.3 then strengthens the result by dropping the assumption that $T$ is sound in favour of the assumption of omega-consistency, together with the usual assumption that $T$ is (primitive) recursively axiomatized and includes a small amount of arithmetic (e.g. contains Robinson arithmetic Q -- the crucial thing is being strong enough to represent the (primitive) recursive functions). Then we can prove that $T$ is incomplete, going via the unsolvability of the Halting Problem (it's a half-page proof in detail, so forgive me for not reproducing it here)!
Peter Smith
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