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According to the Second incompleteness theorem $Con(PA)$ is independent of $PA$. So if $PA$ is consistent $PA + \neg Con(PA)$ is also consistent which means that there exist a number $t$ which codes a proof of $1=0$. But $t$ isn't a "regular" natural number because $PA + \neg Con(PA)$ is a nonstandard model of $PA$.

My question is, how do we know that the Gödel's sentence of $PA + \neg Con(PA)$ exists (its code is a "regular" natural number)?

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  • We don't know if $Con(PA)$ is independent of $PA$. Second incompleteness theorem only guarantees that $Con(PA)$ is not provable from $PA$ ($\neg Con(PA)$ might still be provable), which isn't quite what "independent" means. – Wojowu Oct 30 '15 at 16:38
  • Because one has been explicitly constructed, or at least a precise recipe for constructing it has been given. {Except that the word "it" hides some complications.) – André Nicolas Oct 30 '15 at 16:39
  • (Sorry, the downvote was mine - I misclicked.) – Noah Schweber Oct 30 '15 at 16:40
  • @Wojowu How could PA prove $\neg$ Con(PA)? In the standard model, Con(PA) is true so PA cannot prove its negation. Sorry if I misunderstood your remark. – hot_queen Oct 30 '15 at 20:07
  • @hot_queen It might be the case that only nonstandard models of arithmetic exist. – Wojowu Oct 30 '15 at 20:09
  • @Wojowu Can you elaborate? Isn't it clear that $(\omega, +, .)$ models PA? – hot_queen Oct 30 '15 at 20:12
  • @hot_queen I have just now realized that if PA proves $\neg Con(PA)$, then ZFC would necessarily be inconsistent, so if you believe ZFC is consistent, then indeed what I say is incorrect :P – Wojowu Oct 30 '15 at 20:12
  • @Wojowu Doesn't it say more? Suppose PA proves $\neg$ Con(PA). Then we can search for a witness in the standard model and decode it to a standard proof of $0 = 1$ implying that PA is inconsistent. – hot_queen Oct 30 '15 at 20:15
  • @hot_queen Your argument is sound on the grounds of ZFC, where you can prove existence of the standard model. So in ZFC, you can actually prove that $PA$ does not prove $\neg Con(PA)$. However, the same cannot be shown on the grounds on $PA$ itself. i.e. there is a (necessarily nonstandard) model of $PA$ in which "$PA$ proves $\neg Con(PA)$". – Wojowu Oct 30 '15 at 20:21
  • @Wojowu I wasn't working in a a formal system but I guess I now understand what you mean: Working in PA + Con(PA) we cannot prove Con(PA + Con(PA)). – hot_queen Oct 30 '15 at 20:55
  • @hot_queen Yes, that's essentially it. – Wojowu Oct 30 '15 at 21:06

2 Answers2

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I think there's a bit of confusion here.

First, "$PA+\neg Con(PA)$ is a nonstandard model of $PA$" is false. "$PA+\neg Con(PA)$" is a set of sentences. It can be true, or false, in nonstandard models, but it is not a model itself.

As to how we know the sentence "$\neg Con(PA)$" (which is the only weird sentence in $PA+\neg Con(PA)$) exists: we can actually explicitly write it down! Godel's original paper gives a recipe for how to do this. Actually carrying the recipe out all the way is lengthy, but has been done - see e.g. What does a Godel sentence actually look like?. And there are much more efficient ways to do it, too.

EDIT: Just to be clear, that example Godel sentence is NOT MINE - it was constructed by Hagen von Eitzen.

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Noah Schweber
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The construction of the Gödel for any theory that includes PA depends only on what the axioms of that theory are, textually. It does not depend on having a particular model of the theory, or what the properties of that model is, but proceeds entirely in the metatheory, where we assumes the integers are just the standard ones.

This holds in particular for $\mathrm{PA}+\neg\mathit{Con}(\mathrm{PA})$.

hmakholm left over Monica
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