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Gödel's second incompleteness theorem can be proven in PA + COH, where PA is Peano Arithmetic and COH the consistency statment. By this I mean : $PA+COH\vdash \neg \square COH$ where $\square P$ means 'P is provable in PA', so that $COH=\neg \square \bot$.

But PA+COH does not prove $\neg \square\neg COH$. However, we know meta-theoratically (in ZFC ?) that COH is undecidable in PA, so PA does not prove $\neg COH$.

So which extra-feature from ZFC are we using that PA+COH does not have in the proof of undecidability of COH ?

user1162101
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The additional assumption is a soundness principle for PA. Perhaps surprisingly, PA does not prove "If I prove $\varphi$, then $\varphi$ is true" for every sentence $\varphi$! In fact, if it does, then it proves $\varphi$ outright - this is Lob's theorem.

A soundness principle is a statement of the form "If PA proves a sentence $\varphi$ of the form [blah], then $\varphi$ is true." Such statements can have lots of strength over PA. In this case, we need soundness for $\Sigma^0_1$ sentences - sentences of the complexity of "there is a proof of...". Goedel originally used a stronger assertion, namely $\omega$-consistency (which, despite the name, is a soundness principle).

However, if we replace "PA is consistent" with the related Rosser sentence $R$, then this discrepancy goes away: PA + "PA is consistent" proves that PA neither proves nor disproves $R$.

Noah Schweber
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  • Thanks for your answer. So you suggest to take $PA+\square \phi \rightarrow \phi$ for a given class of sentences $\phi$ as a theory ? btw, do you mean that Gödel did not prove $PA+COH \vdash \neg \square COH$, but rather $PA+\omega\text{-consistency}\vdash \neg \square COH$ ? Which one is stronger than the other ? – user1162101 Mar 27 '17 at 15:10
  • And I have an additional question : what is a minimal extension of PA that makes it as powerful as ZFC when it comes to arithmetic ? – user1162101 Mar 27 '17 at 15:13
  • @user1162101 (I'm still unsure why you're writing consistency the way you are - where does the "H" come from? I'll write "$Con(PA)$" below.) Goedel proved $PA+\omega Con(PA)\vdash \neg\Box_{PA} Con(PA).$ In fact, he could have used something less than $\omega Con$ - the weaker property $1Con$, though still stronger than mere $Con$, would have been enough. Note that any amount of soundness immediately implies consistency: if a theory is inconsistent, it proves every sentence, even quantifier-free ones like $0=1$. So all soundness principles considered are stronger than consistency. (cont'd) – Noah Schweber Mar 29 '17 at 21:39
  • So the most natural extension of PA to use is $PA+1Con(PA)$: this is $PA$ + "Every $\Sigma_1$-sentence provable in $PA$ is true." This is basically the weakest natural theory that does the job. Meanwhile towards your second question, that's a great question - but no snappy answer is currently known (obviously PA + ${$arithmetic consequences of ZFC$}$ does the job, but that's silly). – Noah Schweber Mar 29 '17 at 21:42
  • Very interesting ! I write COH for coherence because in French, this is a synonymous of consistency. What do you mean exactly by any amount of soundness ? I guess you don't mean, for any arbitrary set $X$ of formulae, $PA+{ \square P \rightarrow P \mid P \in X } \vdash Con$ ? – user1162101 Mar 31 '17 at 07:36
  • For the second question, why not encode the provability of ZFC in PA (which looks feasible, doesn't it ?), and then add the axioms $\square' P \rightarrow P$ for any formula P of ZFC which is only about arithmetics, where $\square' P$ means that P is provable in ZFC ? Wouldn't it be as powerful as ZFC as far as arithmetics is concerned ? – user1162101 Mar 31 '17 at 07:40