Correct me if I am wrong at any point!

Godel's incompleteness theorem allows us to express "PA is consistent" in the language of Peano arithmetic, and shows that this is not provable in PA. Let's call this sentence $P$.

Godel's completeness theorem tells us that $P$ does not hold in every model of PA; it also tells us that is holds in some models of PA, since otherwise PA would prove $\neg P$ and so PA would be inconsistent.

However, we have a "standard model" of PA, i.e. $\mathbb{N}$, which we can prove things about using ZFC or second-order arithmetic. So my question is, does $P$, thought of as a statement about natural numbers, hold in $\mathbb{N}$? Or have I misunderstood a subtlety (or not-so-subtlety) somewhere?

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    $\begingroup$ Yes, the standard model satisfies $ P $. $\endgroup$ – Andrés E. Caicedo Feb 4 '15 at 16:10
  • $\begingroup$ @AndresCaicedo: Hmm is that the actual question? $P$ is not provable in PA, and so a proof of $P$ for $\mathbb{N}$ must invoke something beyond the PA axioms applied to $\mathbb{N}$. $\endgroup$ – user21820 Feb 4 '15 at 16:24
  • $\begingroup$ @user21820 it was the question. I'd be interested to see a proof, though. $\endgroup$ – Christopher Feb 4 '15 at 16:31
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    $\begingroup$ You can see the proof in George Tourlakis, Lectures in Logic and Set Theory. Volume 1 : Mathematical Logic (2003), page 205-315. $\endgroup$ – Mauro ALLEGRANZA Feb 4 '15 at 16:51
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    $\begingroup$ A non-technical (i.e crappy) "proof" that $\mathbb{N}\models P$ is to observe that the provability predicate used in the definition of $P$, accurately models provability, when you are working in $\mathbb{N}$. So the statement $P$, which says, roughly "I am not provable" is true (in $\mathbb{N}$), beacuse the first incompleteness theorem holds, and, hence, $\neg(PA \vdash P)$ is true, yet, this is precisely the content of $P$. (All this, assuming $PA$ is consistent). $\endgroup$ – James Feb 4 '15 at 18:11

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