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We know that PA proves consistency of $I\Sigma_{n}$ for any $n$. But does PA prove the sentence: $\forall n (con(I\Sigma_{n}))$?

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    $\begingroup$ I'm quite certain that sentence in question implies consistency of PA. $\endgroup$
    – Wojowu
    Jan 11, 2015 at 19:57
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    $\begingroup$ I might be missing something, but isn't it clear that that's in fact equivalent to the consistency of PA? $\endgroup$ Jan 11, 2015 at 20:18

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No. Given any $\varphi\in \mathcal{L}_{PA}$, if $PA\vdash \varphi$ then there exists $n$ such that $I\Sigma_n\vdash \varphi$ (by finitarity of proofs). So $\forall n con(I\Sigma_n)$ implies for any $n$ $I\Sigma_n \not \vdash 0=1$ therefore $PA\not \vdash 0=1$, i.e. $con(PA).$

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  • $\begingroup$ Dear Zhang, it is not obvious that your argument can be formulated inside PA or not; we know (in metalanguage) that if PA⊢φ then there exists n such that IΣn⊢φ, but dose PA proves arithmetized version of this fact ? $\endgroup$ Jan 13, 2015 at 12:56
  • $\begingroup$ Suppose you could formalize $\forall n I\Sigma_n^0$ in PA in the first place, then for any model of PA, $\{n: I\Sigma_n^0\vdash \varphi\}$ is an arithmetic set in the model. Take the least element and it would be a real natural number. $\endgroup$
    – Jing Zhang
    Jan 13, 2015 at 17:31
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    $\begingroup$ @PayamSeraji Yes, it is obvious that PA proves "for all $\phi$, if PA proves $\phi$ then so does $I\Sigma_n$ for some $n$." Proof formalizable in PA: Assume PA proves $\phi$ and let $n$ be the Gödel number of a proof $p$ of $\phi$ in PA. Then $n$ (vastly) exceeds the number of quantifiers in $p$. So $p$ is also a proof of $\phi$ in $I\Sigma_n$. $\endgroup$ Jul 21, 2019 at 12:08

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