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



Does PA prove a sentence asserting that all of I-sigma(n) theories are consistent?
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. $con(PA).$


