This is probably a really basic result that I'm forgetting but if $M \models \text{PA}$ and $M \models \phi$ for some $\Sigma^0\_1$ sentence $\phi$ such that $\mathbb{N} \models \lnot \phi$ does it follow that there is some consistent sentence $\psi$ such that $M \models \lnot \text{Con}(\psi)$? I vaguely feel like it ought to be possible to find a way to code up any $\Sigma^0\_1$ claim as some kind of claim about consistency but I also know that this kind of intuition is very often wrong.


If yes, does the answer change if I insist that $\psi$ not only be consistent but $PA+\psi$ be consistent? So if I'm looking at some consistent extension $T$ of PA that proves some false $\Sigma^0\_1$ claim must it prove that some consistent extension of $T$ is in fact inconsistent?


(I'm kinda hoping the answer is really complex since I feel I should know the answer but it's probably just been too long since grad school :-))



Model of PA with false $\Sigma^0_1$ sentence but no false Con sentence?
This essentially boils down to the provable $\Sigma^0\_1$-completeness of PA. Suppose $\phi$ is a $\Sigma^0\_1$ sentence false in $\mathbb N$ but true in some other model $M$ of PA. Then, by $\Sigma^0\_1$-completeness, $M$ also satisfies "$\phi$ is provable in PA" and therefore also satisfies $\neg\text{Con}\_{PA}(\neg\phi)$. On the other hand, since $\mathbb N$ satisfies $\neg\phi$ and all the axioms of PA, the theory PA$\,+\,\neg\phi$ is consistent and therefore the standard model satisfies $\text{Con}\_{PA}(\neg\phi)$. So $\neg\phi$ can serve as the $\psi$ in your question.


