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May 30, 2017 at 18:19 comment added aws I don't know a reference for it, but after thinking some more, I think it should be fairly straightforward to prove. In Grothendieck toposes the n.n.o. is just the coproduct of $\mathbb{N}$ copies of $1$ with "standard" zero and successor. So if $\forall n\,\phi(n)$ is a true $\Pi^0_1$, it suffices to show for every $n$, $\phi(\bar{n})$ holds (with $\bar{n}$ the $n$th successor of zero). But this holds in general when $\phi(\bar{n})$ is true and primitive recursive.
May 30, 2017 at 15:06 comment added user71137 Thanks, especially for the observation about $\Pi^0_1$ sentences. Do you have a source for that claim? I think I agree after playing around with it, but I'm not super confident in my work.
May 29, 2017 at 10:14 history answered aws CC BY-SA 3.0