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    $\begingroup$ I don't understand the question in the box -- in most of mathematics, $P(\mathbb N)$ is treated as a set and not a class. So it sounds like you're simply asking whether ordinary mathematics can prove false arithmetic statements. But I get the impression from the surrounding discussion that you're trying to ask something specific about Grothendieck universes... I don't know what you're asking though. $\endgroup$ Commented Jun 19, 2019 at 19:05
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    $\begingroup$ Ok. So there are two things going on in your question: 1.) You're asking whether "conceptualist mathematics" might prove different arithmetical statements then ZFC, say -- a mathematical question. 2.) You're asking whether there are reasons to think that the arithmetical consequences of one theory are more "true" than another--a philosophical question. To answer either question would require a much more precise specification of which theories, exactly we're discussing. FWIW I've never heard of "conceptualist mathematics" but generally restricting the powerset axiom is called "predicativism". $\endgroup$ Commented Jun 19, 2019 at 19:34
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    $\begingroup$ "my understanding is that this dependence can be completely removed due to the fact that Fermat's Last Theorem has small quantifier complexity." No, the use of Grothendieck universes can be removed because they are unnecessary for treating the necessary topos theory for relatively small sites, not due to the complexity of the statement of FLT. $\endgroup$ Commented Jun 20, 2019 at 1:23
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    $\begingroup$ @HagenvonEitzen Consistency and truth are two different things. For instance, $PA + Con(PA)$ and $PA + \neg Con(PA)$ are both consistent, but at most one of them can be true -- after all, one of them asserts $Con(PA)$ while the other asserts $\neg Con(PA)$! Depending on your philosophy, you might deny that it is ultimately meaningingful to ask whether / assert that a mathematical theory is "true". But in this case, you can still "model" the situation as follows: when somebody makes such assertions, interpret them as being relative to some (unspecified) "ultimate metatheory". $\endgroup$ Commented Jun 20, 2019 at 13:31
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    $\begingroup$ Re: Fermat's last theorem, you're conflating the issue of universes (which can only be removed by actually looking at the proof - no general metatheorem applies) with choice (where we have a metatheorem, namely absoluteness, which automatically removes choice based purely on the syntactic complexity of the theorem itself). $\endgroup$ Commented Jun 21, 2019 at 2:50