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I know we can't formalize the Kunen inconsistency as an assertion in the first-order language of set theory. But the wikipedia say "Con(MK + The proper class ordinal is a measurable cardinal.)$\leftrightarrow$ Con(NFU + Infinity + Large Ordinals + Small Ordinals)", So I think it's no problem for NFU.

NFU is a non-well founded set theory. NFU had Universal set V, Universal set cardinal and automorphism j, NFU's axiom schema of comprehension is very different from MK/TG, They are very important in Kunen's inconsistency theorem. So a interesting question is:

Can we use Kunen's inconsistency theorem in NFU + AC, to proof NFU + AC + Reinhardt/super Reinhardt/totally Reinhardt/limit club Berkeley cardinal $\to 0=1$?

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    $\begingroup$ This question would be better with 1) a link to the Wikipedia article, 2) a more standard choice of which paragraphs to highlight, 3) a clearer statement of why this inconsistency can’t be formalized in the usual way, and 4) a main question whose first presentation is not distracted with slashes, and which leaves the alternative questions for a separate paragraph. $\endgroup$
    – user44143
    Commented Oct 1, 2019 at 12:00
  • $\begingroup$ @MattF. Thank you, link add finish $\endgroup$
    – Ember Edison
    Commented Oct 1, 2019 at 13:20

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In general, NFU + AC is going to have exactly the same possible large cardinals as ordinary set theory. NFU + AC has the same stratified mathematics as ordinary set theory. There will be no interesting results of this kind, unless they involve unstratified assertions in the language of NFU.

I should be slightly more explicit. NFU + AC has the same strength as bounded Zermelo set theory. However, there will be models of NFU + AC associated with any of the usual levels of strength we are used to.

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