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In "Mathematical Logic, by Joseph R. Shoenfield" we find an exercise to prove it from ZFC. Would it be possibly to prove the Tukey lemma also from ZC only, i.e. without making use of the axiom schema of replacement.

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You are not really using Replacement. You are using transfinite recursion over a long enough well-order.

But Replacement is not necessary for Hartogs' theorem, which then only gives you a well-ordered set, just not a von Neumann ordinal. And Replacement is not necessary for phrasing transfinite recursion over an existing well-order.

So yes, Zermelo + Choice is enough to prove the Teichmüller–Tukey lemma.

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  • $\begingroup$ Well what confuses me, metamath has usually shortest proofs, and proofs with least assumptions. Their proof is already >3 years old, from 2015. But I didn't find an alternative proof without ax-rep. So its not based on some mathematical argument, more an assumption about the state of the art of published proof. ProofWiki seems to also not use ax-rep. But there the proof is more informal. Didn't research more yet, started inquiry only yesterday. $\endgroup$ – j4n bur53 May 25 at 16:48

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