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This 2007 paper presents a 5-quantifier $(\in, =)$-expression that is ZF-equivalent to the axiom of choice, but leaves open the 4-quantifier case:

Thus the gap is reduced to the undecided case of a 4 quantifier sentence ZF-equivalent to AC.

Has this gap been closed?

Comments from the same paper:

All this poses the next question: “Can AC be stated by an $(\in, =)$-sentence with only 4 quantifiers?”. I personally believe this isn’t possible. This believe comes from the fact that all known (to me) sentences ZF-equivalent to AC have a form like $\forall x \ldots \exists y \phi(x, y)$. The dots allow for some premises which must be satisfied by $x$ and $\phi(x, y)$ must, in some way, express that $y$ represents a maximal decision relative to $x$. Usually this $\phi$ can’t be stated with 2 or less quantifiers. The only exception I am aware of is a reworked version of Zorn’s Lemma. But in this case the premises on $x$ become quantifier-loaded.

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  • $\begingroup$ See also mathoverflow.net/q/178747/1946. Possible duplicate? $\endgroup$
    – Joel David Hamkins
    Commented Nov 24 at 18:06
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    $\begingroup$ @JoelDavidHamkins I don't see an answer there. This question is more specific. $\endgroup$
    – user76284
    Commented Nov 24 at 18:59
  • $\begingroup$ The point is that the same question was asked there, even if the answer is that no further progress was made. $\endgroup$
    – Joel David Hamkins
    Commented Nov 24 at 19:02
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    $\begingroup$ @JoelDavidHamkins "Shortness" need not mean quantifier count. It could mean formula length or quantifier complexity. Unfortunately, that question doesn't clarify which one it's referring to. $\endgroup$
    – user76284
    Commented Nov 24 at 19:33
  • $\begingroup$ Ah, that's fine, then. $\endgroup$
    – Joel David Hamkins
    Commented Nov 24 at 23:19

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