Zermelo–Fraenkel set theory

Yes

... given the rest of the ZF axioms.

And I, I think, I mean, there is a lot of convincing and cheating there involved.

Um, if, if I think about this version, the infinite product of non-empty sets is, uh, non-empty, this sounds so obvious.

Whereas the other way, choice does imply excluded middle.

So, so an, an axiom of choice is, is the, the stronger brutalist version of, uh, of classical logic.

Maybe one final remark, I think from my side, would be that classically speaking, choice is rather tame because it doesn't change the consistency threa-strength of ZF.

Yeah.