The axiom of regulatity is equivalent to the statement that there are no downward infinite membership chains. Such a result implies that no set contains itself. I was wondering if, covnersely, the statement $\forall x(x\notin x)$ implies the axiom of regularity (assuming the rest of the axioms of ZFC).
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No, it is not equivalent. For instance it’s consistent that there are double Quine atoms ($x\ne y$ with $x\in y\in x$) but no Quine atoms (sets with $x\in x$.). – spaceisdarkgreen Nov 25 '24 at 14:09