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In set theory, the axiom of regularity can be used to prove

$$ \tag{*}\label{*} \forall x:x\notin x. $$

Can the direction be reversed, i.e., can $(\ref{*})$ prove regularity in ZF$\setminus$R?

By ZF$\setminus$R I mean the standard axioms of ZF, without the axiom of regularity. Stated differently, as in the title, is $(\ref{*})$ equivalent to the axiom of regularity in ZF$\setminus$R?

WillG
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    It's not equivalent, since (I believe) it is consistent that sets $x,y$ such that $x\in y\in x$ exist, which still fails regularity (the set ${x,y}$ would have no $\in$-minimal element). – Vsotvep Jun 23 '21 at 18:42
  • @Vsotvep But you still have to argue that the remaining $\mathsf{ZF}$-axioms can't build a self-containing set from such an $x,y$. (I think that is in fact the case, but it isn't quite immediate.) – Noah Schweber Jun 23 '21 at 18:50
  • @NoahSchweber Indeed, I think I've read something like this before, but I can't find where and don't remember why... – Vsotvep Jun 23 '21 at 18:53
  • The axiom of extensionality on its own does not prevent having $x\ne y$ with $x={y}$ and $y={x}$, but the other axioms require having so many sets in any model that it's a difficult job to show that if ZF is consistent then even such a minimal quirk has a model. –  Jun 23 '21 at 18:59
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    Here is a simple construction: Let $(V,\in)$ be a model of ZF and then let $F:V\to V$ be the permutation swapping $\emptyset$ and ${{\emptyset}}$ and leaving everything else fixed. Then let $E$ be the relation on $V$ defined by $xEy$ if and only if $x\in F(y).$ $(V,E)$ is a model of ZF-R (this part requires a little work... cf "Rieger-Bernays permuation model") and in $(V,E),$ we have $\emptyset ;E;{\emptyset};E;\emptyset$ and there are no sets with $x E x.$ – spaceisdarkgreen Jun 23 '21 at 20:19

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