## [SOLVED] Simple set theory question (wellfoundedness)

I'm having some trouble convincing myself of this statement (Kunen, 99):

"If x = {y}, y = {x} and x != y, then y notin WF but set membership is well-founded (in fact empty) on y."

So then z in x implies z = y, and z in y implies z = x. Finally, set membership on y well-founded implies every non-empty subset of y = {x} (i.e. {x}) has an element w such that there is no v in w (i.e. w is empty). But w must be x, so x is hence empty. But then {x} = 0, which implies x is in 0, a contradiction.

Any help?

We're using ZF-.