Epsilon Induction implies Axiom Of Foundation (Or Regularity)

I know that, given the other Zermelo-Fraenkel Axioms, Epsilon Induction is equivalent to the Axiom Of Foundation.

I've proven that Axiom Of Foundation implies Epsilon Induction, but i can't prove the other implication. Any suggest?

This is similar to Brian's answer but with fewer negations. Prove, by $\in$-induction on $x$ that "Every set $u$ that contains $x$ (as an element) has an $\in$-minimal member, i.e., there is $y\in u$ with $y\cap u=\varnothing$." The point is that, if $x$ itself isn't a minimal member of $u$, then there is some $x'\in x\cap u$, and you can apply the induction hypothesis to $x'$.

The contrapositive of $\in$-induction says that if $\varphi(x)$ fails for some set $x$, then there is a set $x$ such that $\varphi(y)$ holds for each $y\in x$, but $\varphi(x)$ fails. Now let $a$ be any set, and let $\varphi_a(x)$ be the formula $x\notin a$. Use the contrapositive to show that either $a=0$, or there is an $x\in a$ such that $x\cap a=0$.

Define $Unending \ chain$: $$ Unending \ chain (c) \iff \forall m \in c \ [\exists n \in m \ (n \in c)]$$

Define $WF(x)$ $$WF(x) \iff \neg \exists c \ (Unending \ chain (c) \wedge x \in c)$$

Now take any arbitrary set $a$, if every member of $a$ is not a member of some Unending chain (i.e. all are well founded), then clearly $a$ would be well founded! since if otherwise we'll have an Unending chain $c$ with $a \in c$, but by then there will exist an element $n$ of $a$ such that $n \in c$, but $c$ is an Unending chain, and the existence of such an element $n$ clearly contradicts the condition that all elements of $a$ are not elements of unending chains.

So the antecedent of Epsilon induction is fulfilled for the property $WF$, so all sets fulfill $WF$. Which is Regularity! that is "every Unending chain is empty", i.e. every set is well founded.