TITLE: Epsilon Induction implies Axiom Of Foundation (Or Regularity)
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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.