Axiom of regularity

Every nonempty set $A$ has an element $x\in A$ such that $x\cap A=\varnothing$.

A descending membership sequence is a function on the natural numbers with each next value a member of the preceding value. Foundation excludes it: the range is a nonempty set with no member disjoint from that range.

Foundation implies that no descending membership sequence exists. Conversely, with dependent choice, failure of foundation gives a nonempty set in which every element contains another element of that set. Successively selecting such members produces a descending membership sequence. Thus the two conditions are equivalent with dependent choice; full choice is sufficient.

Membership induction is the schema displayed, with set parameters permitted in $\phi$. The Axiom of foundation proves it by choosing a minimal counterexample in the transitive closure $\mathrm{TC}(\{x\})$ of the singleton containing a counterexample $x$. Conversely, if a nonempty set $a$ had no membership-minimal element, the property $x\notin a$ would be progressive, so induction would force $a$ to be empty. Thus over ZF without Axiom of foundation, this schema implies Axiom of foundation. It also implies that every set lies in the cumulative hierarchy: bound the stage indices of all members of a set by replacement, then place that set in the next power-set stage.

The principle of epsilon induction says that every progressive class is universal: ifthen $\phi (x)$ holds for every set $x$. Over the other axioms of ZF, it is equivalent to the Axiom of foundation.

Given a definable operation $G(x,h)$ on a set $x$ and a function $h$ with domain $x$, there is a unique class function $F$ satisfyingfor every set $x$. Existence and uniqueness follow by epsilon induction, because the value at $x$ depends only on values at members of $x$.