 13I

The Axiom of foundation states that every nonempty set $A$ has a member $a$ with $a\cap A=\varnothing$. The principle of epsilon induction states that a formula $\varphi$ satisfying $\forall x[(\forall y\in x\varphi (y))\Rightarrow \varphi (x)]$ holds for every set $x$.

Assume Axiom of foundation and suppose a progressive formula has a counterexample $a$. In the transitive closure of $\{a\}$, use axiom schema of separation to collect the counterexamples into a nonempty set $B$. Foundation supplies $b\in B$ having no member in $B$. Since the closure is transitive, every member of $b$ satisfies $\varphi$, contradicting progressiveness. Conversely, if a nonempty set $A$ had no foundation-minimal member, every $x\in A$ would have a member $y\in x\cap A$. The formula $x\notin A$ is then progressive: if every member of $x$ lies outside $A$, $x$ cannot belong to $A$. Epsilon induction would make $A$ empty, a contradiction. This proves equivalence in the presence of the other Zermelo–Fraenkel set theory.

To obtain the cumulative hierarchy, perform epsilon induction on the assertion that $x$ belongs to some $V_{\alpha }$. If this holds for each member $y$, Axiom schema of replacement collects their least such ordinals into a set. Choose an ordinal $\beta$ above them all. Then $x\subseteq V_{\beta }$ and $x\in V_{\beta +1}$. Thus every set occurs in the hierarchy. Use the rank of a set convention $\mathrm{rk}(x)={\sup }_{y\in x}(\mathrm{rk}(y)+1)$, equivalently the least $\alpha$ with $x\subseteq V_{\alpha }$.