= Solution
The <Axiom of foundation> says that every nonempty set $A$ contains an $x$ with $x\cap A=\varnothing$. The principle of <epsilon induction> says that if a formula $\varphi$ is progressive,
$$
\forall x\left[
(\forall y\in x,\ \varphi(y))\Longrightarrow\varphi(x)
\right],
$$
then $\varphi$ holds for every set.
Assume foundation and let $\varphi$ be progressive. If $\varphi(a)$ failed for some $a$, use <axiom schema of specification>[separation] to form the nonempty set of failures inside $\operatorname{TC}(\{a\})$. Foundation supplies an $\in$-minimal failure $x$. Every member of $x$ lies in the transitive closure and is not a failure, so progressiveness gives $\varphi(x)$, a contradiction. Hence epsilon induction holds.
Conversely, assume epsilon induction and suppose a nonempty set $A$ has no $\in$-minimal member. Let
$$
\varphi(x)\quad\Longleftrightarrow\quad x\notin A.
$$
If every member of $x$ satisfies $\varphi$ but $x\in A$, then $x\cap A=\varnothing$, making $x$ an $\in$-minimal member of $A$, contrary to the assumption. Thus $\varphi$ is progressive. Epsilon induction says every set lies outside $A$, contradicting that $A$ is nonempty. Foundation follows.
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