Axiom of Foundation (Strong Form)/Proof 2
Let $B$ be a class.
Suppose $B$ is non-empty.
Then $B$ has a $\in$-minimal element.
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Let $x \in B$.
Let $x'$ be the transitive closure of $x$.
Let $L = x' \cap B$.
Then $x \in L$, so $L$ is not empty.
Since $x'$ is a set, so is $L$, by the axiom of subset.
Thus by the Axiom of Foundation, $L$ has an $\in$-minimal element $m$.
By the definition of intersection, $m \in B$.
Suppose for the sake of contradiction that there is an element $b \in B$ such that $b \in m$.
Then since $m \in x'$ and $x'$ is transitive, $b \in x'$.
Thus $b \in L$, contradicting the minimality of $m$.
So $m$ is an $\in$-minimal element of $B$.