Axiom of Foundation (Strong Form)/Proof 2

Theorem
Let $B$ be a class.

Suppose $B$ is not empty.

Then $B$ has a $\in$-minimal element.

Proof
Let $x \in B$.

Let $x'$ be the transitive closure of $x$ by the alternative definition of transitive closure.

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 $B$.