Axiom of Foundation (Strong Form)/Proof 2

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Theorem

Let $B$ be a class.

Suppose $B$ is non-empty.

Then $B$ has a strictly minimal element under $\in$.

Proof

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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, $m \in B$.

Aiming for a contradiction, suppose 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 $B$.

$\blacksquare$