# Axiom of Foundation (Strong Form)/Proof 2

## Theorem

Let $B$ be a class.

Suppose $B$ is non-empty.

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

## Proof

This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.

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 $B$.

$\blacksquare$