# Axiom of Foundation (Strong Form)

## Theorem

Let $B$ be a class.

Suppose $B$ is non-empty.

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

## Proof 1

*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 see any proofs that link to this page, please insert this template at the top.*

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

By Epsilon Relation is Strictly Well-Founded, $\Epsilon$, the epsilon relation, is a strictly well-founded relation on $B$.

By Epsilon Relation is Proper, $\struct {\mathbb U, \Epsilon}$ is a proper relational structure, where $\mathbb U$ is the universal class.

By Well-Founded Proper Relational Structure Determines Minimal Elements, $B$ has a strictly minimal element under $\Epsilon$.

$\blacksquare$

## Proof 2

*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 see any proofs that link to this page, please insert this template at the top.*

*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 $B$.

$\blacksquare$