Epsilon Relation is Strictly Well-Founded

Theorem
Let $\Epsilon$ denote the epsilon relation.

Then $\Epsilon$ is a foundational relation on every class $A$.

Proof
By the axiom of foundation:


 * $\forall S: \paren {\exists x: x \in S \implies \exists y \in S: \forall x \in S: \neg x \in y}$

That is, by Nonempty Class has Members:


 * $\forall S: \paren {S \ne \O \implies \exists y \in S: \forall x \in S: \neg x \in y}$

This holds for all sets $S$ whose construction is based on the Zermelo-Fraenkel axioms.

We can weaken the antecedent of the above statement with this statement:


 * $\forall S: \paren {\paren {S \ne \O \land S \subseteq A} \implies \exists y \in S: \forall x \in S: \neg x \in y}$

Note that this step does not require that $A$ be a set: it can be any class, even a proper class.

By definition, it follows that $\Epsilon$ is a foundational relation on every class $A$.