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:


 * $\forall S: \left({\exists x: x \in S \implies \exists y \in S: \forall x \in S: \neg x \in y }\right)$

that is:
 * $\forall S: \left({S \ne \varnothing \implies \exists y \in S: \forall x \in S: \neg x \in y }\right)$

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:


 * $\forall S: \left({ \left({ S \ne \varnothing \land S \subseteq A }\right) \implies \exists y \in S: \forall x \in S: \neg x \in y }\right)$

Note that this step does not require that $A$ be a set: it can be any class, even a proper class. Finally, by applying the definition of a foundational relation, then it follows that $\Epsilon$ is a foundational relation on every class $A$.