Epsilon Induction

Theorem
Let $A$ be a class.

Let $\Bbb U$ denote the universe.


 * $\left({\forall x: \left({x \subseteq A \implies x \in A}\right)}\right) \implies A = \Bbb U$

Proof
Suppose that
 * $\forall x: \left({x \subseteq A \implies x \in A}\right)$

Consider $\Bbb U \setminus A$.

Aiming for a contradiction, suppose that:
 * $\Bbb U \setminus A \ne \varnothing$

Then by Axiom of Foundation (Strong Form), we have that:


 * $\exists x \notin A: \left({x \cap \left({\Bbb U \setminus A}\right)}\right) = \varnothing$

But:

Thus:
 * $x \setminus A = \varnothing$

So from Set Difference with Superset is Empty Set:
 * $x \subseteq A$

Thus by hypothesis:
 * $x \in A$

contradicting the fact that $x \notin A$.

Therefore we can conclude that
 * $\left({\Bbb U \setminus A}\right) = \varnothing$

and so from Set Difference with Superset is Empty Set:
 * $\Bbb U \subseteq A$

Furthermore, by definition of universe:
 * $A \subseteq \Bbb U$

so by definition of set equality:
 * $A = \Bbb U$