Cowen's Theorem/Proof

Lemma $7$
Recall:
 * From Lemma $2$:
 * $\O \in M$


 * From Lemma $7$:
 * $M$ is closed under $g$ relative to $x$


 * From Lemma $4$:
 * $M$ is closed under chain unions.

Hence by definition:
 * $M$ is superinductive under $g$.

Lemma $8$
We have that $M$ is superinductive under $g$.

Let $A$ be superinductive under $g$.

Let $x \in M$.

Then:
 * $x \in M_x$

and from Lemma $8$:
 * $M_x \subseteq A$

Hence:
 * $x \in A$

As $x$ is arbitrary, it follows by definition of subclass that:
 * $M \subseteq A$.

Hence by definition $M$ is minimally superinductive under $g$.