Cowen's Theorem/Lemma 6

Lemma for Cowen's Theorem
Let $g$ be a progressing mapping.

Let $x$ be a set.

Let $\powerset x$ denote the power set of $x$.

Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.

Let $M$ be the class of all $x$ such that $x \in M_x$.

We have that:
 * $\forall z: M_z \subseteq M$

Proof
Let $x \in M_z$.

Let $y = x \cup z$.

From Set is Subset of Union:
 * $z \subseteq y$

Hence by Lemma $3$:
 * $M_z \subseteq M_y$

Hence:
 * $x \in M_y$

Also, we have:
 * $x \subseteq y$

and so from Lemma $5$:
 * $M_y \subseteq M_x \cup \paren {\powerset y \setminus \powerset x}$

Hence:
 * $x \in M_x \cup \paren {\powerset y \setminus \powerset x}$

But we have that:
 * $x \notin \powerset y \setminus \powerset x$

Hence:
 * $x \in M_x$

Thus:
 * $x \in M$

and so:
 * $M_z \subseteq M$

and the result is shown to hold.