Sandwich Principle/Corollary 2

Theorem
Let $A$ be a class.

Let $g: A \to A$ be a mapping on $A$ such that:
 * for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.

Let $g$ be a progressing mapping.

Let $x \subseteq y$.

Then:
 * $\map g x \subseteq \map g y$

Proof
Let $x \subseteq y$.

Suppose $x = y$.

Then $\map g x \subseteq \map g y$ and the result holds.

Suppose that $x \ne y$.

Then $x \subset y$

It follows from Corollary 1 that:
 * $\map g x \subseteq y$

As $g$ is a progressing mapping on $A$:
 * $y \subseteq \map g y$

Hence by Subset Relation is Transitive:
 * $\map g x \subseteq \map g y$

So in either case:
 * $\map g x \subseteq \map g y$