G-Tower is G-Ordered

Theorem
Let $M$ be a class.

Let $g: M \to M$ be a progressing mapping on $M$.

Let $M$ be a $g$-tower.

Then $M$ is $g$-ordered.

Proof
Recall the definition of a $g$-ordered class:

$M$ is a $g$-ordered class $M$ is well-ordered by the subset relation such that:

So, let $M$ be a $g$-tower.

From $g$-Tower is Well-Ordered under Subset Relation:
 * $M$ is well-ordered under the subset relation.

From $g$-Tower is Well-Ordered under Subset Relation: Empty Set:
 * $\O$ is the smallest element of $M$.

From $g$-Tower is Well-Ordered under Subset Relation: Successor of Non-Greatest Element:
 * Let $x \in M$ such that $x$ is not the greatest element of $M$.
 * Then the immediate successor of $x$ is $\map g x$.

From $g$-Tower is Well-Ordered under Subset Relation: Union of Limit Elements:
 * Let $x \in M$ be a limit element of $M$.
 * Then:
 * $x = \ds \bigcup x^\subset$
 * where $\ds \bigcup x^\subset$ denotes the union of the lower section of $x$.

The result follows.