G-Tower is Well-Ordered under Subset Relation

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 well-ordered under the subset relation.

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

By $g$-Tower is Nest, $M$ is a nest.

Hence $\subseteq$ is a total ordering on $M$.

It remains to be shown that $\subseteq$ is a well-ordering ordering, by showing the following:

Let $A \subseteq M$ be an arbitrary non-empty subclass of $M$.

Let $L$ be the set of all elements $x$ which are proper subsets of all elements of $A$:
 * $L = \set {x \in M: \forall z \in A: x \subsetneqq z}$

It is to be shown that the following conditions hold:


 * $(1): \quad$ If $L$ is empty, then $\O$ is the smallest element of $A$.


 * $(2): \quad$ If $L$ is non-empty and contains a greatest element $x$, then $\map g x$ is the smallest element of $A$.


 * $(3): \quad$ If $L$ is non-empty and contains no greatest element, then its union $\ds \bigcup L$ is the smallest element of $A$.


 * Case $(1)$ -- $L$ is empty

As $L$ is empty:
 * $\O \notin L$

Hence $\O$ is not a proper subset of every element of $A$.

However, $\O$ is a subset (although not necessarily proper) of every element of $A$.

Hence $\O$ must be one of the elements of $A$..

Hence $\O$ is by definition the smallest element of $A$.