Progressing Function Lemma

Theorem
Let $A$ be a class.

Let $g$ be a progressing mapping on $A$.

Let $\RR$ be the relation defined as:
 * $\map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$

where $\lor$ denotes disjunction (inclusive "or").

Then:
 * $(1): \quad \forall y \in \Dom g: \map \RR {y, \O}$
 * $(2): \quad \forall x, y \in \Dom g: \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y}$

where $\land$ denotes conjunction ("and").

Proof
In the following, let $x$ and $y$ be arbitrary elements of $A$ in the domain of $g$.

From Empty Set is Subset of All Sets:
 * $\O \subseteq y$

Thus by the Rule of Addition:
 * $\map g y \subseteq \O \lor \O \subseteq y$

and so it is seen that $\map \RR {y, \O}$.

That is, $(1)$ holds.

Let $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold.

Thus by definition of $\RR$, both of the following hold:
 * $(\text a): \quad \map g x \subseteq y$ or $y \subseteq x$


 * $(\text b): \quad \map g y \subseteq x$ or $x \subseteq y$

We are to show that $\map \RR {x, \map g y}$ holds.

That is:
 * $\map g x \subseteq \map g y$ or $\map g y \subseteq x$

Let $\map g x \subseteq y$.

Because $g$ is a progressing mapping:
 * $y \subseteq \map g y$

and so:
 * $\map g x \subseteq \map g y$

Thus $\map \RR {x, \map g y}$ holds.

Let $\map g y \subseteq x$.

Then by definition $\map \RR {x, \map g y}$ holds.

If neither $\map g x \subseteq y$ nor $\map g y \subseteq x$, then by $(\text a)$ and $(\text b)$:
 * $y \subseteq x$ and $x \subseteq y$

and so by definition of set equality:
 * $x = y$

Hence:
 * $\map g x = \map g y$

Thus, again by definition of set equality:
 * $\map g x \subseteq \map g y$

and so again, by definition, $\map \RR {x, \map g y}$ holds.

Thus it has been shown that in all cases, if $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold, then $\map \RR {x, \map g y}$ holds.

Thus condition $(2)$ has been satisfied.

Hence the result.