Double Superinduction Principle/Lemma

Lemma for Double Superinduction Principle
Let $M$ be a minimally superinductive class under a mapping $g$.

Let $\RR$ be a relation on $M$ which satisfies:

Let $x$ be a right normal element of $M$ with respect to $\RR$.

Then $x$ is also a left normal element of $M$ with respect to $\RR$.

Proof
The proof proceeds by Proof by Superinduction.

Let $x \in M$ be right normal with respect to $\RR$.

Let $\map P y$ be the proposition:
 * $\map \RR {x, y}$ holds.

Basis for the Induction
By condition $\text D_1$ of the definition of $\RR$:
 * $\map \RR {x, \O}$

for all $x \in M$.

Thus $\map P \O$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P y$ is true, where $x \in M$, then it logically follows that $\map P {\map g y}$ is true.

So this is the induction hypothesis:
 * $\map \RR {x, y}$ holds

from which it is to be shown that:
 * $\map \RR {x, \map g y}$ holds

Induction Step
This is the induction step:

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

As $x$ is right normal with respect to $\RR$:
 * $\map \RR {y, x}$ holds.

Thus by condition $\text D_2$ of the definition of $\RR$:
 * $\map \RR {x, \map g y}$ holds.

So $\map P x \implies \map P {\map g x}$ and it follows by Proof by Superinduction that $\map \RR {x, y}$ holds for every $x, y \in M$.

Closure under Chain Unions
This is the closure under chain unions:

Now suppose that $C$ is a chain of elements of $M$ such that $\map \RR {x, y}$ holds for all $y \in C$.

Then by $\text D_3$ we have $\map \RR {x, \bigcup C}$.

Therefore:
 * $\forall x \in M$: if $x$ is a right normal element of $M$ with respect to $\RR$, then $x$ is a left normal element of $M$ with respect to $\RR$.