Double Induction Principle/Proof 1

Proof
The proof proceeds by general induction.

Let an element $x$ of $M$ be defined as:
 * left normal with respect to $\RR$ $\map \RR {x, y}$ for all $y \in M$
 * right normal with respect to $\RR$ $\map \RR {y, x}$ for all $y \in M$.

Let the hypothesis be assumed.

First we demonstrate a lemma:

Lemma
We now show by general induction that every $x \in M$ is right normal with respect to $\RR$.

It then follows from the lemma that $\map \RR {x, y}$ for all $x, y \in M$.

Basis for the Induction
$\map P \O$ is the case:

From condition $\text D_1$ of the definition of $\RR$, we have immediately that:
 * $\map \RR {x, \O}$

for all $x \in M$.

That is, that $\O$ is right normal with respect to $\RR$.

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 $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true.

So this is the induction hypothesis:
 * $x$ is right normal with respect to $\RR$

from which it is to be shown that:
 * $\map g x$ is right normal with respect to $\RR$.

Induction Step
This is the induction step:

Let $x \in M$ be right normal with respect to $\RR$:
 * $\forall y \in M: \map \RR {y, x}$

By the lemma we have that $x$ is left normal with respect to $\RR$.

That is:
 * $\forall y \in M: \map \RR {x, y}$

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

That is, $\map g x$ is right normal with respect to $\RR$.

So $\map P x \implies \map P {\map g x}$ and by the Principle of General Induction:


 * $\forall x \in M$: $x$ is right normal with respect to $\RR$.

The result follows.