# Double Induction Principle/Lemma

## Theorem

Let $M$ be a minimally inductive class under a mapping $g$.

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

 $(\text D_1)$ $:$ $\ds \forall x \in M:$ $\ds \map \RR {x, \O}$ $(\text D_2)$ $:$ $\ds \forall x, y \in M:$ $\ds \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y}$

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 general induction.

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 the result follows by the Principle of General Induction.

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 $\RR$

$\blacksquare$