# Double Induction Principle/Proof 1

## Theorem

Let $M$ be a class.

Let $g: M \to M$ be a mapping on $M$.

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

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

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

Then $\map \RR {x, y}$ holds for all $x, y \in M$.

## Proof

The proof proceeds by general induction.

Let an element $x$ of $M$ be defined as:

left normal with respect to $\RR$ if and only if $\map \RR {x, y}$ for all $y \in M$
right normal with respect to $\RR$ if and only if $\map \RR {y, x}$ for all $y \in M$.

Let the hypothesis be assumed.

First we demonstrate a lemma:

### Lemma

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 $\RR$.

$\Box$

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.

$\blacksquare$