# Double Induction Principle

## 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 $x, y \in M$.

### Minimally Closed Class

The Double Induction Principle can be made more general by applying it to a minimally closed class:

Let $M$ be a class which is closed under a progressing mapping $g$.

Let $b$ be an element of $M$ such that $M$ is minimally closed under $g$ with respect to $b$.

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

\((\text D_1)\) | $:$ | \(\ds \forall x \in M:\) | \(\ds \map \RR {x, b} \) | |||||

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

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

## Proof 1

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.

$\blacksquare$

## Proof 2

By definition, a minimally inductive class under $g$ is a minimally closed class under $g$ with respect to $\O$.

Recall the Double Induction Principle for Minimally Closed Class:

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

\((\text D_1)\) | $:$ | \(\ds \forall x \in M:\) | \(\ds \map \RR {x, b} \) | |||||

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

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

In this context:

- $b = \O$

and the result follows.

$\blacksquare$