# Principle of General Induction

## Theorem

Let $M$ be a class.

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

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

Let $P: M \to \set {\T, \F}$ be a propositional function on $M$.

Suppose that:

\((1)\) | $:$ | \(\ds \map P \O \) | \(\ds = \) | \(\ds \T \) | |||||

\((2)\) | $:$ | \(\ds \forall x \in M:\) | \(\ds \map P x \) | \(\ds = \) | \(\ds \T \implies \map P {\map g x} = \T \) |

Then:

- $\forall x \in M: \map P x = \T$

## Proof

We are given that $M$ is a minimally inductive class under $g$.

That is, $M$ is an inductive class under $g$ with the extra property that $M$ has no proper class which is also an inductive class under $g$.

Let $P$ be a propositional function on $M$ which has the properties specified:

- $(1): \quad \map P \O = \T$

- $(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$

Thus by definition, the class $S$ of all elements of $M$ such that $\map P x = \T$ is an inductive class under $g$.

But because $M$ is minimally inductive under $g$, $S$ contains all elements of $M$.

That is:

- $\forall x \in M: \map P x = \T$

as we were to show.

$\blacksquare$

## Terminology

### Basis for the Induction

The step that shows that the propositional function $P$ holds for $\O$ is called the **basis for the induction**.

### Induction Hypothesis

The assumption made that $\map P x$ is true for some $x \in M$ is called the **induction hypothesis**.

### Induction Step

The step which shows that $\map P x = \T \implies \map P {\map g x} = \T$ is called the **induction step**.

## Also see

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications