# Principle of General Induction/Minimally Closed Class

## Theorem

Let $M$ be a class.

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

Let $b \in M$ such that $M$ is minimally closed under $g$ with respect to $b$.

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

Suppose that:

$(1): \quad \map P b = \T$
$(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$

Then:

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

## Proof

We are given that $M$ is minimally closed under $g$ with respect to $b$.

That is, $M$ is closed under $g$ with the extra property that $M$ has no proper class containing $b$ which is also closed under $g$.

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

$(1): \quad \map P b = \T$
$(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$

Let $S \subseteq M$ be the subclass of $M$ defined as:

$S = \set {x \in M: \map P x}$

We have:

 $\ds \map P b$ $=$ $\ds \T$ $\ds b$ $\in$ $\ds M$ $\ds \leadsto \ \$ $\ds b$ $\in$ $\ds S$

By definition, the class $S$ of all elements of $M$ such that $\map P x = \T$ is closed under $g$.

But because $M$ is minimally closed under $g$ with respect to $b$, if $b \in S$ then $S$ contains all elements of $M$.

That is:

$S = M$

and so:

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

as we were to show.

$\blacksquare$

## Terminology

### Basis for the Induction

The step that shows that the propositional function $P$ holds for the distinguished $b \in M$ 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.