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:


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


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

Then:


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

Proof
We are given that $M$ is 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.

Also see

 * Principle of Mathematical Induction
 * Principle of Finite Induction