Principle of Superinduction

Theorem
Let $A$ be a class.

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

Let $A$ be minimally superinductive under $g$.

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

Suppose that:

Then:


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

Proof
We are given that $A$ is a minimally superinductive class under $g$.

That is, $A$ is a superinductive class under $g$ with the extra property that for $A$ has no proper class which is also superinductive under $g$.

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

Then by definition, the class $S$ of all elements of $A$ such that $\map P x = \T$ is a superinductive class under $g$.

But because $A$ is minimally superinductive under $g$, $S$ contains all elements of $A$.

That is:
 * $\forall x \in A: \map P x = \T$

as we were to show.

Also see

 * Proof by General Induction