Double Induction Principle/Proof 2

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 $x, y \in M$.

Proof

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.

$\blacksquare$