General Double Induction Principle/Proof

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 which satisfies the following conditions:

$({\text D'}_1)$	$:$			$\ds \forall x \in M:$		$\ds \map \RR {x, 0} \land \map \RR {0, x} $
$({\text D'}_2)$	$:$			$\ds \forall x, y \in M:$		$\ds \paren {\map \RR {x, y} \land \map \RR {x, \map g y} \land \map \RR {\map g x, y} } $	$\ds \implies $	$\ds \map \RR {\map g x, \map g y} $

Then:

$\forall x, y \in M: \map \RR {x, y}$

Proof

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Sources

2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional: Exercise $9.1$

\(({\text D'}_1)\)	$:$			\(\ds \forall x \in M:\)		\(\ds \map \RR {x, 0} \land \map \RR {0, x} \)
\(({\text D'}_2)\)	$:$			\(\ds \forall x, y \in M:\)		\(\ds \paren {\map \RR {x, y} \land \map \RR {x, \map g y} \land \map \RR {\map g x, y} } \)	\(\ds \implies \)	\(\ds \map \RR {\map g x, \map g y} \)