User:Dfeuer/Double Induction Principle

Theorem
Let $A$ be a class.

Let $\mathcal R$ be a relation on $A$.

Let $g$ be a mapping whose domain includes $A$.

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

Suppose that:


 * $\forall x: (x \in a \implies x \mathrel{\mathcal R} 0)$
 * $\forall x: \forall y: (x \mathrel{\mathcal R} y \land y \mathrel{\mathcal R} x \implies x \mathrel{\mathcal R} g(y) )$

Then for all $x, y \in A$, $x \mathrel{\mathcal R} y$

Lemma
Let $y \in A$.

Then:


 * If $x \mathrel{\mathcal R} y$ for all $x \in A$, then $y \mathrel{\mathcal R} x$ for all $x \in A$.

Proof
Suppose that $x \mathrel{\mathcal R} y$ for all $x \in A$.

Let $B$ be the class of all sets $z$ such that $y \mathrel{\mathcal R} z$.

By the premise, $\varnothing \in B$.

Suppose that $z \in B$.

Then $y \mathrel{\mathcal R} z$.

By assumption, $z \mathrel{\mathcal R} y$.

Thus by the premise: $