Transfinite Induction/Principle 1/Proof 2

Theorem
Let $\On$ denote the class of all ordinals.

Let $A$ denote a class.

Suppose that:
 * For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$.

Then $\On \subseteq A$.

Proof
Suppose for the sake of contradiction that $\neg \On \subseteq A$.

Then:
 * $\paren {\On \setminus A} \ne \O$

From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.

By Ordinal Class is Strongly Well-Ordered by Subset, $\On \setminus A$ must have a smallest element $y$.

Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$.

This contradicts the fact that $y$ is an element of $\On \setminus A$.

Therefore $\On \subseteq A$.