Transfinite Induction/Principle 1/Proof 2

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

Let $A$ denote a class.

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

Then $\operatorname{On} \subseteq A$.

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

Then:
 * $\left({\operatorname{On} \setminus A}\right) \ne \varnothing$

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

By Ordinal Class is Strongly Well-Ordered by Subset, $\operatorname{On} \setminus A$ 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 $\operatorname{On} \setminus A$.

Therefore $\operatorname{On} \subseteq A$.