Transfinite Induction/Schema 1

Theorem
Let $\phi \left({x}\right)$ be a property

Suppose that:
 * If $\phi \left({x}\right)$ holds for all ordinals $x$ less than $y$, then $\phi \left({y}\right)$ also holds.

Then $\phi \left({x}\right)$ holds for all ordinals $x$.

Proof
The statement:
 * $\forall x \in \operatorname{On}: x \in y \implies \phi \left({x}\right)$

is equivalent to:
 * $y \subseteq \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\}$

Since $\forall x \in \operatorname{On}: \left({x \in y \implies \phi \left({x}\right)}\right) \implies \phi \left({y}\right)$:
 * $y \subseteq \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\} \implies y \in \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\}$

By the Principle of Transfinite Induction (above):
 * $\operatorname{On} \subseteq \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\}$

Therefore:
 * $x \in \operatorname{On} \implies \phi \left({x}\right)$

for all $x$.