Transfinite Induction/Schema 1

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

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

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

Proof
Since an ordinal $x$ is less than an ordinal $y$ if and only if $x \in y$, the premise of the theorem can be written


 * $\forall y \in \operatorname{On}: \left({\left({\forall x \in \operatorname{On}: \left({x \in y \implies P \left({x}\right)}\right)}\right) \implies P \left({y}\right)}\right)$.

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

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

Since $\left({\forall x \in \operatorname{On}: \left({x \in y \implies P \left({x}\right)}\right)}\right) \implies P \left({y}\right)$,
 * $y \subseteq \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\} \implies y \in \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\}$ by the above equivalence.

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

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

for all $x$.