Transfinite Induction/Schema 1/Proof 2

Proof
It should be noted that for any two ordinals, $x \lt y \iff x \in y$.

Let $P \left({x}\right)$ be a property that satisfies the above conditions.

Aiming for contradiction, let $y$ be an ordinal such that $\neg P \left({y}\right)$.

Let $S$ be the set defined as:
 * $S = \left\{{x \in y^+: \neg P \left({x}\right)}\right\}$

It is seen that this set is non-empty because by Set is Element of Successor:
 * $y \in y^+$

$y^+$ is an ordinal because of Successor Set of Ordinal is Ordinal.

By Element of Ordinal is Ordinal, $S$ is a set of ordinals.

By the Rule of Transposition, it follows that for every $x \in S$, there exists an ordinal $\alpha \lt x \iff \alpha \in x$ such that $\neg P \left({\alpha}\right)$.

But by transitivity:
 * $\alpha \in x \land x \in y^+ \implies \alpha \in y^+$

So the epsilon restriction $\in_S$ is a right-total relation on $S$.

Therefore by the Axiom of Dependent Choice there exists a sequence $\left\langle{x_n}\right\rangle_{n \in \N}$ in $S$ such that:
 * $\forall n \in \N: x_{n+1} \in x_n$

But this contradicts No Infinitely Descending Membership Chains.

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

Hence the result.