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 $\neg P \left({y}\right)$.

By the Rule of Transposition, it follows that if $y$ is an ordinal such that $\neg P \left({y}\right)$, then there exists an ordinal $x \lt y \iff x \in y$ such that $\neg P \left({x}\right)$.

If $y$ is an ordinal such that $\neg P \left({y}\right)$, then let $\phi_y$ be the smallest ordinal in $y$ such that $\neg P \left({x}\right)$.

Let $\operatorname {Ord} \ S$ be true $S$ is an ordinal.

Let $g$ be a weakly defined function defined as:
 * $g \left({x}\right) = \begin{cases}

\phi_x & : \operatorname {Ord} \ x \land \neg P \left({x}\right)\\ \varnothing & : \text{otherwise} \end{cases}$

Let $f$ be a recursive sequence defined as:
 * $f \left({n}\right) = \begin{cases}

y & : n = 0 \\ g \left({f \left({n-1}\right)}\right) & : n \gt 0 \end{cases}$

Let $\operatorname {Im} \left({f}\right)$ denote the image of $f$.

It is to be shown that $\forall x \in \operatorname {Im} \left({f}\right): \neg P \left({x}\right) \land \operatorname {Ord} \ x$.

This will be done using induction.

Basis for Induction
Remember that $y$ is an ordinal such that $\neg P \left({y}\right)$.

$f \left({0}\right) = y$, so we have $\neg P \left({f \left({0}\right)}\right)$.

Induction Hypothesis
Let $n \in \N$.

Suppose that $f \left({n}\right)$ is an ordinal such that $\neg P \left({f \left({n}\right)}\right)$.

Induction Step
Then $f \left({n+1}\right) = \phi_{f \left({n}\right)}$.

And so $f \left({n+1}\right)$ is an ordinal such that $\neg P \left({f \left({n+1}\right)}\right)$.

We have constructed a function $f$ such that:
 * $\forall n \in \N: f \left({n+1}\right) \in f \left({n}\right)$

But this contradicts No Infinitely Descending Membership Chains.

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

Hence the result.