Transfinite Induction/Schema 2/Proof 1
Theorem
Let $\map \phi x$ be a property satisfying the following conditions:
- $(1): \quad \map \phi \O$ is true
- $(2): \quad$ If $x$ is an ordinal, then $\map \phi x \implies \map \phi {x^+}$
- $(3): \quad$ If $y$ is a limit ordinal, then $\paren {\forall x < y: \map \phi x} \implies \map \phi y$
where $x^+$ denotes the successor of $x$.
Then, $\map \phi x$ is true for all ordinals $x$.
Proof
It should be noted that for any two ordinals, $x \lt y \iff x \in y$.
Let $\map \phi x$ be a property that satisfies the above conditions.
Aiming for a contradiction, suppose $y$ be an ordinal such that $\neg \map \phi y$.
It is noted that $y \ne \O$.
Therefore $y$ must be either a successor ordinal or a limit ordinal.
If $y$ is a successor ordinal, then let $x$ be the ordinal such that $x^+ = y$.
By the Rule of Transposition, it is seen that $\neg \map \phi x$.
By Set is Element of Successor, it follows that $x \in y$, and so $x \lt y$.
If $y$ is a limit ordinal, then by the Rule of Transposition there exists an ordinal $x \lt y \iff x \in y$ such that $\neg \map \phi x$.
It has been shown that if $y$ is an ordinal such that $\neg \map \phi y$, then there exists an ordinal $x \in y$ such that $\neg \map \phi x$.
The rest follows from the proof of Schema 1.
$\blacksquare$