Transfinite Induction/Schema 2

Theorem
Let $\phi(x)$ be a property satisfying the following conditions:


 * $\phi(\varnothing)$ is true.
 * $\phi(x) \implies \phi(x^+)$ is true.
 * $( \forall x < y: \phi(x) ) \implies \phi(y)$.

Then, $\phi(x)$ is true for all ordinals $x$.

Proof
Take the class $\{ x : \phi(x) \}$. We shall set $A$ equal to this class. Then, $\phi(x)$ is equivalent to the statement that $x \in A$. The three conditions in the hypothesis become:


 * $\varnothing \in A$
 * $x \in A \implies x^+ \in A$
 * $( \forall x < y: x \in A ) \implies y \in A$

These are precisely the conditions for the class $A$ in the second principle of transfinite induction. Therefore, $\operatorname{On} \subseteq A$. Thus, $\phi(x)$ holds for all $x \in \operatorname{On}$.

Remark
$\phi(\varnothing)$ is called the base case.

$\phi(x) \implies \phi(x^+)$ is called the inductive case.

$( \forall x < y: \phi(x) ) \implies \phi(y)$ is called the limit case.

Then, this formulation of transfinite induction says that if the base case, inductive case, and limit case are satisfied, then the statement holds for all ordinals.