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 $x \in \operatorname{On}$.

Basis for the Induction
In proofs by transfinite induction using this particular schema, the proposition $\phi(\varnothing)$ is called the basis for the induction.

Induction Step
The proposition $\phi(x) \implies \phi(x^+)$ is called the inductive step. It says that the property will pass from an ordinal number to its successor.

Limit Case
$( \forall x < y: \phi(x) ) \implies \phi(y)$ is called the limit case. It is essentially proving that the proposition will hold for limit ordinals.

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