Suppose we are interested in showing the property $\Phi(\alpha)$ holds for all ordinals $\alpha$ using transfinite induction. The proof basically involves three steps:

1. (first ordinal) show that $\Phi(0)$ holds;
2. (successor ordinal) if $\Phi(\beta)$ holds, then $\Phi(S\beta)$ holds;
3. (limit ordinal) if $\Phi(\gamma)$ holds for all $\gamma<\beta$ and $\beta=\sup\lbrace \gamma\mid \gamma<\beta\rbrace$ , then $\Phi(\beta)$ holds.

Below is an example of a proof using transfinite induction.

Proof. Let $\Phi(\alpha)$ be the property: $0+\alpha=\alpha$ . We follow the three steps outlined above.

1. Since $0+0=0$ by definition, $\Phi(0)$ holds.
2. Suppose $0+\beta=\beta$ . By definition $0+S\beta=S(0+\beta)$ , which is equal to $S\beta$ by the induction hypothesis, so $\Phi(S\beta)$ holds.
3. Suppose $\beta=\sup\lbrace\gamma\mid \gamma<\beta\rbrace$ and $0+\gamma=\gamma$ for all $\gamma<\beta$ . Then $$0+\beta = 0+\sup\lbrace\gamma\mid \gamma<\beta\rbrace = \sup \lbrace 0+\gamma\mid \gamma<\beta\rbrace.$$ The second equality follows from definition. Furthermore, the last expression above is equal to $\sup \lbrace \gamma\mid \gamma<\beta\rbrace =\beta$ by the induction hypothesis. So $\Phi(\beta)$ holds.

Therefore $\Phi(\alpha)$ holds for every ordinal $\alpha$ , which is the statement of the theorem, completing the proof.