Now \(T'\) has \(k\) vertices, so by the inductive hypothesis, has \(k-1\) edges. What can we say about \(T\text{?}\) Well, it has one more edge than \(T'\text{,}\) so it has \(k\) edges. But this is exactly what we wanted: \(v=k+1\text{,}\) \(e=k\) so indeed \(e = v-1\text{.}\) This completes the inductive case, and the proof.
in-context