Since we can build any graph using a combination of these two moves, and doing so never changes the quantity \(v - e + f\text{,}\) that quantity will be the same for all graphs. But notice that our starting graph \(P_2\) has \(v = 2\text{,}\) \(e = 1\) and \(f = 1\text{,}\) so \(v - e + f = 2\text{.}\) This argument is essentially a proof by induction. A good exercise would be to rewrite it as a formal induction proof.
in-context