proof of Veblen’s theorem

The proof is very easy by induction on the number of elements ofthe set $E$ of edges.If $E$ is empty, then all the vertices have degree zero, which is even.Suppose $E$ is nonempty.If the graph contains no cycle, then some vertex has degree $1$, which is odd.Finally, if the graph does contain a cycle $C$, then every vertex hasthe same degree mod $2$ with respect to $E-C$, as it has with respectto $E$, and we can conclude by induction.