Proof of Euler's theorem Part 2:

Sufficiency, on the other hand, can be shown through the following tour construction argument. We begin at some initial node k0 and draw a circuit through G (thus eventually returning to k0). Let this circuit be denoted C0. If C0 happens to be an Euler circuit, this is fine; we stop. If C0 is not an Euler circuit, then if we remove from G all edges used by circuit C0, there must be some edges left over. Moreover, at least two of these edges must be incident on some node k1 through which circuit C0 has passed. This must be so since, by assumption, G is, first, connected and, second, all its nodes are of even degree (and C0 has only used up an even number of edges which are incident on k1). Thus, it is possible to draw another circuit C1 originating and terminating at k1, which uses only edges of G', the graph left after we eliminate the edges of C0 from G.

This procedure may now be continued until eventually, say after the nth step, there will be no edges left uncovered. At that time, an Euler circuit will also have been obtained which will be a combination of circuits C0, C1, C2, . . ., Cn.