A has r 1, and B has s 1 edges. T is connected and has n - 1 edges. Let e be any edge of T . Remove e and you get a graph with n vertices and n - 2 edges. As e was an arbitrary edge, then every edge of T disconnects T .
T is connected and every edge is an isthmus.
Let v and w be any two vertices of T . T is connected, so there is a simple path from v to w . Assume that there are 2 or more such paths. Then the removal of any edge of the second path that is not part of the first path does not disconnect the graph, T . Therefore, that edges removed was not an isthmus, giving a contradiction.