<P> Proof: Assume the contrary, i.e. that e belongs to an MST T1 . Then deleting e will break T1 into two subtrees with the two ends of e in different subtrees . The remainder of C reconnects the subtrees, hence there is an edge f of C with ends in different subtrees, i.e., it reconnects the subtrees into a tree T2 with weight less than that of T1, because the weight of f is less than the weight of e . </P> <P> For any cut C of the graph, if the weight of an edge e in the cut - set of C is strictly smaller than the weights of all other edges of the cut - set of C, then this edge belongs to all MSTs of the graph . </P> <P> Proof: Assume that there is a MST T that does not contain e . Adding e to T will produce a cycle, that crosses the cut once at e and crosses back at another edge e' . Deleting e' we get a spanning tree T \ (e') U (e) of strictly smaller weight than T . This contradicts the assumption that T was a MST . </P> <P> By a similar argument, if more than one edge is of minimum weight across a cut, then each such edge is contained in some minimum spanning tree . </P>

Is every minimum spanning tree also a minimax spanning tree