Prove that if G is a graph with spanning tree T and e is an edge of G that is not in T, then the graph obtained by adding e to T contains one and only one set of edges that forms a circuit.

Prove that if G is a graph with spanning tree T and e is an edge of G that is not in T, then the graph obtained by adding e to T contains one and only one set of edges that forms a circuit.

Given info:

G is a graph with spanning tree T and e is an edge of G that is not in T.

Proof:

We have:

G is a graph with spanning tree T.

e is an edge in G that is not in T.

G′ is the graph obtained by adding the edge e to the spanning tree T.

T is a spanning tree, thus T is a tree and this implies that T does not contain any simple circuits.

Let the endpoints of the edge e in G be v and w. Since T is a spanning tree, the vertices v and w are also vertices of T.

By the previous exercise, we also know that there exists a unique path P from v to w and let E ( P ) be the set of edges contained in the unique path