WebDFS-tree of G has V − 1 edges. Therefore since E ≥ V there will be at least an edge in G which is not in the DFS-tree of G. This edge gives a cycle in G. If the graph G is not connected: If G has 2 connected components G1 = (V1,E1) and G2 = (V2,E2). Then it is easy to prove, by contradiction, that E ≥ V implies http://gandvcontracts.com/
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