Let u and v be distinct vertices of a graph.Prove that if there exists at least two distinct paths in the graph from u to v ,then the graph contains a simple circuit.
.I have started by defining a few things A simple circuit is a circuit in which except for the first and last vertices which are the same,there are no repeated vertices.
The two paths which start from u and end at v are said to be distinct if they do not have the same internal vertex in common or the same internal edge in common.
With the new information received,I would start by assuming that this graph contains a simple circuit with vertices a,b,c,d,e,f it has edges ab,bc,cd,de,ea which make up the simple circuit . where a=U and d=V. If there existed only one distinct path,there would be no way back from d to a which would give a simple circuit therefore a graph would have to have at least two distinct paths to have a simple circuit.
I would like my proof to be scrutinized.