Prove, if possible, if an edge lies on a circuit in $G$, then the edge also lies on a cycle in $G$.
I attempted to find a counterexample but it seems pretty evident that this statement is true. The trouble I am having is how to show that it is in fact true.
The strategy I was attempting to use was to assume it did not lie on a cycle and to show a contradiction. I can't seem to come up with anything concrete in order to use this method (if this is even a viable method).
Path - A walk that does not include any vertex twice.
Trail - A walk that does not pass the same edge twice.
Circuit - A trail that begins and ends on the same vertex.
Cycle - A path that begins and ends on the same vertex.
Any help would be greatly appreciated!
If I understand your definitions correctly then generating a counterexample is fairly simple. I assume circuit to be an eulerian circuit which must touch all nodes and a cycle to be a closed trail that need not touch all the nodes.
Start with a circle with 6 nodes and number them 1-2-3-4-5-6, then 1-3-5-1 is a cycle that does not have any edge in common with the circle.