(1) If G is connected graph, and for every edge $e$, there are cycles $C_1$ and $C_2$ such that $E(C_1) \bigcap E(C_2) = \{e\},$ then $G$ is 3 edge connected.
I'm trying to figure out how to prove this statement. Not sure how. I've been told that it suffices to show that no set of two edges disconnects $G$.
My question is why? Why does showing that no set of two edges disconnects $G$ prove (1)?
Take any edge $e$ of the graph. Denote the vertices connected by this edge $u$ and $v$.
Since $e\ = \ u-v\ $ is in two cylces, there are two disjoint paths from $u$ to $v$ not using the edge $e$ forming a cycle. Denote it by $C$. In particular, there is still a path between the vertices $u$ and $v$.
Delete a second edge. If it is in $C$, then the graph remains connected because in a cycle there are always two paths between two vertices.
If it is not in $C$, then you can use the argument for the first edge again.