Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail such that every edge of $G$ lies on one of these trails. Let $G$ be a connected graph with exactly 2 odd vertices $u$ and $v$ such that $deg(u) \geq 3$ and $deg(v) \geq 3$. Prove or disprove if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ trail in $G$ such that every edge of $G$ lies on one of these trails.
I know that if if $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ path in $G$. But the rest bother me, I feel like they want to say $G$ contains an Eulerian trail. I tried to find a counter example, but got no luck.
I don't know if this is correct, but let's try it. Because $G$ is 2-edge-connected then there exist 2 edges disjoint $u-v$ paths in $G$ called $p_1$ and $p_2$. But if one of these trail , says $p_1$, contain every edge of $G$ then it also contain edges in $p_2$, so $p_1$ and $p_2$ can't be edge-disjoint paths. So this is false because it contradict itself, right?

If a connected graph has exactly two odd vertices it has an eulerian trail between them. Why? Call those vertices $uv$. If $uv$ is in the graph remove it, if it is not in the graph add it. You now have a graph where all the vertices have even degree, thus an eulerian circuit, that circuit becomes a path when you take the graph back to its original form.
