Finding optimal minimum rooted k-edge disjoint connected paths to build a feasible solution for the undirected version Suppose we have an undirected weighted graph G with a root node, and the goal is to find minimum cost k-edge disjoint paths from the root node to each vertex. Now, suppose we create a directed version of G called G' by adding two edges (u, v), (v, u) to the G' graph for each undirected edge (u, v) in G, with the same cost as the corresponding undirected edge.
Suppose we solve the minimum rooted k-edge disjoint path problem on G', and we want to find a feasible (not optimal) solution for G using this solution. We consider each edge (u, v) in G to be in the solution if (u, v) or (v, u) is in the solution for G'. I want to see if this strategy can lead to a feasible solution for G. It's obvious that we would indeed have the k paths for each (r, v) pair in the undirected graph, but I can not prove that the paths would be disjoint, or not necessarily disjoint by a counterexample. What do you think? Are there any theorems that can help in this case?

If you have two edge disjoint paths in $G'$ from $s$ to $t$ such that one use the edge $(u,v)$ and the other the edge $(v,u)$ (so one path is $s\rightarrow \text{subpath}_{1,1}\rightarrow u\rightarrow v\rightarrow \text{subpath}_{1,2} \rightarrow t$ and the other is $s\rightarrow \text{subpath}_{2,1}\rightarrow v\rightarrow u\rightarrow \text{subpath}_{2,2} \rightarrow t$, then it is easy to see that we can re-route these paths such that none use  $(u,v)$ and $(v,u)$ ($s\rightarrow \text{subpath}_{1,1}\rightarrow u \rightarrow \text{subpath}_{2,2} \rightarrow t$ and $s\rightarrow \text{subpath}_{2,1}\rightarrow v\rightarrow \text{subpath}_{1,2} \rightarrow t$). Do this for each edge used in both ways, and you get a solution in the undirected case.
