I want to prove that the hardness of the Steiner tree problem lies in determining the optimal subset of Steiner vertices that need to be included in the tree and I need to show this by proving that if this set is provided, then the optimal Steiner tree can be computed in polynomial time.
It is clear that we can find an MST on the union of this set and the set of required vertices with Kruskal algorithm, and by contradiction proof, can be said that if this union of sets didn't give the MST so the subset of Steiner vertices is not the optimal subset. Any hints or tips would be appreciated.