MHB NP Problems: Weighted Vertex Cover vs Weighted Independent Set

[JohnDoe2013]

There is a Theorem that states that if G = (V, E) is a graph, then S is an independent set $\Leftarrow\Rightarrow$ V - S is a vertex cover.

Suppose the vertices have positive integer weights. Does it follow from the theorem that:

S is an independent set with maximum weight $\Leftarrow\Rightarrow$ V - S is a vertex cover with minimum weight?

 

[Evgeny.Makarov]

I think so. If $S\subseteq V$, let $w(S)$ denote the sum of weights of vertices is $S$, and let $W=w(V)$. Suppose that $S$ is an independent set with maximum weight, but $V-S$ is not minimal, i.e., there exists a $T\subseteq V$ such that $T$ is a vertex cover and $w(T)<w(V-S)=W-w(S)$. Then $w(S)<W-w(T)=w(V-T)$ and $V-T$ is an independent set, a contradiction with maximality of $S$. The other direction is shown similarly.