Let $G = (W, E)$ be an undirected graph, where $W = \{(v_i,v_j) \in V \times V : v_i > v_j\}$ and $E$ is a set of $2$-element subsets of $W$ such that, given two edges $e_1 = (w_1, w_2)$ and $e_2 = (w_1, w_3)$, $(w_2, w_3) \in E$ necessarily if there exists some $v$ belonging to $w_1$, $w_2$ and $w_3$. (For example, if $((v_1, v_2), (v_1, v_3)) \in E$ and $((v_1, v_2), (v_1, v_4)) \in E$, then $((v_1, v_3), (v_1, v_4)) \in E$.)
In such a scenario, we want to remove the minimum number of vertices in $W$ to disconnect the graph, taking into consideration that whenever we remove a vertex $w = (v_1, v_2)$, the rest of the vertices containing $v_1$ or $v_2$ must be removed as well.
My question is whether this is actually the vertex cover problem in disguise or, somehow, the restrictions applied above make it easier or more difficult to solve it.
As an example, consider the following graph, where it's clear that removing $v_2$, $v_3$ or $v_4$ leads to the disconnection of the majority of vertices in $W$.