\wedge \{v,v'\} \cap \{u,u'\} = \emptyset \} \cup \{uv \mid u'v \in A(H)\}  \\
             & \cup \{vu \mid vu' \in A(H)\} \cup \{u'v \mid uv \in A(H)\} \cup \{vu' \mid vu \in A(H)\}.   
  \end{aligned}$$ Then, we can assume that $N^+_H(u) \subseteq N^+_H(u')$ and $N^-_H(u)\subseteq N^-_H(u')$ hold (otherwise we can add the missing arcs to $H$ without additional cost). Then, it follows from that $G[V(G')]$ is a solution of $I'$. Its cost is at most $\operatorname{OPT}(I) - \min_{i \in [x]} p_{r_i}$ because $u$ pays at least $\min_{i \in [x]} p_{r_i}$ in $H$. This shows that $\operatorname{OPT}(I) \ge \operatorname{OPT}(I') + \min_{i \in [x]} p_{r_i}$. For the other direction, suppose that $H'$ is an optimal solution of $I'$. Let $u' \in \mathcal{P}_{r_1, \dots, r_{2x}} \setminus \{ u \}$ be a vertex and let $v$ be a vertex with $w(u'v) = \min_{i \in [x]} p_{r_i}$. Since $H'$ is strongly connected, there exists a vertex $v' \in X$ with $v' u' \in A(H')$. Due to , $H' + uv + v'u$ is strongly connected. Observe that the cost for $u$ is $\min_{i \in [x]} p_{r_i}$ and the cost for $v'$ remains unchanged because $v'$ pays at least $w(v'u') = w(v'u)$ in $H'$. Hence, $\operatorname{OPT}(I) = 