\geq f'(U \cup V) - q > f(U \cup V) - n + \Delta \geq f'(U \cup V) - f'(V \backslash N_G(v)) =
f'(N_{G'}(v))$. Therefore, $f'$ is an additive coloring of $G'$.

In order to prove optimality, note first that any two vertices in $U$ are true twins of $G'$. By Observation \[LOWERBOUNDTWINS\], $\eta(G') \geq q$. In addition, suppose that $\eta(G') < \eta(G)$. Hence, there exists an additive $k$-coloring $f'$ of $G'$ with $k = \eta(G)-1$. Let $f$ be the labeling of $G$ satisfying $f(v) = f'(v)$ for all $v \in V$. We have $f(N_G(v)) = f'(N_{G'}(v)) - f'(U) \neq f'(N_{G'}(v')) - f'(U) = f(N_G(v'))$ for any $(v,v') \in E$. Therefore, $f$ is an additive $k$-coloring of $G$ which leads to a contradiction.

When Theorem \[UNIVERTICES\] is applied one must keep in mind that the size of a complete graph that can be joined to a graph is limited by $n-\Delta-1$. In fact, if one chooses $q = n-\Delta$, $\eta(G \lor K_q) = \max \{ \eta(G), q \}$ does no longer hold. For instance, let $G$ be the graph of Figure \[CONTRAEJEMPLO\] and $q = 2$. It can be proven that $\eta(G)=2$ and $\eta(G \lor K_2)=3$. On the other hand, 