So the following exercise is a proof:
$\text{Let } G \text{ a graph such that for } x \in V(G) \text{ } \delta_G(x) = \delta(G).\text{ Show that } \delta_{G^c}(x)=\Delta(G^c).$
Proof: If $\delta_G(x) = \delta(G)$ we know that the neighborhood of the vertex $x$ is given with the minimum cardinality. So by definition of neighborhood: $$N_{G^c}(x)= \{y \in V(G) : xy \notin E(G)\}$$ Thus, the cardinality of this set must be de maximum. Therefore $\delta_{G^c}(x) = \Delta(G^c)_\square$.
I'm not sure if the argument of the neighborhood is valid, or I must to justify it. Either way, if there is a more convincing proof, i'd accept it.