vertex in $S$ has degree ${\Delta}$ in $G$ with at most one edge among $S$ and $\overline{S}$, we have $$N(S,S) + N(S,\overline{S}) = \frac{|S| {\Delta}}2 + \frac{N(S,\overline{S})}2 \le \frac{|S| {\Delta}}2 + \frac{|S|}2 = \frac{|S|({\Delta}+1)}2.$$ Besides, since $|\overline{S}|=n-|S|$, we have $$N(\overline{S},\overline{S}) \le \frac{(n-|S|)(n-|S|-1)}2.$$ If $m$ denotes the size of $G$, then $$\begin{aligned}
0 & = 2\left(N(S,S) + N(S,\overline{S}) + N(\overline{S},\overline{S})\right) - 2m \\
&\le |S|({\Delta}+1) + (n-|S|)(n-|S|-1) - n{\Delta}\\
& = |S|^2 + |S|({\Delta}+2-2n) + n^2 - n - n{\Delta}.
\end{aligned}$$ Define $P(x):=x^2 + x({\Delta}+2-2n) + n^2 - n - n{\Delta}$; then $P(|S|)\ge0$. Since $$\begin{aligned}
P\left(\frac{n}2\right) &= \frac{n^2}4 + \frac{n}2 ({\Delta}+2-2n) + n^2 - n - n{\Delta}\\
&= \frac{n^2}4 + \frac{n{\Delta}}2 + n -n^2 + n^2 - n - n{\Delta}\\
&= \frac{n}4(n-2{\Delta})<0
\end{aligned}$$ and $$\begin{aligned}
P(n-2)&= (n-2)^2 + (n-2)({\Delta}+2-2n) + n^2 - n - n{\Delta}\\
&= (n-2)^2 + (n-2)({\Delta}-n) - (n-2)^2 + n^2 - n - n{\Delta}\\
&= n-2{\Delta}< 0,
\end{aligned}$$ we obtain that $P(|S|)<0$. This is the contradiction we were looking for, so, there not exists an exact defensive (${\Delta}-2$)-alliance $S$ with $|S|\le n-2$. This finishes the proof since $V$ is an exact defensive ${\Delta}$-alliance.

A *clique* in a graph $G = (V, E)$ is a subset $C$ of the 