$2{\Delta}>n-2$.

First we prove that $G$ does not have cut vertices. If $n=2{\Delta}$, then $G$ is a Hamiltonian graph by Theorem \[Dirac\]. If $n=2{\Delta}+1$, seeking for a contradiction assume that there is a cut vertex $w$ in $G$. Let $S_1,S_2 \subset V$ with $S_1\cup S_2\cup \{w\}=V$ such that $\langle S_1\rangle$ and $\langle S_2\rangle$ are disjoint. Without loss of generality we can assume that $|S_1|\le {\Delta}\le |S_2|$. Since ${\delta}_{S_1}(w),{\delta}_{S_2}(w)\ge1$, ${\delta}_{S_1}(w)+{\delta}_{S_2}(w)={\Delta}$ and ${\delta}_{S_1}(u)\le |S_1|-1 \le {\Delta}-1$ for all $u\in S_1$, we have ${\delta}_{S_1}(w)=|S_1|$ and ${\delta}_{S_1}(u)={\Delta}-1$ for all $u\in S_1$. Then, we obtain that $|S_1|={\Delta}$, but this is a contradiction since ${\delta}_{S_1}(w)={\Delta}-{\delta}_{S_2}(w)\le {\Delta}-1<{\Delta}=|S_1|={\delta}_{S_1}(w)$. Then, $G$ does not have cut vertices.

By Theorem \[t:PropRegular\] i), we have that $A_{{\Delta}-2}(G)$ is the number of connected induced subgraphs of $G$ with minimum degree ${\Delta}-1$; thus, any exact defensive (${\Delta}-2$)-alliance $S$ in $G$ verifies $|S|\ge {\Delta}$. Let us consider $u\in V$ and denote by $S_u:=V\setminus\{u\}$. Since $G$ does not have cut vertices, $\langle S_u\rangle$ is connected. Besides, we have ${\delta}_{S_u}(v)\ge {\Delta}-1 \ge {\delta}_{\overline{S_u}}(v) + {\Delta}-2$ for all $v\in S_u$ and the equality holds for every $v\in N(u)$; so, $S_u$ is an exact defensive (${\Delta}-2$)-alliance in $G$. Thus, $A_{{\Delta}-2}(G)\ge n$.

Let us consider $u_1,u_2\in V$ with $u_1\neq u_2$ 