of $G^*$.

Assume first that ${\Delta}=2$. By Theorem \[t:PolyReg\] we have $A(G;x)= n x^{n-2} + A_0(G) x^n + A_2(G) x^{n+2}$, thus, by Theorem \[p:AlliPoly\] iii) the degree sequence of $G^*$ has at most two different values. If $G^*$ is regular then Theorem \[t:Regulars\] gives the result. Therefore, seeking for a contradiction assume that the degree sequence of $G^*$ has exactly two different values (i.e., $G^*$ is bi-regular). By Theorems \[p:AlliPoly\] iv) and \[t:PropRegular\] vi) we have ${\delta}_1 \equiv {\Delta}_1 (\text{mod } 2)$. By Theorems \[p:AlliPoly\] v) and \[t:PropRegular\] ii) we have $A_{-{\Delta}_1}(G^*)=A_2(G)=n<n_1$ and $n-2=n_1-{\Delta}_1$, so, we have ${\Delta}_1>2$. By Theorems \[p:AlliPoly\] vii) and \[t:PropRegular\] iii) we have $n_1+{\delta}_1 \le n+2$, so, we obtain $0\le {\delta}_1 \le 1$. If ${\delta}_1=0$, then there is a connected component $G'$ of $G^*$ which is ${\Delta}_1$-regular. So, $k_{V(G')}={\Delta}_1$ and $\operatorname{Deg}\big(A(G^*;x)\big)=n_1+{\Delta}_1 > n+2$, which is a contradiction. Thus, we can assume that ${\delta}_1=1$. Then, we have $n_1=n+1$; and so, ${\Delta}_1=3$. We prove now that $A_1(G^*)\ge n$. Let $u_0,v_0$ be the vertices of $G^*$ with ${\delta}(u_0)=1$ and $v_0\sim u_0$. If $G^*$ is not connected, then it has a $3$-regular connected component $G_0^*$; since $V(G_0^*)$ is an exact defensive $3$-alliance, then $\operatorname{Deg}\big( A(G^*;x)\big) \ge n_1+3 > n+2 = 