and, minimum and maximum degrees ${\delta}_1$ and ${\Delta}_1$, respectively. If $A(G^*;x)=A(G;x)$, then $G^*$ is a connected graph with exactly $n$ vertices of degree ${\Delta}_1={\Delta}+n_1-n$, $n_1\ge n$, ${\Delta}_1\ge{\Delta}$ and ${\delta}_1 \equiv {\Delta}_1 (\text{mod } 2)$.

Furthermore, if $n_1 > n$, then the following inequalities hold: $$\label{eq:Poly2}
       \frac{{\Delta}_1+{\delta}_1+2}2\le {\Delta}.$$ $$\label{eq:Poly1}
       {\delta}_1+2 < {\Delta}< {\Delta}_1,$$ $$\label{eq:Poly3}
       {\Delta}+ 1 \le {\Delta}_1 \le 2{\Delta}-3,$$ $$\label{eq:Poly4}
       {\delta}_1+4 \le {\Delta}_1.$$

Since $A(G^*;x)=A(G;x)$ is a symmetric polynomial by Theorem \[t:PropRegular\] vi), we conclude that ${\delta}_1 \equiv {\Delta}_1 (\text{mod } 2)$ by Theorem \[p:AlliPoly\] iv). By Theorems \[p:AlliPoly\] v) and \[t:PropRegular\] ii), $G^*$ has $n$ vertices of maximum degree ${\Delta}_1$, so, $n_1 \geq n$; besides, $n_1 - {\Delta}_1 = n-{\Delta}$. Note that if $n_1=n$ then $G^*$ is a ${\Delta}$-regular graph with $A_{{\Delta}}(G^*)=1$, so, Theorem \[c:ConComp\] gives that $G^*$ is a connected graph.

Assume that $n_1 > n$. Denote by $t:=n_1-n={\Delta}_1 -{\Delta}$. Let $v_1,\ldots,v_n \in V(G^*)$ be the vertices in $G^*$ with degree ${\Delta}_1$ and define $S:=\{v_1,\ldots,v_n\}$. Note that for any $v\in S$ we have ${\delta}_{S}(v) \ge {\Delta}_1 - t = t + ({\Delta}_1-2t) \ge {\delta}_{\overline{S}}(v) + {\Delta}_1-2t$; hence, $S$ contains a defensive (${\Delta}_1-2t$)-alliance $S_1$ and $k_{S_1} \ge {\Delta}_1-2t$. Therefore, there is at least one term of degree greater or 