equal than $n_1+{\Delta}_1-2t$ in $A(G^*;x)$. Since $x^{n_1+{\Delta}_1-2t}=x^{n+{\Delta}}$, $S_1$ is an exact defensive (${\Delta}_1-2t$)-alliance in $G^*$. Finally, note that if $\langle S\rangle$ is not a connected subgraph (i.e., $S_1\neq S$), then in $A(G^*;x)$ appear at least two terms $x^{n+{\Delta}}$, but this is a contradiction since $A(G;x)$ is a monic polynomial by Theorem \[p:AlliPoly\] vi). Hence, $\langle S\rangle$ is connected. Since the degree of $A(G^*;x)$ is $n+{\Delta}=n_1+{\Delta}_1-2t$, then $S$ is an exact defensive (${\Delta}_1-2t$)-alliance in $G^*$; therefore, there exists $1\le j\le n$ such that ${\Delta}_1={\delta}(v_j)=2{\delta}_{\overline{S}}(v_j)+{\Delta}_1-2t$, and we have ${\delta}_{\overline{S}}(v_j)=t$. Since $|S|=n=n_1-t$ and $|\overline{S}|=t$, $\overline{S}\subseteq N(v_j)$ and $G^*$ is a connected graph.

Also, since $G^*$ is connected, $A(G^*;x)=A(G;x)$, $k_S={\Delta}_1-2t$ and $k_{V(G^*)}={\delta}_1$, we have ${\delta}_1\le {\Delta}_1-2t$. We are going to prove ${\delta}_1 < {\Delta}_1 - 2t$; seeking for a contradiction assume that ${\delta}_1 = {\Delta}_1 - 2t$. Since $G^*$ is connected, $k_{V(G^*)}={\delta}_1= {\Delta}_1 - 2t = k_{S}$ and this contradicts that $A(G^*;x)$ is a monic polynomial. Therefore, ${\delta}_1 < {\Delta}_1 - 2t$. But, since ${\delta}_1\equiv {\Delta}_1 (\text{mod } 2)$ we obtain ${\delta}_1+2\le {\Delta}_1-2({\Delta}_1-{\Delta})=2{\Delta}-{\Delta}_1$, so holds.

Besides, since ${\Delta}_1>{\Delta}$, gives ${\delta}_1+2 <{\Delta}$, and so, holds. Furthermore, we have ${\Delta}+1\le {\Delta}_1$ and gives , since ${\delta}_1\ge1$. Finally, since ${\Delta}\le {\Delta}_1-1$, provides .

Alliance polynomials of regular 