$x=0$, and its multiplicity is $n-{\Delta}$.]{}

We prove each item separately.

i)  [Let us consider $S\subset V$ with $S$ an exact defensive ($2i-{\Delta}$)-alliance in $G$. Then, we have for all $v\in S$ $$2{\delta}_S(v)\ge {\delta}(v)+2i-{\Delta}= {\Delta}+2i-{\Delta}\quad \Leftrightarrow \quad {\delta}_S(v)\ge i,$$ besides, the equality holds at some $w\in S$. We have the result since $A_{-{\Delta}+2i}(G)$ is the number of exact defensive ($2i-{\Delta}$)-alliance in $G$.]{}

ii) [One can check directly that if $S$ is a single vertex, then $S$ is an exact defensive ($-{\Delta}$)-alliance; furthermore, it is clear that any $S\subseteq V$ with $\langle S\rangle$ connected and more than one vertex is not an exact defensive ($-{\Delta}$)-alliance, since for any $v\in S$ we have $$\label{eq:i}
          {\delta}_{S}(v) - {\delta}_{\overline{S}}(v) \geq  1 - ({\Delta}- 1) = -{\Delta}+ 2.$$ Consequently $A_{-{\Delta}}(G) = n$, since $G$ is a ${\Delta}$-regular graph.]{}

iii) [The maximum value in $\mathcal{K}$ is ${\Delta}$, so $Deg\big(A(G;x)\big)$ is at most $n+{\Delta}$. We have that each connected component of $G$ is an exact defensive ${\Delta}$-alliance since ${\delta}(v)={\Delta}$ for any vertex $v$. Then, $A_{{\Delta}}(G)>0$ and $Deg\big(A(G;x)\big)= n+{\Delta}$. Besides, the other results are consequences of the well known fact $2m=n{\Delta}$ and the previous results.]{}

iv) [By item i), $A_{{\Delta}}(G)$ is the number of connected induced subgraphs of 