mds of $G$. The domination property of $D$ is obvious. Furthermore, every vertex of $S$ is its own private neighbor (possible among others) and for every vertex $v\in D'$ it holds that $pn(v,D')=pn(v,D)$. That means it is not possible to delete one vertex and remain dominating. Hence, for every stable set $S$ there exists an mds $D$ with $S\subseteq D$ and $D\cap N(S)=\emptyset$, which we will often use in the proofs of the upcoming lemmas.

In the following lemmas we examine whether different vertices of an equidominating graph can have equal weights or not. As we will see, two vertices can only have the same weight if they lie in the same twin class or are adjacent. That means for one thing, that when trying to construct an equidominating structure one has to consider fewer combinatorial possibilities. And for another thing, that for a given number of weights to be allocated one can bound the diameter of an equidominating graph.

\[lem:verschiedenegewichteabstandzwei\] Let $G=(V,E)$ be an equidominating graph with equidominating structure $({\omega},t)$ and let $x,\,y \in V$ be two vertices of different twin classes with $dist(x,y)\geq 2$. Then it holds that ${\omega}(x)\neq{\omega}(y)$.

Suppose there are such two vertices $x, y\in V$ 