equidominating graph with equidominating structure $({\omega},t)$ and let $S\subseteq V$ be a stable set class and $C\subseteq V$ a clique class that see each other. Then for all $x\in S$ and for all $y\in C$ it holds that ${\omega}(x)\neq{\omega}(y)$.

Suppose there are vertices $x\in S$ and $y\in C$ with ${\omega}(x)={\omega}(y)$. The other vertices of $S$ must have a different weight than $x$, since otherwise one could extend $S$ to an mds $D$ such that $D\cap N(S)=\emptyset$ and exchange two vertices of $S$ for two vertices of $C$ while maintaining the weight $t$, a contradiction. So, extend $S$ to an mds $D$ such that $D\cap N(S)=\emptyset$. The set $D'\coloneqq D-x+y$ has weight $t$ and, therefore, is also an mds. Now $D'-x'+x$ ($x'\in S,\, x'\neq x$) has a weight different than $t$, but since the swapped vertices lie in one twin class it still must be an mds, again a contradiction.

Further, a vertex of stable set class and an adjacent singleton class cannot have the same weight in an equidominating structure.

\[lem:SSCundsingleton\] Let $G=(V,E)$ be an equidominating graph with equidominating structure $({\omega},t)$. Let $S\subseteq V$ be a stable set class and let $\{y\}$ be a singleton class with $y\in N(S)$. Then 