and $s_2$ are not since there is no mds containing only one of the two vertices.

As the following observations show, the twin relation is a very helpful instrument with regard to equidomination.

\[obs:keinereineroderalleinSSC\] For every mds $D$ and every stable set class $S$, we have $|D\cap S|\in\{0,1,|S|\}$.

\[obs:einknotenproCC\] For every mds $D$ and every clique class $C$, we have $|D\cap C|\leq 1$.

\[obs:constantaufklassen\] Since the vertices of a clique class are pairwise mds-exchangeable, every equidominating function must be constant on every clique class. Analogously, if there exists an mds containing exactly one vertex of a stable set class, every equidominating function must be constant on that stable set class.

\[obs: SSC-maximal-t-knoten\] Since every maximal stable set is also an mds, every (non-maximal) stable set can be extended to an mds. Therefore, for every stable set $S$ and for every equidominating structure $({\omega},t)$ it holds that ${\omega}(S) \leq t$ and hence $|S|\leq t$.

Such an extension of a stable set to an mds can happen in different ways. One way is the following: for a given stable set $S\subseteq V$ of a graph $G=(V,E)$ we determine an mds $D'$ of $G[V\setminus N[S]]$. Then, the union $D\coloneqq S\cup D'$ is an 