for all $x\in S$ it holds that ${\omega}(x)\neq{\omega}(y)$.

Let $x,x'\in S$, $y\in N(S)$ an adjacent singleton class and suppose that ${\omega}(x)={\omega}(y)$. The set $N(S)\setminus N[y]$ cannot be empty, since otherwise one could extend $\{y\}$ to an mds $D$ such that $D\cap N(y)=\emptyset$, but $D-y+x$ does not dominate $x'$, which contradicts ${\omega}(x)={\omega}(y)$. Furthermore, there must be a vertex $v\in N(S)\setminus N[y]$ with $N[v]\subseteq N[S]$ (otherwise $\{y\}\cup (V\setminus N[S])$ is a dominating set, which contains an mds $D$ with $y\in D$, but $D-y+x$ does not dominate $x'$).

So, extend $\{v,y\}$ to an mds $D$ such that $D\cap N(\{v,y\})=\emptyset$. Since $D'\coloneqq D-y+x$ has weight $t$ and thus is an mds, $pn(v,D')=S\setminus \{x\}$. It follows that $D'-v+y$ is also an mds, which implies ${\omega}(v)={\omega}(y)$, a contradiction to Lemma \[lem:verschiedenegewichteabstandzwei\].

The proof of Lemma \[lem:SSCundsingleton\] suggests that if $v$ and $y$ lie in one twin class, $x$ and $y$ could have the same weight. As the next Lemma shows this is indeed possible, but only in a specific situation.

\[lem:SSCmitgleichemgewicht\] Let $G=(V,E)$ be an equidominating graph with equidominating structure $({\omega},t)$ and let $S_1,S_2\subseteq V$ be two adjacent stable set classes with two vertices $x\in S_1,\, y \in S_2$ of the same weight. Then the following 