with ${\omega}(x)={\omega}(y)$. First, let one of the vertices be in a clique class $C$, say $x\in C$. Extend $\{x,y\}$ to an mds and then exchange another vertex of $C$ for $y$. Doing so, one gets a subset with weight $t$, which is not an mds (cf. Observation \[obs:einknotenproCC\]), a contradiction.

So, let both $x$ and $y$ be either elements of singleton classes or stable set classes. As they are in different classes, w.l.o.g. let $v$ be a vertex only seen by $x$. Extend $\{y,v\}$ to an mds $D$ such that $D\cap N(\{y,v\})=\emptyset$ and exchange $y$ for $x$. Again we get a subset (namely $D-y+x$) of weight $t$, that is not an mds since $y$ is no longer dominated.

Note that in the previous lemma the two mentioned vertices must be from different twin classes. Two elements of a stable set class of course can have the same weight while always having distance at least two.

In the following we take a closer look at adjacent vertices, where we find a slightly more complicated situation. We begin by showing that vertices of stable set classes and clique classes that see each other cannot have the same weight.

Let $G=(V,E)$ be an 