we prove the correctness. Since the algorithm checks both combinations (cf. line 1), the following argumentation holds for $v_1=x,\,v_2=y$ as well as for $v_1=y,\,v_2=x$. If the output is NO, the stated set in line 3 contains an mds $D$ with $v'\in pn(v_1,D)$ and $v'\notin N[v_2]$. So, $v'$ is not dominated by $D-v_1+v_2$ and thus $x$ and $y$ cannot be mds-exchangeable. If the output is YES and the algorithm reaches line 8, there is no vertex $v'$ in $N(v_1)$, which is not adjacent (or equal) to $v_2$ and which is a private neighbor of $v_1$ in any mds that does not contain $v_2$. So, we cannot find an mds $D$ with $v_1\in D$, $v_2\notin D$ and $pn(v_1,D)\cap N[v_2]\neq\emptyset$. Further, since $v_1$ and $v_2$ are adjacent, there are mds that contain only $v_1$ and that contain only $v_2$. Therefore, $x$ and $y$ are mds-exchangeable.

It is clear that the algorithm terminates for a finite graph. There are $\mathcal O(n)$ vertices in $N(v_1) \setminus N[v_2]$. For each vertex $v' \in N(v_1) \setminus N[v_2]$ it is sufficient to check if the vertices of $N(v')\setminus\{v_1\}$ are dominated since every other vertex is an element of the set stated in line 3. With the use of 