cuts $F$ and $F'$ of $G({\ensuremath{\mathcal{P}}})$ are not. Then, there exist $\{\{u, v\}, \{p, q\}\} \subseteq F'$ where $\{u, v\}$ and $\{p, q\}$ are present in different components of $G({\ensuremath{\mathcal{P}}})-F$. Since $F$ and $F'$ are parallel separators in $L({\ensuremath{\mathcal{P}}})$, and $F$ does not contain $\{u, v\}$ and $\{p, q\}$, there exists a path between vertices $\{u, v\}$ and $\{p, q\}$ in $L({\ensuremath{\mathcal{P}}})-F$. Then, by Observation \[obs:path\] there also exists a path between vertices $u$ and $q$ in $G({\ensuremath{\mathcal{P}}})-F$. Thus, edges $\{u, v\}$ and $\{p, q\}$ are in the same connected component of $G({\ensuremath{\mathcal{P}}})-F$ which is a contradiction.

The other direction can be proved similarly using Observation \[obs:path\].

The next lemma follows from the definition of restricted triangulation and is from [@Gysel2012].

\[lm:two\_cliques\] Let $H$ be a restricted triangulation of $L({\ensuremath{\mathcal{P}}})$ and let $(T, B)$ be a clique tree of $H$. Then, for every vertex $e = \{u, v\}$ in $L({\ensuremath{\mathcal{P}}})$, there does not exist a node $x \in V(T)$ where $B(x)$ contains vertices from both $\hat{K}(u)\setminus e$ and $\hat{K}(v)\setminus e$.

A set ${\ensuremath{\mathcal{F}}}$ of pairwise parallel legal minimal cuts of $G({\ensuremath{\mathcal{P}}})$ is *complete*, if for every input tree $T \in {\ensuremath{\mathcal{P}}}$ and for every internal edge $e$ of $T$, there 