\emptyset$, $f \notin F$. Thus, $e$ is the only vertex of $L(T)$ in $F$.

\[lm:maximal\_cuts\_equals\_maximal\_seps\]

If there exists a complete set of pairwise parallel legal minimal cuts for $G({\ensuremath{\mathcal{P}}})$, then there exists a maximal set ${\ensuremath{\mathcal{F}}}$ of pairwise parallel minimal separators of $L({\ensuremath{\mathcal{P}}})$ where every separator in ${\ensuremath{\mathcal{F}}}$ is legal.

Consider any complete set of pairwise parallel legal minimal cuts ${\ensuremath{\mathcal{F}}}'$ of $G({\ensuremath{\mathcal{P}}})$. By Theorem \[thm:cuts\_seps\] and Lemma \[lm:parallel\], ${\ensuremath{\mathcal{F}}}'$ is a set of pairwise parallel legal minimal separators of $L({\ensuremath{\mathcal{P}}})$. There exists a maximal set ${\ensuremath{\mathcal{F}}}$ of pairwise parallel minimal separators where ${\ensuremath{\mathcal{F}}}' \subseteq {\ensuremath{\mathcal{F}}}$. Assume that there exists a minimal separator $F$ in ${\ensuremath{\mathcal{F}}}\setminus {\ensuremath{\mathcal{F}}}'$ which is not legal. Using an argument similar to the proof of Theorem \[thm:cuts\_seps\], it can be shown that $F$ is a cut of $G({\ensuremath{\mathcal{P}}})$, every connected component of $G({\ensuremath{\mathcal{P}}})-F$ has at least one edge, and $F$ is also a minimal cut of $G$.

Since, by assumption, minimal separator $F$ of $L({\ensuremath{\mathcal{P}}})$ is not legal, there exists a tree $T \in {\ensuremath{\mathcal{P}}}$ where at least two nonincident edges of $T$ are in $F$. Let $e_1=\{x, y\}$ and $e_2=\{x', y'\}$ be those nonincident edges. Consider the any internal edge $e_3$ in $T$ where $e_1$ 