such component. Then, $\hat{K}(u) \subseteq F$. But, $F$ is a minimal separator of $L({\ensuremath{\mathcal{P}}})$ and by Lemma \[lm:no\_loner\], $\hat{K}(u) \not \subseteq F$ which is a contradiction. Thus, $F$ is a legal cut of $G({\ensuremath{\mathcal{P}}})$.

Lastly, we show that $F$ is a minimal cut of $G({\ensuremath{\mathcal{P}}})$. Assume the contrary. Then, there exists $F' \subset F$ where $G({\ensuremath{\mathcal{P}}})-F'$ is disconnected but $L({\ensuremath{\mathcal{P}}})-F'$ is connected. Since $F' \subset F$ and every connected component of $G({\ensuremath{\mathcal{P}}})-F$ has at least one edge, every connected component of $G({\ensuremath{\mathcal{P}}})-F'$ also has at least one edge. Let $\{u,v\}$ and $\{p, q\}$ be the edges in different components of $G({\ensuremath{\mathcal{P}}})-F'$. Since, $L({\ensuremath{\mathcal{P}}})-F'$ is connected, there exists an path between $\{u,v\}$ and $\{p, q\}$ in $L({\ensuremath{\mathcal{P}}})-F'$. Then, by Observation \[obs:path\] there also exists a path between vertices $u$ and $p$ in $G({\ensuremath{\mathcal{P}}})-F'$. Hence, edges $\{u, v\}$ and $\{p, q\}$ are in the same connected component of $G-F'$ which is contradiction. Thus, $F$ is also minimal cut of $G({\ensuremath{\mathcal{P}}})$.

\[lm:parallel\] Two legal minimal separators $F$ and $F'$ of $L({\ensuremath{\mathcal{P}}})$ are parallel if and only the if legal minimal cuts $F$ and $F'$ are parallel in $G({\ensuremath{\mathcal{P}}})$.

Assume that legal minimal separators $F$ and $F'$ of $L({\ensuremath{\mathcal{P}}})$ are parallel, but legal minimal 