is a legal minimal separator of $L({\ensuremath{\mathcal{P}}})$ if and only if $F$ is a legal minimal cut of $G({\ensuremath{\mathcal{P}}})$.

We will prove that if $F$ is a legal minimal separator of $L({\ensuremath{\mathcal{P}}})$ then $F$ is a legal minimal cut of $G({\ensuremath{\mathcal{P}}})$. The proof for the other direction is similar and is omitted.

First, we show that $F$ is a cut of $G({\ensuremath{\mathcal{P}}})$. Assume the contrary. Let $\{u, v\}$ and $\{p, q\}$ be vertices in different components of $L({\ensuremath{\mathcal{P}}})-F$. Since, $G({\ensuremath{\mathcal{P}}})-F$ is connected, there exists a path between vertices $u$ and $q$. Also, $\{u,v\} \notin F$ and $\{p,q\} \notin F$. Thus, by Observation \[obs:path\] there also exists a path between vertices $\{u, v\}$ and $\{p, q\}$ of $L({\ensuremath{\mathcal{P}}})-F$ and are in the same connected component of $L({\ensuremath{\mathcal{P}}})-F$ which is a contradiction. Thus $F$ is a cut of $G({\ensuremath{\mathcal{P}}})$.

Next we show that separator $F$ of $L({\ensuremath{\mathcal{P}}})$ is a legal cut of $G({\ensuremath{\mathcal{P}}})$. For every $T \in {\ensuremath{\mathcal{P}}}$ all the vertices of $L(T)$ in $F$ form a clique in $L(T)$. Thus, all the edges of $T$ in $F$ are incident on a common vertex. Assume that $G({\ensuremath{\mathcal{P}}})-F$ has a connected component with no edge and let $u$ be the vertex in one 