e'$. Thus, $e \cap e_1 = v$ and $e \cap e_2 =v$. Let $\pi = a, \cdots, e_1, e, e_2, \cdots, b$. Then $\pi' = a, \cdots, e_1, e_2, \cdots, b$ is also a path from $a$ to $b$. Since $F$ is a separator, there should exist a vertex in path $\pi'$ which is also in $F$. Thus, there exists two vertices of $F$ in path $\pi$, contradicting the assumption that $\pi$ contains exactly one vertex of $F$. Thus, neither such a minimal separator $F$ nor such a vertex $u$ can exist.

A cut $F$ of display graph $G({\ensuremath{\mathcal{P}}})$ is *legal*, if it satisfies the following:

1.  For every tree $T \in {\ensuremath{\mathcal{P}}}$, the edges of $T$ in $F$ are incident on a common vertex.

2.  There is at least one edge in each of the connected components of $G({\ensuremath{\mathcal{P}}})-F$.

We make use of the following simple observation in some of the proofs in this section.

\[obs:path\] Let $I$ be a set of edges of $G({\ensuremath{\mathcal{P}}})$. Then, there exists a path $v_1, v_2, \cdots, v_m$ in $G({\ensuremath{\mathcal{P}}})-I$ where $m \geq 2$, if and only if there exists a path $\{v_1, v_2\}, \cdots, \{v_{m-1}, v_m\}$ in $L({\ensuremath{\mathcal{P}}})-I$.

\[thm:cuts\_seps\] A set $F$ 