where $\hat{K}(v) \subseteq B(q)$. Since $(S, B)$ is a clique tree of $L({\ensuremath{\mathcal{P}}})_{{\ensuremath{\mathcal{F}}}}$, there will always exist such vertices $p$ and $q$. Also, by Lemma \[lm:two\_cliques\], $p \neq q$, $B(p) \cap \{\hat{K}(v) \setminus e\} = \emptyset$ and $B(q) \cap \{\hat{K}(u) \setminus e\} = \emptyset$

Let $\pi = p, x_1, x_2, \cdots, x_m, q$ be the path from $p$ to $q$ in $S$ where $m \geq 0$. Let $x_0=p$ and $x_{m+1}=q$. Let $x_i$ be the vertex nearest to $p$ in path $\pi$ where $i \in [m+1]$ and $B(x_i) \cap \{\hat{K}(u) \setminus e\} = \emptyset$. Let $F = B(x_{i-1}) \cap B(x_i)$. Then by Theorem \[thm:parallel\_minimal\_ct\], $F \in {\ensuremath{\mathcal{F}}}$. Since $\hat{K}(u) \cap \hat{K}(v) = e$, by the coherence property of the clique tree, $e \in B(x_j)$ for every $j \in [m]$. Thus, $e \in F$. By Lemma \[lm:two\_cliques\], $B(x_{i-1}) \cap \{\hat{K}(v) \setminus e\} = \emptyset$. Since $B(x_i) \cap \{\hat{K}(u) \setminus e\} = \emptyset$, $F \cap \hat{K}(u) = e$ and $F \cap \hat{K}(v) = e$. Thus, for every vertex $e' \in L(T)$ where $e \neq e'$ and $e \cap e' \neq \emptyset$, $e' \notin F$. Also, since every separator in ${\ensuremath{\mathcal{F}}}$ is legal, for every vertex $f \in L(T)$ where $f \cap e = 