if $\{e, e'\} \subseteq E(T)$ for some input tree $T$ then $e \cap e' \neq \emptyset$. Thus, both $e$ and $e'$ are not from $L(T)$ for any input tree $T$. Hence, every fill-in edge in $L({\ensuremath{\mathcal{P}}})_{{\ensuremath{\mathcal{F}}}}$ is valid, and $L({\ensuremath{\mathcal{P}}})_{{\ensuremath{\mathcal{F}}}}$ is a restricted triangulation.

For any vertex $u$ of an input tree, $\hat{K}(u)$ represents the set of all vertices of $L({\ensuremath{\mathcal{P}}})$ where for every $e \in \hat{K}(u)$, $u \in e$.

\[lm:no\_loner\] Let $F$ be any minimal separator of $L({\ensuremath{\mathcal{P}}})$ and $u$ be any vertex of any input tree. Then, $\hat{K}(u) \not \subseteq F$.

Suppose $F$ is a minimal $a$-$b$ separator of $L({\ensuremath{\mathcal{P}}})$ and $u$ is a vertex of some input tree such that $\hat{K}(u) \subseteq F$. Consider any vertex $e \in \hat{K}(u)$. Then, there exists a path $\pi$ from $a$ to $b$ in $L({\ensuremath{\mathcal{P}}})$ where $e$ is the only vertex of $F$ in $\pi$. If such a path $\pi$ did not exist, then $F-e$ would still be a $a$-$b$ separator, and $F$ would not be minimal, a contradiction. Let $e_1$ and $e_2$ be the neighbors of $e$ in $\pi$ and let $e = \{u, v\}$. Since $\hat{K}(u) \subseteq F$, $\pi$ does not contain any other vertex $e'$ where $u \in 