${\ensuremath{\mathcal{P}}}$ where $\Sigma(S) = \Sigma({\ensuremath{\mathcal{F}}})$.

\(i) The statement follows from Lemma \[lm:parallel\_implies\_compatible\] and the splits equivalence theorem.

\(ii) Let $T$ be an input tree of ${\ensuremath{\mathcal{P}}}$ and let $S' = S|{\ensuremath{\mathcal{L}}}(T)$. We will show that $S'$ displays $\Sigma(e)$ for every internal edge $e$ of $T$. Let $\Sigma(e) = A|B$. There exists a cut $F \in {\ensuremath{\mathcal{F}}}$ where $e$ is the only edge of $T$ in $F$. Since $F$ is a minimal cut, the endpoints of $e$ are in different components of $G({\ensuremath{\mathcal{P}}})-F$. Thus, if $\Sigma(F) = A'|B'$ then up to relabeling of sets we have $A \subseteq A'$ and $B \subseteq B'$. Because $S$ displays $\Sigma(F)$, $S'$ also displays $\Sigma(e)$. Since $S'$ displays all the splits of $T$, $T$ can be obtained from $S'$ by contraction of zero or more edges [@SempleSteel03]. Thus, $S$ displays $T$. Since $S$ displays every tree in ${\ensuremath{\mathcal{P}}}$, $S$ is a compatible tree of ${\ensuremath{\mathcal{P}}}$.

\(iii) This is a consequence of the well-known fact (see, e.g., [@SempleSteel03]) that if $X$ is a set of compatible splits, there exists a tree $T$ where $\Sigma(T) = X$.

For the cuts of the display graph in Fig. \[fig:example\] given in Example 1, we have $\Sigma(F_1) =abc | defg$, 