$\{L_1, L_2\}$ is a split of ${\ensuremath{\mathcal{L}}}(T)$. We denote the split corresponding to edge $e$ of $T$ by $\Sigma_e(T)$ and we denote by $\Sigma(T)$ the set of all splits corresponding to all internal edges of $T$.

A tree $T$ *displays* a split $X$ if there exists an internal edge $e$ of $T$ where $\Sigma_e(T) = X$. Then, we also say $T$ is *compatible* with $X$. A set of splits is compatible if there exists a tree which displays all the splits in the set. Two splits $A_1|A_2$ and $B_1|B_2$ are compatible if and only if at least one of $A_1 \cap B_1$, $A_1 \cap B_2$, $A_2 \cap B_1$ and $A_2 \cap B_2$ is empty [@SempleSteel03]. By the Splits Equivalence Theorem [@Buneman71; @SempleSteel03], a collection of splits is compatible if and only if every pair is compatible.

Let $F$ be a legal minimal cut of $G({\ensuremath{\mathcal{P}}})$ and let $G_1$ and $G_2$ be the two connected components of $G({\ensuremath{\mathcal{P}}})-F$. Then, ${\ensuremath{\mathcal{L}}}(G_1)| {\ensuremath{\mathcal{L}}}(G_2)$ is a split of ${\ensuremath{\mathcal{L}}}({\ensuremath{\mathcal{P}}})$.

Consider $G_i$ for any $i \in \{1,2\}$. We will show that ${\ensuremath{\mathcal{L}}}(G_i)$ is non-empty. Since $F$ is a legal minimal cut, $G_i$ contains at least one edge $e$ of $G({\ensuremath{\mathcal{P}}})$. If $e$ is a non 