construction, the elements in $C$ are pairwise-disjoint triples of distinct elements in $A$ and, so, the three sets $A$, $B$, and $C$ form an instance of [Intermezzo]{}.

Now, we show how the pairs and triples in $B \cup C$ can be partitioned into sets $B_\ell \cup C_\ell$ with $B_\ell \subseteq
B$, $C_\ell \subseteq C$, and $1 \leq \ell \leq 4$ such that the elements in $B_\ell\cup C_\ell$ are pairwise disjoint. Recalling that $C$ is a set of pairwise-disjoint triples, we start by setting $B_1 = \emptyset$ and $C_1 = C$. Furthermore, we set $$\begin{aligned}
B_2 &= \{(u_{k,1},\,\bar{u}_{k,3}),\,(\bar{u}_{k,1},\,u_{k,3}) \mid 1 \leq k \leq n\} \cup {}\\
&\hphantom{{} = {}}\{(c_{i,j \oplus 1}^1,\,c_{i,j}^3) \mid 1 \leq i \leq m \wedge 1 \leq j \leq 3\}\end{aligned}$$ and $C_2 = \emptyset$. By construction, it is easy to check that the pairs in $B_2$ are pairwise disjoint. Lastly, consider the remaining pairs $$B \setminus B_2 = \{(c_{i,j,2}, c_{i,j}^1), (c_{i,j}^2, c_{i,j,1}) \mid 1 \leq i \leq m \wedge 1 \leq j \leq 3\}$$ and observe that the only possibility for two pairs in $B \setminus B_2$ to have a non-empty intersection is to have an element $c_{i,j,l}$ with $l\in\{1,2\}$ in common. Now, since each $c_{i,j,l}$ is equal 