$\newcommand\calF{\mathcal{F}} \def\cupdot {\stackrel{\bullet}{\cup}} \def\minusdot {\stackrel{\bullet}{\setminus}}$This post presents a conjecture that we have with some colleagues. It is about reordering inclusion–exclusion to use only the intersections that do not cancel and the operations of disjoint union and subset complement. We are posting it here in case someone has an idea or know combinatorialists that might be interested (we are from TCS).
I write $[n] = \{1,\dotsc,n\}$. Let $\calF = \{S_1,\dotsc,S_n\}$ be a finite family of pairwise incomparable (for inclusion) finite sets. For $T \subseteq [n]$, $T\neq \emptyset$, define $S_T = \bigcap_{j\in T} S_j$. By inclusion–exclusion we have $$\left\vert\bigcup_{i=1}^n S_i\right\vert = \sum_{T\subseteq [n],T\neq \emptyset} (-1)^{|T|+1} |S_T|.$$
It can happen that $S_T = S_{T'}$ for $T \neq T'$, so that some of the intersections disappear from the above sum. Call such an intersection cancelling, the other intersections being non-cancelling.
Example:
Take $\calF = \{S_1,S_2,S_3\}$ with $S_1=\{a,b,d\}$, $S_2=\{a,b,c,e\}$, $S_3=\{a,c,f\}$. Then: $$ \begin{align} |S_1 \cup S_2 \cup S_3| ={} &|S_1| + |S_2| + |S_3| \\\\ & - (|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|) \\\\ & + |S_1 \cap S_2 \cap S_3|. \end{align} $$
Since we have $S_1 \cap S_3 = S_1 \cap S_2 \cap S_3$, we obtain \begin{align*} |S_1 \cup S_2 \cup S_3| &= |\{a,b,d\}| + |\{a,b,c,e\}| + |\{a,c,f\}|- |\{a,b\}| - |\{a,c\}|. \end{align*} The non-cancelling intersections are the ones that remain, i.e., $\{a,b,d\}$, $\{a,b,c,e\}$, $\{a,c,f\}$, $\{a,b\}$, and $\{a,c\}$, while $\{a\}$ ($=S_1 \cap S_3 = S_1 \cap S_2 \cap S_3$) is a cancelling term.
The conjecture is that we can always express the union $\bigcup_{i=1}^n S_i$ (not its size, but the set itself) from the non-cancelling intersections, using only the operations of disjoint union and subset complement. Formally, for two sets $A$, $B$ such that $A\cap B = \emptyset$, define the disjoint union $A\cupdot B = A \cup B$. For two sets $A,B$ such that $B\subseteq A$, define the subset complement $A\minusdot B = A \setminus B$.
Conjecture: For any finite family of pairwise incomparable finite sets $\calF = \{S_1,\dotsc,S_n\}$, we can express $\bigcup_{i=1}^n S_i$ using only the non-cancelling intersections and the operations of disjoint union and subset complement.
Example: Continuing the example, we can express~$S_1 \cup S_2 \cup S_3 = \{a,b,c,d,e,f\}$ with $ \bigl[\bigr((\{a,b,d\} \minusdot \{a,b\}) \cupdot \{a,b,c,e\}\bigr) \minusdot \{a,c\}\bigr] \cupdot \{a,c,f\}$: the reader can easily check that each $\cupdot$ (resp., each $\minusdot$) is a valid disjoint union (resp., subset complement), and that we have only used the non-cancelling intersections. Note that this is not the only valid expression, for instance we can also obtain the union with the expression $[\{a,b,d\} \minusdot \{a,b\}] \cup [\{a,b,c,e\} \minusdot \{a,c\}] \cup \{a,c,f\}$.
What we know: We have tried on millions of examples by a bruteforce approach and this always seems to be true. We have some partial results in a note on arXiv (calling this conjecture the “Non-Cancelling Intersections Conjecture”) concerning a reformulation of this conjecture, but we do not have a general solution so far.
$that ends the "preamble" must appear immediately adjacent to the opening text of the post, or else the post itself will start with spurious blank space. TeX note: In{align}environments, even-numbered columns start with atoms, making $\begin{aligned}a&=b\end{aligned}$\begin{align} a & = b \end{align}space nicely but requiring you to insert the atom yourself in $\begin{aligned}a=&b\end{aligned}$\begin{align} a = & b \end{align}, making it $\begin{aligned}a={}&b\end{aligned}$\begin{align} a ={} & b \end{align}, for proper spacing. I edited both accordingly. $\endgroup$