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(Def.) A lattice of sets is a class (="sets of sets") $\bf{L}$ such that $\emptyset\in\bf{L}$ and if $E,F\in\bf{L}$ then $E\cup F,E\cap F\in\bf{L}$.

Assume that $E_i\subset F_i$, $E_i,F_i\in\bf{L}$, $i=1,2$ and $(F_1-E_1)\cap(F_2-E_2)=\emptyset$. Does it follow that $(F_1-E_1)\cup(F_2-E_2)$ can be written in the form $H-G$, where $G\subset H$ and $G,H\in \bf{L}$? I conjecture that not, but I have no counterexample. (I preferred those where every sets have infinite cardinality.)

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Take five disjoint (infinite) sets ($A, B, C, D, E$) and the lattice generated by them (which is isomorphic to Boolean algebra $2^5$. Then take $F_1 = A ∪ B ∪ C$, $E_1 = A$, $F_2 = C ∪ D ∪ E$, $E_2 = C ∪ D$. Then $(F_1 \setminus E_1) ∪ (F_2 \setminus E_2) = (B ∪ C) ∪ E$. The sublattice generated by $E_1, F_1, E_2, F_2$ is a counterexample. You have to add $F_2$ to get $E$ and $F_1$ to get $B ∪ C$. Then you have to remove $E_1$ to get rid of $A$ and $E_2$ to get rid of $D$, but that also discards $C$. There is no way to include $C$ but exclude $D$. Try to draw a picture.

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