If you have a distinction that is made by \\(A\wedge B\\) you know more then that distinction is made by A and that it is made by B, there must be a common partition into two parts, that both A and B refine.

Conversely if you use the atoms, (partitions into exactly two parts) as your element analog, then an atom refined by \\(A \vee B\\) might not be refined by either A or B.

You can see this in the lattice of partitions of the 3 element set. Both {1|2,3} and {1,2|3} distinguish between 1 and 3, but their meet, is the blob: {1,2,3} which makes no distinguishments.

Similarly, the join of {1|2,3} and {1,2|3} is {1|2|3} also refines the atom {1,3|2}.

Which all has to do with transitively closing the equivalence relation that a partition is equivalent too. (by the closure adjunction of normal logic, a function that gives you the membership function of the elements in the same box has type A->(A->Bool) while an equivalence relation has type (A,A) -> Bool, though in practice haskell almost always uses A->B->C instead of (A,B)->C because the syntax ends up being way less noisy.