containers: intersections for Set, along with Semigroup newtype

Eric Mertens emertens at
Sun Dec 6 18:58:24 UTC 2020

First consider how `and` and `or` work for booleans.

and (x ++ y) == and x && and y

For this to work we need `and [] == True`

or (x ++ y) == or x || or y

For this to work we need `or [] == False`

and and or are duals of each other.

There’s an analogue here to union and intersection which are also duals of each other.

We have: unions (x ++ y) == union (unions x) (unions y)

This requires `union [] == []` since any list xs could be split as `[] ++ xs`

We'd like to have: intersections (x ++ y) == intersect (intersections x) (intersections y)

For this kind of splitting property to make sense for intersections we’d need `intersections [] == listOfAllElementsOfThisType`, but it’s not easy to construct that list of all elements in general.
So instead we punt on the problem and refuse to define intersections on an empty list.


> On Dec 6, 2020, at 10:50 AM, Sven Panne <svenpanne at> wrote:
> Am So., 6. Dez. 2020 um 07:20 Uhr schrieb Reed Mullanix <reedmullanix at <mailto:reedmullanix at>>:
> [...]
>   intersections :: Ord a => NonEmpty (Set a) -> Set a
>   intersections (s :| ss) = Foldable.foldl' intersection s ss
> [...]
> Why NonEmpty? I would expect "intersections [] = Set.empty", because the result contains all the elements which are in all sets, i.e. none. That's at least my intuition, is there some law which this would violate? 
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