containers: intersections for Set, along with Semigroup newtype

Carter Schonwald carter.schonwald at gmail.com
Mon Dec 21 16:12:27 UTC 2020


why are we equating the lattice operators for True and false with the
lattice operators for set? (for both structures, we have the dual partial
order is also a lattice, so unless we have )
(i'm going to get the names of these equations wrong, but )

the "identity" law is going to be  max `intersect` y == y ,  min `union` y
=== y

the "absorbing" law is going to be   min `intersect` y == min , max `union`
y == max

these rules work the same for (min = emptyset, max == full set, union ==
set union, intersect == set intersecct)
OR for its dual lattice (min == full set, max == emtpy set, union = set
intersection, intersect == set union)

at some level arguing about the empty list case turns into artifacts of
"simple" definitions

that said, i suppose a case could be made that for intersect :: [a] -> a ,
that as the list argument gets larger the result should be getting
*smaller*, so list intersect of lattice elements should be "anti-monotone",
and list union should be monotone (the result gets bigger). I dont usually
see tht

either way, I do strongly feel that either way, arguing by how we choose to
relate the boolean lattice and seet lattices is perhaps the wrong choice...
because both lattices are equivalent to theeir dual lattice

On Mon, Dec 21, 2020 at 5:12 AM Tom Ellis <
tom-lists-haskell-cafe-2017 at jaguarpaw.co.uk> wrote:

> On Sun, Dec 20, 2020 at 11:05:39PM +0100, Ben Franksen wrote:
> > Am 06.12.20 um 19:58 schrieb Sven Panne:
> > > To me it's just the other way around: It violates aesthetics if it
> doesn't
> > > follow the mathematical definition in all cases, which is why I don't
> like
> > > NonEmpty here.
> >
> > I think you've got that wrong.
> >
> >   x `elem` intersections []
> > = {definition}
> >   forall xs in []. x `elem` xs
> > = {vacuous forall}
> >   true
> >
> > Any proposition P(x) is true for all x in []. So the mathematical
> > definition of intersections::[Set a]-> Set a would not be the empty set
> > but the set of all x:a, which in general we have no way to construct.
>
> Yes, and to bring this back to Sven's original claim
>
> | 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?
>
> the correct definition of "intersections []" should be "all elements
> which are in all of no sets" i.e. _every_ value of the given type!
>
> Tom
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