haskell - No se pudo deducir KnownNat en dos existenciales con respecto a la biblioteca singletons

dependent-type existential-type

Question

Estaba experimentando con la biblioteca de singletons y encontré un caso que no entiendo.

{-# LANGUAGE GADTs, StandaloneDeriving, RankNTypes, ScopedTypeVariables,
FlexibleInstances, KindSignatures, DataKinds, StandaloneDeriving #-}
import Data.Singletons.Prelude
import Data.Singletons.TypeLits
data Foo (a :: Nat) where
Foo :: Foo a
deriving Show
data Thing where
Thing :: KnownNat a => Foo a -> Thing
deriving instance Show Thing
afoo1 :: Foo 1
afoo1 = Foo
afoo2 :: Foo 2
afoo2 = Foo
athing :: Thing
athing = Thing afoo1
foolen :: forall n. KnownNat n => Foo n -> Integer
foolen foo =
case sing of (SNat :: Sing n) -> natVal (Proxy :: Proxy n)
minfoo :: forall a b c. (Min a b ~ c, KnownNat c) => Foo a -> Foo b -> Integer
minfoo _ _ =
let c = case sing of (SNat :: Sing c) -> natVal (Proxy :: Proxy c)
in natVal (Proxy :: Proxy c)
thinglen :: Thing -> Integer
thinglen (Thing foo) = foolen foo 

Podría usar esto para obtener el mínimo de dos cosas

minthing :: Thing -> Thing -> Integer
minthing (Thing foo1) (Thing foo2) = min (foolen foo1) (foolen foo2)

Pero ¿por qué no puedo hacer esto?

minthing'' :: Thing -> Thing -> Integer
minthing'' (Thing foo1) (Thing foo2) = minfoo foo1 foo2
• Could not deduce (KnownNat
(Data.Singletons.Prelude.Ord.Case_1627967386
a
a1
(Data.Singletons.Prelude.Ord.Case_1627967254
a a1 (GHC.TypeLits.CmpNat a a1))))

Answer 1

Necesitas hacer un poco de demostración de teorema para comprobar que, dado KnownNat a y KnownNat b , puedes obtener KnownNat (Min ab) . Una posible solución:

import Data.Constraint

(...)

theorem :: forall a b. (KnownNat a, KnownNat b) =>
Sing a -> Sing b -> Dict (KnownNat (Min a b))
theorem sa sb = case sCompare sa sb of
SLT -> Dict
SEQ -> Dict
SGT -> Dict
fooSing :: forall a. KnownNat a => Foo a -> Sing a
fooSing _ = sing
minthing'' :: Thing -> Thing -> Integer
minthing'' (Thing foo1) (Thing foo2) =
case theorem (fooSing foo1) (fooSing foo2) of
Dict -> minfoo foo1 foo2

Answer 2

Siento que el comentario de user3237465 debe ser inmortalizado porque elimina la dependencia de la biblioteca de contraint y es bastante prolijo.

minthing'' :: Thing -> Thing -> Integer
minthing'' (Thing foo1) (Thing foo2) =
theorem (fooSing foo1) (fooSing foo2) $ minfoo foo1 foo2
where
fooSing :: KnownNat a => Foo a -> Sing a
fooSing = const sing
theorem :: forall a b c. (KnownNat a, KnownNat b) =>
Sing a -> Sing b -> (KnownNat (Min a b) => c) -> c
theorem sa sb c = case sCompare sa sb of
SLT -> c
SEQ -> c
SGT -> c