Hello, I have written the below proof as an exercise. I want to explicitly annotate the proof with type variables.....but I cant get a line to work...
{-# LANGUAGE DataKinds #-} {-# LANGUAGE ExplicitForAll #-} {-# LANGUAGE FlexibleContexts #-} {-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE GADTs #-} {-# LANGUAGE MultiParamTypeClasses #-} {-# LANGUAGE PolyKinds #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE TypeFamilies #-} {-# LANGUAGE TypeOperators #-} {-# LANGUAGE UndecidableInstances #-} {-# LANGUAGE ScopedTypeVariables #-}
import Prelude hiding (head, tail, (++), (+), replicate) import qualified Prelude as P
data Nat where Z :: Nat S :: Nat -> Nat
data SNat (a :: Nat) where SZ :: SNat 'Z SS :: SNat a -> SNat ('S a)
here a nice plus
type family (n :: Nat) :+ (m :: Nat) :: Nat type instance Z :+ m = m type instance (S n) :+ m = S (n :+ m)
lets try to do
data val1 :== val2 where Refl :: val :== val
If I prove its abelian then we get it... the "proof" works if we remove the type annotation but I want the annotation to convince myself I know whats going on.
theoremPlusAbelian :: SNat a -> SNat b -> (a :+ b) :== (b :+ a) theoremPlusAbelian (SZ :: SNat a) (SZ :: SNat b) = -- forall 0 and 0 (Refl :: (a :+ b) :== (b :+ a)) theoremPlusAbelian ((SS a) :: SNat a) (SZ :: SNat b) = -- forall a + 1 and 0 case theoremPlusAbelian (a :: (a ~ 'S a1) => SNat a1) (SZ :: SNat b) of (Refl :: (a1 :+ b) :== (b :+ a1)) -> (Refl :: (a :+ b) :== (b :+ a)) theoremPlusAbelian (SZ :: SNat a) ((SS a) :: SNat b) = -- forall 0 and a + 1 case theoremPlusAbelian (SZ :: SNat a) (a :: (b ~ 'S b1) => SNat b1) of (Refl :: (a :+ b1) :== (b1 :+ a)) -> (Refl :: (a :+ b) :== (b :+ a)) -- cant seem to prove this... theoremPlusAbelian ((SS a) :: (SNat a)) ((SS b) :: (SNat b)) = -- forall a+1 and b+1 -- 1st prove (a + 1) + b = b + (a + 1)...from above case theoremPlusAbelian ((SS a) :: (SNat a)) (b :: (b ~ 'S b1) => SNat b1) of
THE COMMENTED LINE FAILS. Could not deduce ((b1 :+ 'S a1) ~ (a2 :+ 'S a1)) from the context (a ~ 'S a1) bound by a pattern with constructor SS :: forall (a :: Nat). SNat a -> SNat ('S a), in an equation for 'theoremPlusAbelian' at Cafe2.lhs:57:24-27 or from (b ~ 'S a2) bound by a pattern with constructor SS :: forall (a :: Nat). SNat a -> SNat ('S a), in an equation for 'theoremPlusAbelian' at Cafe2.lhs:57:45-48 NB: ':+' is a type function, and may not be injective The type variable 'b1' is ambiguous Expected type: (a :+ a2) :== (a2 :+ a) Actual type: (a :+ b1) :== (b1 :+ a) Relevant bindings include b :: SNat a2 (bound at Cafe2.lhs:57:48) a :: SNat a1 (bound at Cafe2.lhs:57:27) In the pattern: Refl :: (a :+ b1) :== (b1 :+ a) In a case alternative: (Refl :: (a :+ b1) :== (b1 :+ a)) -> case theoremPlusAbelian a (SS b) of { Refl -> case theoremPlusAbelian a b of { Refl -> Refl } } In the expression: case theoremPlusAbelian ((SS a) :: SNat a) (b :: b ~ S b1 => SNat b1) of { (Refl :: (a :+ b1) :== (b1 :+ a)) -> case theoremPlusAbelian a (SS b) of { Refl -> case theoremPlusAbelian a b of { Refl -> ... } } }
-- (Refl :: (a :+ b1) :== (b1 :+ a)) -> Refl -> -- now prove a + (b + 1) = (b + 1) + a...from above case theoremPlusAbelian a (SS b) of Refl -> -- now prove a + b = b + a case theoremPlusAbelian a b of -- which seems to have proved a + b = b + a Refl -> Refl
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