[Haskell-cafe] using scoped type variables in proofs...

[Nicholls, Mark]

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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