Actually, I was wrong there. The paper "Lexically-scoped type variables" from 2004 claims that variables only get bound in value bindings, but GHC has changed its design since. The docs say expression signatures do bind type variables: http://bit.ly/1GZ7VQw On 11/05/15 15:43, Nicholls, Mark wrote:
That worked!....
But doesn’t seem to be true of all the previous cases!
i.e.
The b1 here is bound quite happily
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))
yet
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 (Refl :: (a :+ b1) :== (b1 :+ a)) ->
Complains
There seems little difference?
GHC tells you that b1 is ambiguous, so that should tell you that it isn't getting bound in the case expression as you'd think.
The reason is that type variables only get bound at value bindings; but you attached the signature to an expression.
To fix it, move the annotation to the pattern like this:
theoremPlusAbelian ((SS a) :: (SNat a)) ((SS (b :: SNat b1)) :: (SNat b)) =
On 11/05/15 13:33, Nicholls, Mark wrote:
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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