Type equality unification shortcoming

The following program demonstrates cases where the equality f a ~ g b is not enough for the type checker to deduce f ~ g and a ~ b.

{-# LANGUAGE GADTs, TypeOperators, KindSignatures, PolyKinds, TypeFamilies, RankNTypes #-}
data a :=: b where Refl :: a :=: a
type family Arg x
type instance Arg (f x) = x
type family Func (x :: *) :: * -> *
type instance Func (f x) = f

congArg :: x :=: y -> Arg x :=: Arg y
congArg Refl = Refl
congFunc :: x :=: y -> Func x :=: Func y
congFunc Refl = Refl
inversion :: forall (f :: * -> *)(g :: * -> *)(a :: *)(b :: *). 
             f a :=: g b -> (f :=: g , a :=: b)
--inversion Refl = (Refl, Refl) -- Doesn't type check!
inversion eq = (congFunc eq , congArg eq) -- Does type check

-- Even simpler example that doesn't type check
-- foo :: f a ~ g b => f () -> g ()
-- foo x = x

Trac metadata

Trac field	Value
Version	7.6.1
Type	Bug
TypeOfFailure	OtherFailure
Priority	normal
Resolution	Unresolved
Component	Compiler
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