Yes, that's right. If (f a ~ g b) and `f` and `g` have syntactically different kinds, then we would hope that those kinds are in fact equal. But, GHC has no way of doing this currently -- though it does store kind equalities, it is an invariant that all such equalities must be simply reflexivity. Hopefully, this will change soon. :)
Richard On Dec 19, 2013, at 11:11 AM, Gábor Lehel <glaebho...@gmail.com> wrote: > Does this boil down to the fact that GHC doesn't have kind GADTs? I.e. > given `(f a ~ g b)` there's no possible way that `a` > and `b`, resp. `f` and `g` might have different kinds (how could you > ever have constructed `f a ~ g b` if they did?), but GHC isn't > equipped to reason about that (to store evidence for it and retrieve > it later)? > > On Thu, Dec 19, 2013 at 4:01 PM, Richard Eisenberg <e...@cis.upenn.edu> wrote: >> Let me revise slightly. GHC wouldn't guess that f3 would be f just because f >> is the only well-kinded thing in sight. >> >> Instead, it would use (f2 i ~ a) to reduce the target equality, (f3 i2 ~ a), >> to (f3 i2 ~ f2 i). It would then try to break this down into (f3 ~ f2) and >> (i2 ~ i). Here is where the kind problem comes in -- these equalities are >> ill-kinded. So, GHC (rightly, in my view) rejects this code, and reports an >> appropriate error message. Of course, more context in the error message >> might be an improvement, but I don't think the current message is wrong. >> >> As for Thijs's comment about lack of decomposition in GHC 7.6.3: You're >> right, that code fails in GHC 7.6.3 because of an attempt (present in GHC >> 7.6.x) to redesign the Core type system to allow for unsaturated type >> families (at least in Core, if not in Haskell). There were a few cases that >> came up that the redesign couldn't handle, like Thijs's. So, the redesign >> was abandoned. In GHC HEAD, Thijs's code works just fine. >> >> (The redesign was to get rid of the "left" and "right" coercions, which >> allow decomposition of things like (f a ~ g b), in favor of an "nth" >> coercion, which allows decomposition of things like (T a ~ T b).) >> >> Good -- I feel much better about this answer, where there's no guess for the >> value of f3! >> >> Richard >> >> On Dec 18, 2013, at 11:30 PM, Richard Eisenberg wrote: >> >>> I'd say GHC has it right in this case. >>> >>> (f a ~ g b) exactly implies (f ~ g) and (a ~ b) if and only if the kinds >>> match up. If, say, (f :: k1 -> *), (g :: k2 -> *), (a :: k1), and (b :: >>> k2), then (f ~ g) and (a ~ b) are ill-kinded. In Gabor's initial problem, >>> we have (with all type, kind, and coercion variables made explicit) >>> >>>> data InnerEq (j :: BOX) (k :: BOX) (i :: j) (a :: k) where >>>> InnerEq :: forall (f :: j -> k). f i ~ a => InnerEq j k i a >>>> >>>> class TypeCompare (k :: BOX) (t :: k -> *) where >>>> maybeInnerEq :: forall (j :: BOX) (f :: j -> k) (i :: j) (a :: k). >>>> t (f i) -> t a -> Maybe (InnerEq j k i a) >>>> >>>> instance forall (j :: BOX) (k :: BOX) (i :: j). TypeCompare k (InnerEq j k >>>> i) where >>>> maybeInnerEq :: forall (j2 :: BOX) (f :: j2 -> k) (i2 :: j2) (a :: k). >>>> InnerEq j k i (f i2) -> InnerEq j k i a -> Maybe (InnerEq >>>> j2 k i2 a) >>>> maybeInnerEq (InnerEq (f1 :: j -> k) (co1 :: f1 i ~ f i2)) >>>> (InnerEq (f2 :: j -> k) (co2 :: f2 i ~ a)) >>>> = Just (InnerEq (f3 :: j2 -> k) (co3 :: f3 i2 ~ a)) >>> >>> GHC must infer `f3` and `co3`. The only thing of kind `j2 -> k` lying >>> around is f. So, we choose f3 := f. Now, we need to prove `f i2 ~ a`. Using >>> the two equalities we have, we can rewrite this as a need >>> to prove `f1 i ~ f2 i`. I can't see a way of doing this. Now, GHC complains >>> that it cannot (renaming to my variables) deduce (i ~ i2) from (f1 i ~ f >>> i2). But, this is exactly the case where the kinds *don't* match up. So, I >>> agree that GHC can't deduce that equality, but I think that, even if it >>> could, it wouldn't be able to type-check the whole term.... unless I've >>> made a mistake somewhere. >>> >>> I don't see an immediate way to fix the problem, but I haven't thought much >>> about it. >>> >>> Does this help? Does anyone see a mistake in what I've done? >>> >>> Richard >>> >>> On Dec 18, 2013, at 6:38 PM, Gábor Lehel <glaebho...@gmail.com> wrote: >>> >>>> Hello, >>>> >>>> The upcoming GHC 7.8 recently gave me this error: >>>> >>>> Could not deduce (i ~ i1) >>>> from the context (f1 i ~ f i1) >>>> >>>> Which is strange to me: shouldn't (f1 i ~ f i1) exactly imply (f1 ~ f, >>>> i ~ i1)? (Or with nicer variable names: (f a ~ g b) => (f ~ g, a ~ >>>> b)?) >>>> >>>> When I inquired about this in #haskell on IRC, a person going by the >>>> name xnyhps had this to say: >>>> >>>>> I've also noticed that, given type equality constraints are never >>>>> decomposed. I'm quite curious why. >>>> >>>> and later: >>>> >>>>> It's especially weird because a given f a ~ g b can not be used to solve >>>>> a wanted f a ~ g b, because the wanted constraint is decomposed before it >>>>> can interact with the given constraint. >>>> >>>> I'm not quite so well versed in the workings of GHC's type checker as >>>> she or he is, but I don't understand why it's this way either. >>>> >>>> Is this a relic of https://ghc.haskell.org/trac/ghc/ticket/5591 and >>>> https://ghc.haskell.org/trac/ghc/ticket/7205? Is there a principled >>>> reason this shouldn't be true? Is it an intentional limitation of the >>>> constraint solver? Or is it just a bug? >>>> >>>> Thanks in advance, >>>> Gábor >>>> >>>> P.S. I got the error on this line: >>>> https://github.com/glaebhoerl/type-eq/blob/master/Type/Eq.hs#L181, >>>> possibly after having added kind annotations to `InnerEq` (which also >>>> gets a less general kind inferred than the one I expect). If it's >>>> important I can try to create a reduced test case. >>>> _______________________________________________ >>>> Glasgow-haskell-users mailing list >>>> Glasgow-haskell-users@haskell.org >>>> http://www.haskell.org/mailman/listinfo/glasgow-haskell-users >>>> >>> >>> _______________________________________________ >>> Glasgow-haskell-users mailing list >>> Glasgow-haskell-users@haskell.org >>> http://www.haskell.org/mailman/listinfo/glasgow-haskell-users >>> >> > _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
