OK, it all makes sense then. Thanks!
On Fri, Dec 20, 2013 at 3:34 PM, Richard Eisenberg <e...@cis.upenn.edu>wrote: > 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 > >>> > >> > > > >
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