On Mon, Dec 30, 2013 at 7:00 PM, Simon Peyton-Jones <simo...@microsoft.com>wrote:
> | 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?) > > Wait. It's quite possible for them to have different kinds. E.g. > > f :: (* -> *) -> * > a :: (* -> *) > g :: * -> * > b :: * > > Then (f a :: *) and (g b :: *), and it'd be quite reasonable to > form the equality (f a ~ g b). > Yes, it's quite possible, given f, a, g, and b of different kinds, to make `f a` and `g b` have the same *kind*. But how could they ever be the same type? Is it not the case that (f a ~ g b) iff (f ~ g) and (a ~ b) (obviously impossible if those are of different kinds)? You would have to worry about this possibility if type constructor variables weren't injective, but they are. > > Simon > > | -----Original Message----- > | From: Glasgow-haskell-users [mailto:glasgow-haskell-users- > | boun...@haskell.org] On Behalf Of Gábor Lehel > | Sent: 19 December 2013 16:12 > | To: Richard Eisenberg > | Cc: glasgow-haskell-users@haskell.org > | Subject: Re: Decomposition of given equalities > | > | 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 >
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