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?
Ah I see. Good point. Anyway, the conclusion is that decomposing a well-kinded (f a ~ g b) into ill-kinded goals is indeed a bad thing, so it's good that GHC doesn't do it. Simon From: Gábor Lehel [mailto:glaebho...@gmail.com] Sent: 30 December 2013 18:11 To: Simon Peyton-Jones Cc: Richard Eisenberg; glasgow-haskell-users@haskell.org Subject: Re: Decomposition of given equalities On Mon, Dec 30, 2013 at 7:00 PM, Simon Peyton-Jones <simo...@microsoft.com<mailto: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-<mailto:glasgow-haskell-users-> | boun...@haskell.org<mailto:boun...@haskell.org>] On Behalf Of Gábor Lehel | Sent: 19 December 2013 16:12 | To: Richard Eisenberg | Cc: glasgow-haskell-users@haskell.org<mailto: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<mailto: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<mailto: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<mailto:Glasgow-haskell-users@haskell.org> | >>> http://www.haskell.org/mailman/listinfo/glasgow-haskell-users | >>> | >> | >> _______________________________________________ | >> Glasgow-haskell-users mailing list | >> Glasgow-haskell-users@haskell.org<mailto:Glasgow-haskell-users@haskell.org> | >> http://www.haskell.org/mailman/listinfo/glasgow-haskell-users | >> | > | _______________________________________________ | Glasgow-haskell-users mailing list | Glasgow-haskell-users@haskell.org<mailto:Glasgow-haskell-users@haskell.org> | http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
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