Decomposition of given equalities
Gábor Lehel
glaebhoerl at gmail.com
Mon Dec 30 18:10:57 UTC 2013
On Mon, Dec 30, 2013 at 7:00 PM, Simon Peyton-Jones
<simonpj at 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-
> | bounces at haskell.org] On Behalf Of Gábor Lehel
> | Sent: 19 December 2013 16:12
> | To: Richard Eisenberg
> | Cc: glasgow-haskell-users at 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 <eir at 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 <glaebhoerl at 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.
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> | >>>
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