Decomposition of given equalities
Richard Eisenberg
eir at cis.upenn.edu
Thu Dec 19 04:30:49 UTC 2013
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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