RE: Fundeps and type equality

The manual for HEAD is always online here http://www.haskell.org/ghc/dist/current/docs/html/users_guide/type-families.... Simon From: Richard Eisenberg [mailto:eir@cis.upenn.edu] Sent: 11 January 2013 03:03 To: Carter Schonwald Cc: Martin Sulzmann; glasgow-haskell-bugs@haskell.org; Simon Peyton-Jones; GHC Users Mailing List Subject: Re: Fundeps and type equality Yes, I finished and pushed in December. A description of the design and how to use the feature is here: http://hackage.haskell.org/trac/ghc/wiki/NewAxioms There's also a section (7.7.2.2 to be exact) in the manual, but building the manual from source is not for the faint of heart. Richard On Jan 10, 2013, at 3:14 PM, Carter Schonwald mailto:carter.schonwald@gmail.com> wrote: so the overlapping type families are in HEAD? Awesome! I look forward to finding some time to try them out :) On Thu, Jan 10, 2013 at 1:56 PM, Richard Eisenberg mailto:eir@cis.upenn.edu> wrote: For better or worse, the new overlapping type family instances use a different overlapping mechanism than functional dependencies do. Class instances that overlap are chosen among by order of specificity; overlapping instances can be declared in separate modules. Overlapping family instances must be given an explicit order, and those that overlap must all be in the same module. The decision to make these different was to avoid type soundness issues that would arise with overlapping type family instances declared in separate modules. (Ordering a set of family instance equations by specificity, on the other hand, could easily be done within GHC.) So, as yet, we can't fully encode functional dependencies with type families, I don't think. Richard On Jan 2, 2013, at 4:01 PM, Martin Sulzmann mailto:martin.sulzmann.haskell@googlemail.com> wrote: I agree with Iavor that it is fairly straight-forward to extend FC to support FD-style type improvement. In fact, I've formalized such a proof language in a PPDP'06 paper: "Extracting programs from type class proofs" (type improvement comes only at the end) Similar to FC, coercions (proof terms) are used to represent type equations (improvement). Why extend FC? Why not simply use type families to encode FDs (and thus keep FC simple and small). It's been a while, but as far as I remember, the encoding is only problematic in case of the combination of FDs and overlapping instances. Shouldn't this now be fixable given that type family instances can be overlapping? Maybe I'm missing something, guess it's also time to dig out some old notes ... -Martin On Wed, Jan 2, 2013 at 10:04 AM, Simon Peyton-Jones mailto:simonpj@microsoft.com> wrote: As far as I understand, the reason that GHC does not construct such proofs is that it can't express them in its internal proof language (System FC). Iavor is quite right It seems to me that it should be fairly straight-forward to extend FC to support this sort of proof, but I have not been able to convince folks that this is the case. I could elaborate, if there's interest. Iavor: I don't think it's straightforward, but I'm willing to be educated. By all means start a wiki page to explain how, if you think it is. I do agree that injective type families would be a good thing, and would deal with the main reason that fundeps are sometimes better than type families. A good start would be to begin a wiki page to flesh out the design issues, perhaps linked from http://hackage.haskell.org/trac/ghc/wiki/TypeFunctions The main issues are, I think: * How to express functional dependencies like "fixing the result type and the first argument will fix the second argument" * How to express that idea in the proof language Simon From: glasgow-haskell-bugs-bounces@haskell.orgmailto:glasgow-haskell-bugs-bounces@haskell.org [mailto:glasgow-haskell-bugs-bounces@haskell.orgmailto:glasgow-haskell-bugs-bounces@haskell.org] On Behalf Of Iavor Diatchki Sent: 26 December 2012 02:48 To: Conal Elliott Cc: glasgow-haskell-bugs@haskell.orgmailto:glasgow-haskell-bugs@haskell.org; GHC Users Mailing List Subject: Re: Fundeps and type equality Hello Conal, GHC implementation of functional dependencies is incomplete: it will use functional dependencies during type inference (i.e., to determine the values of free type variables), but it will not use them in proofs, which is what is needed in examples like the one you posted. The reason some proving is needed is that the compiler needs to figure out that for each instantiation of the type `ta` and `tb` will be the same (which, of course, follows directly from the FD on the class). As far as I understand, the reason that GHC does not construct such proofs is that it can't express them in its internal proof language (System FC). It seems to me that it should be fairly straight-forward to extend FC to support this sort of proof, but I have not been able to convince folks that this is the case. I could elaborate, if there's interest. In the mean time, the way forward would probably be to express the dependency using type families, which I find tends to be more verbose but, at present, is better supported in GHC. Cheers, -Iavor PS: cc-ing the GHC users' list, as there was some talk of closing the ghc-bugs list, but I am not sure if the transition happened yet. On Tue, Dec 25, 2012 at 6:15 PM, Conal Elliott mailto:conal@conal.net> wrote: I ran into a simple falure with functional dependencies (in GHC 7.4.1):

class Foo a ta | a -> ta

foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool foo = (==)

I expected that the `a -> ta` functional dependency would suffice to prove that `ta ~ tb`, but Pixie/Bug1.hs:9:7: Could not deduce (ta ~ tb) from the context (Foo a ta, Foo a tb, Eq ta) bound by the type signature for foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool at Pixie/Bug1.hs:9:1-10 `ta' is a rigid type variable bound by the type signature for foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool at Pixie/Bug1.hs:9:1 `tb' is a rigid type variable bound by the type signature for foo :: (Foo a ta, Foo a tb, Eq ta) => ta -> tb -> Bool at Pixie/Bug1.hs:9:1 Expected type: ta -> tb -> Bool Actual type: ta -> ta -> Bool In the expression: (==) In an equation for `foo': foo = (==) Failed, modules loaded: none. Any insights about what's going wrong here? -- Conal _______________________________________________ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.orgmailto:Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.orgmailto:Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.orgmailto:Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.orgmailto:Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users