RE: Fundeps and type equality
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-boun...@haskell.org [mailto:glasgow-haskell-bugs-boun...@haskell.org] On Behalf Of Iavor Diatchki Sent: 26 December 2012 02:48 To: Conal Elliott Cc: 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:co...@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.org<mailto:Glasgow-haskell-bugs@haskell.org> http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs ___ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
Re: Fundeps and type equality
Hi Iavor, Thanks much for the explanation. Before this experiment with FDs, I was using a type family. I tried switching to FDs, because I wanted the compiler to know that the family is injective in order to assist type-checking. Can we declare type families to be injective? Now I see that I ran into a similar problem almost two years ago: http://haskell.1045720.n5.nabble.com/Injective-type-families-td3385084.html. I guess the answer is still "no", judging by this ticket: http://hackage.haskell.org/trac/ghc/ticket/6018 . -- Conal On Tue, Dec 25, 2012 at 6:47 PM, Iavor Diatchki wrote: > 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 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.org >> http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs >> >> > ___ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
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 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.org > http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs > > ___ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
Fundeps and type equality
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.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
