termination for FDs and ATs

[Manuel M T Chakravarty]

Stefan Wehr:
> Manuel M T Chakravarty <chak at cse.unsw.edu.au> wrote::
> 
> > Martin Sulzmann:
> >> Manuel M T Chakravarty writes:
> >>  > Martin Sulzmann:
> >>  > > A problem with ATs at the moment is that some terminating FD programs
> >>  > > result into non-terminating AT programs.
> >>  > > 
> >>  > > Somebody asked how to write the MonadReader class with ATs:
> >>  > > http://www.haskell.org//pipermail/haskell-cafe/2006-February/014489.html
> >>  > >
> >>  > > This requires an AT extension which may lead to undecidable type
> >>  > > inference:
> >>  > > http://www.haskell.org//pipermail/haskell-cafe/2006-February/014609.html
> >>  > 
> >>  > The message that you are citing here has two problems:
> >>  > 
> >>  >      1. You are using non-standard instances with contexts containing
> >>  >         non-variable predicates.  (I am not disputing the potential
> >>  >         merit of these, but we don't know whether they apply to Haskell'
> >>  >         at this point.)
> >>  >      2. You seem to use the super class implication the wrong way around
> >>  >         (ie, as if it were an instance implication).  See Rule (cls) of
> >>  >         Fig 3 of the "Associated Type Synonyms" paper.
> >>  > 
> >> 
> >> I'm not arguing that the conditions in the published AT papers result
> >> in programs for which inference is non-terminating.
> >> 
> >> We're discussing here a possible AT extension for which inference
> >> is clearly non-terminating (unless we cut off type inference after n
> >> number of steps). Without these extensions you can't adequately
> >> encode the MonadReader class with ATs.
> >
> > This addresses the first point.  You didn't address the second.  let me
> > re-formuate: I think, you got the derivation wrong.  You use the
> > superclass implication the wrong way around.  (Or do I misunderstand?)
> 
> I think the direction of the superclass rule is indeed wrong. But what about
> the following example:
> 
>     class C a
>     class F a where type T a
>     instance F [a] where type T [a] = a
>     class (C (T a), F a) => D a where m :: a -> Int
>     instance C a => D [a] where m _ = 42
> 
> If you now try to derive "D [Int]", you get
> 
>              ||- D [Int]
>     subgoal: ||- C Int        -- via Instance
>     subgoal: ||- C (T [Int])  -- via Def. of T in F
>     subgoal: ||- D [Int]      -- Superclass

You are using `T [a] = a' *backwards*, but the algorithm doesn't do
that.  Or am I missing something?

Manuel