Re: #16365: Inconsistency in quantified constraint solving

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#16365: Inconsistency in quantified constraint solving
-------------------------------------+-------------------------------------
        Reporter:  RyanGlScott       |                Owner:  (none)
            Type:  bug               |               Status:  new
        Priority:  normal            |            Milestone:
       Component:  Compiler          |              Version:  8.6.3
      Resolution:                    |             Keywords:
                                     |  QuantifiedConstraints
Operating System:  Unknown/Multiple  |         Architecture:
 Type of failure:  GHC rejects       |  Unknown/Multiple
  valid program                      |            Test Case:
      Blocked By:                    |             Blocking:
 Related Tickets:                    |  Differential Rev(s):
       Wiki Page:                    |
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Comment (by RyanGlScott):

 I'm still not convinced. You say:

 Replying to [comment:3 simonpj]:
 > From superclass expansion we get
 > {{{
 >   [G] forall z. C (F a z)   -- From superclass expansion
 >   [W] C (F a b)
 > }}}
 > And now we are stuck.  What `z` would make `C (F a z)` match `C (F a
 b)`?  Well, yes, `b` would, but maybe also lots of other things.  GHC
 simply doesn't support matching involving type families.

 I don't really see why `F` being a type family has anything to do with
 this. Recall the definition of `F`:

 {{{#!hs
 type family F a :: Type -> Type
 }}}

 Note that `F`'s second argument is //matchable//. This means that given an
 equality between `F`s, we can easily decompose it into their second type
 arguments, as demonstrated by the fact that this function typechecks:

 {{{#!hs
 f :: Proxy a -> F a b :~: F a c -> b :~: c
 f _ Refl = Refl
 }}}

 Now recall what things we are trying to solve:

 {{{
   [G] forall z. C (F a z)   -- From superclass expansion
   [W] C (F a b)
 }}}

 These two constraints only differ in their second argument. But as we just
 established above, `F` is matchable in its second argument. Therefore, GHC
 should have no trouble concluding that `z` equals `b`. In other words, `b`
 is the only sensible choice here.

-- 
Ticket URL: <http://ghc.haskell.org/trac/ghc/ticket/16365#comment:4>