[GHC] #14333: GHC doesn't use the fact that Coercible is symmetric

[GHC]

#14333: GHC doesn't use the fact that Coercible is symmetric
-------------------------------------+-------------------------------------
        Reporter:  Iceland_jack      |                Owner:  (none)
            Type:  bug               |               Status:  new
        Priority:  normal            |            Milestone:
       Component:  Compiler          |              Version:  8.2.1
      Resolution:                    |             Keywords:  TypeFamilies
Operating System:  Unknown/Multiple  |         Architecture:
                                     |  Unknown/Multiple
 Type of failure:  None/Unknown      |            Test Case:
      Blocked By:                    |             Blocking:
 Related Tickets:                    |  Differential Rev(s):
       Wiki Page:                    |
-------------------------------------+-------------------------------------

Comment (by goldfire):

 This is a good point. The problem is that `(a b) ~R# (c d)` tells you
 nothing at all about the relationship between any of the variables `a`,
 `b`, `c`, and `d`. This isn't a restriction in the solver, but simply a
 truism of representational equality. So when GHC knows `(a b) ~R# (c d)`,
 it is too stupid to figure out `(c d) ~R# (a b)`. I can thus boil down the
 problem even further:

 {{{#!hs
 bad :: Coercible (a b) (c d) => c d -> a b
 bad = coerce
 }}}

 will fail to type-check. (The version with argument & result swapped works
 fine.)

 Could the solver work around this? Maybe, with sufficient cleverness.
 (Unlike my concerns around recursive newtypes, I think this may be an
 engineering issue, not undecidability.) I don't think it would be easy
 though, requiring some kind of flattening logic akin to what GHC does for
 type families.

 Do you have a concrete use case?

-- 
Ticket URL: <http://ghc.haskell.org/trac/ghc/ticket/14333#comment:2>
GHC <http://www.haskell.org/ghc/>
The Glasgow Haskell Compiler