[GHC] #14362: Allow: Coercing (a:~:b) to (b:~:a)

[GHC]

#14362: Allow: Coercing (a:~:b) to (b:~:a)
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        Reporter:  Iceland_jack      |                Owner:  (none)
            Type:  feature request   |               Status:  new
        Priority:  normal            |            Milestone:
       Component:  Compiler          |              Version:  8.2.1
      Resolution:                    |             Keywords:  roles
Operating System:  Unknown/Multiple  |         Architecture:
                                     |  Unknown/Multiple
 Type of failure:  None/Unknown      |            Test Case:
      Blocked By:                    |             Blocking:
 Related Tickets:                    |  Differential Rev(s):
       Wiki Page:                    |
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Comment (by simonpj):

 >> I'm not really sure what this ticket is about.

 It's about the following questions:

 * Is `(a :~: b) ~R# (b :~: a)` sound?
 * And if so, what would be its evidence?

 Remember we are talking only of when a value of type `(a :~: b)` can be
 coerced to one of type `(b :~: a)`, ust as we speak of coercing a value of
 type `[Int]` to one of type `[Age]`.  Curiuosly, the paper doesn't
 actually articulate the circumstances under which such a coercion is OK --
 instead it describes role inference.  My sanity check (again not
 articulated explicitly in the paper) is this: it's sound to coerce a value
 of type `t1` into a value of type `t2`, and vice versa, if

 * I could write code of type `t1 -> t2` and `t2 -> t1`
 * The runtime representations of the two are identical

 Both properties hold for `(a :~: b)` and `(b :~: a)`, regardless of `a`
 and `b`, don't they?  So I claim that `(a :~: b) ~R# (b :~: a)` is sound.

 But `(a :~: a) ~R# (a :~: b)` obviously must ''not'' hold, else I could
 write
 {{{
 good :: forall a. a :~: a
 good = Refl

 bad :: forall a b. a -> b
 bad x = case (coerce (good @ a)) :: a :~: b of
            Refl -> x
 }}}
 (And, returning to the sanity check, I could not write a function of type
 `(a :~: a) -> (a :~: b)`.)

 So how ''could'' we prove `(a :~: b) ~R# (b :~: a)`?
 We have two ways to prove `Coercible`:

 * Decomposition on `(T ts1) ~R# (T ts2)`, using the roles of T.  That
 isn't going to work here because it loses the crucial connection bettween
 `ts1` and `ts2`.

 * Newtype-unwrapping on `(N ts1) ~R# t2`.  And (you are way ahead of me as
 usual), we could do that here if only `:~:` was a newtype.  But, even
 leaving aside that we don't have newtype GADTs (I think we could maybe fix
 that), after decomposing both sides we'd have `(a ~ b) ~R# (b ~ a)`.  And
 now we are back to the original problem: what would be the evidence for
 such an equality?

-- 
Ticket URL: <http://ghc.haskell.org/trac/ghc/ticket/14362#comment:9>
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