Re: Explicit inequality evidence

I've thought about inequality on and off for years now, but it's a hard nut to crack. If we want this evidence to affect closed type family reduction, then we would need evidence of inequality in Core, and a brand-spanking-new type safety proof. I don't wish to discourage this inquiry, but I also don't think this battle will be won easily. Richard

On Dec 13, 2016, at 1:02 AM, David Feuer wrote:

On Tue, Dec 13, 2016 at 12:49 AM, Oleg Grenrus wrote:

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First the bike shedding: I’d prefer /~ and :/~:.

Those are indeed better.

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new Typeable [1] would actually provide heterogenous equality:

eqTypeRep' :: forall k1 k2 (a :: k1) (b :: k2). TypeRep a -> TypeRep b -> Maybe (a :~~: b)

And this one is tricky, should it be:

eqTypeRep' :: forall k1 k2 (a :: k1) (b :: k2). TypeRep a -> TypeRep b -> Either (Either (k1 :/~: k2) (a :/~: b)) (a :~~: b)

i.e. how kind inequality would work?

I don't know. It sounds like some details of how kinds are expressed in TypeRep might still be a bit uncertain, but I'm not tuned in. Maybe we should punt and use heterogeneous inequality? That's over my head.

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I'm not sure about propagation rules, with inequality you have to be *very* careful!

irreflexivity, x /~ x and symmetry x /~ y <=> y /~ x are clear.

I assume that in your rules, variables are not type families, otherwise

x /~ y => f x /~ f y doesn't hold if `f` isn't injective. (e.g. type family F x where F x = ()) other direction is true though.

I was definitely imagining them as first-class types; your point that f x /~ f y => x /~ y even if f is a type family is an excellent one.

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Also:

f x ~ a -> b, is true with f ~ (->) a, x ~ b.

Whoops! Yeah, I momentarily forgot that (->) is a constructor. Just leave out that bogus piece.

Thanks, David Feuer _______________________________________________ ghc-devs mailing list ghc-devs@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/ghc-devs