Explicit inequality evidence

[Richard Eisenberg]

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 <david.feuer at gmail.com> wrote:
> 
> On Tue, Dec 13, 2016 at 12:49 AM, Oleg Grenrus <oleg.grenrus at iki.fi> wrote:
>> First the bike shedding: I’d prefer /~ and :/~:.
> 
> Those are indeed better.
> 
>> 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.
> 
>> 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.
> 
>> 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
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