| type family F a b :: Constraint where
| F a a = ()
|
| eq :: F a b => a :~: b
| eq = Refl
This is rejected, and it's not in the least bit puzzling! You have evidence for (F a b) and you need to prove (a~b) -- for any a and b. Obviously you can't.
But how could you possibly have evidence for (F a b) without (a ~ b)? You can't.
And in Core you could write (eq @Int @Bool (error "urk")) and you jolly well don’t want it to return Refl.
> eq = /\ a b. \ (d :: F a b). case d of Something (co :: a ~# b) -> Refl @a @b co
This would obviously require extensions to Core and Haskell, but it's not a priori wrong to do so.