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Problem with Apply instance #22
Description
The following code demonstrates the problem:
works :: (Apply Eq1 r, Eq v) => Sum r v -> Sum r v -> Bool
works x y = x == y
-- Could not deduce (Apply Eq1 (Const () : r))
fails :: (Apply Eq1 r, Eq v) => Sum r v -> Sum r v -> Bool
fails x y = weaken @_ @_ @(Const ()) x == weaken y
Of course, if I try to provide the missing instance it complains about overlap. So how do I convince it that if Apply Eq1 r, then clearly Apply Eq1 (Const () : r), since there is both Eq1 for Const and Eq for ()?
Activity
patrickt commentedon Feb 8, 2022
Hm. This very much feels like it should work. I have no idea why it doesn't.
robrix commentedon Feb 24, 2022
The key detail here is that it's complaining that it lacks the
Applyinstance, notEq1orEq. Lacking theApplyinstance (or rather, a constraint bringing it into scope) is the fault of the sad fact thatApplyisn't inductive, so knowing either ofApply Eq1 rorApply Eq1 (Const : r)is insufficient to deduce the other. That's unfortunately precisely whyfastsumexists: by not defining properties inductively, we avoidghc's problems with constraints under these conditions, which grow (IIRC) quadratically.You can add a constraint for the instance you want, which in this example could replace the instance for
r, but it is admittedly a bit gross to expose implementation details like that. I think what I used to do in cases like this was to operate on the un-weakened things wherever possible—weakenlast, in other words—or equivalently to operate on the things to be inserted directly as long as possible before making a union. Of course, if I'm not in need ofweakening, then there's really no problem.Let us know how it goes!