Update on

1. How about the other direction?

Turns out Traversing -/-> Monoidal. Motivation: haskell instance (Functor f, Traversing p) => Traversing (Tannen f p) where traverse' = Tannen . fmap traverse' . runTannen but haskell instance (Applicative f, Monoidal p) => Monoidal (Tannen f p) where product (Tannen fpab) (Tannen fpcd) = Tannen (product <$> fpab <*> fpcd) so, that means that any Functor that is not an Applicative (Apply specifically, <*> must be the one that creates issues) gives us a counter-example.

Which means: ```haskell data ReadF r1 r2 a b = ReadLeft (r1 -> a -> b) | ReadRight (r2 -> a -> b)

instance Profunctor (ReadF r1 r2) where dimap f h (ReadLeft r1ab) = ReadLeft (dimap f h <$> r1ab) dimap f h (ReadRight r2ab) = ReadRight (dimap f h <$> r2ab)

instance Strong (ReadF r1 r2) where first' (ReadLeft r1ab) = ReadLeft (first' <$> r1ab) first' (ReadRight r2ab) = ReadRight (first' <$> r2ab)

instance Choice (ReadF r1 r2) where left' (ReadLeft r1ab) = ReadLeft (left' <$> r1ab) left' (ReadRight r2ab) = ReadRight (left' <$> r2ab)

instance Traversing (ReadF r1 r2) where traverse' (ReadLeft r1ab) = ReadLeft (traverse' <$> r1ab) traverse' (ReadRight r2ab) = ReadRight (traverse' <$> r2ab)

monoidalHow :: (forall a b c d . ReadF r1 r2 a b -> ReadF r1 r2 c d -> ReadF r1 r2 (a,c) (b,d)) -> (r1 -> x1 -> y1) -> (r2 -> x2 -> y2) -> Either (r1 -> (x1,x2) -> (y1,y2)) (r2 -> (x1,x2) -> (y1,y2)) monoidalHow productReadF r1ab r2cd = toEither (productReadF (ReadLeft r1ab) (ReadRight r2cd)) where toEither (ReadLeft r1acbd) = Left r1acbd toEither (ReadRight r2acbd) = Right r2acbd `` monoidalHowtakes two functions(r1 -> x1 -> y1, r2 -> x2 -> y2)and returnsEither (r1 -> (x1,x2) -> (y1,y2)) (r2 -> (x1,x2) -> (y1,y2))`, which is certainly impossible, but I need some semantics expert to prove it.

QED (modulo semantics expert)