In particular, (~+) = liftA2 (+), and similarly for (~-). This uses the ((->) e) instance of Applicative.

Or by using '<$>' and '<*>', which is a bit more general: (+) <$> f <*> g funcWith3Args <$> f <*> g <*> h Greetings, Daniel

So function application itself is the functor, i.e., the box-like thing, right? Lifting a function of values to work on boxes containing those values corresponds with lifting a function of values to work on functions yielding those values. Right? In which case, fishing something out of the box corresponds with applying the function. Why not? With record notation you use a function to fish out a field anyway. That's very interesting. Now I have this: type Amount = Float type Moment = Day -- from Data.Time.Calendar data Period = Period Moment Moment type Flowfunc = Period -> Amount f0 = const 0 -- Add and subtract flowfuncs infixl 6 ~+, ~- (~+), (~-) :: Flowfunc -> Flowfunc -> Flowfunc f ~+ g = (+) <$> f <*> g f ~- g = (-) <$> f <*> g But here I still seem to be being naive again... -- Multiply flowfunc by a float infixl 7 ~* (~*) :: Float -> Flowfunc -> Flowfunc f ~* n = \p -> n * (f p) -- or with the params the other way around I guess I could write: f ~* n = (*) <$> f <*> const n but it seems clunky. Should it be a monad? How about: -- Sum a list of them sumf :: [Flowfunc] -> Flowfunc sumf l = foldl (~+) f0 l TIA, Adrian.