Hi, I have a question about functor applicate. I know that: pure (+1) <*> Just 2 produce: Just 3 because pure (+1) produce Just (+1) and then Just (+1) <*> Just 2 produce Just (2+1) but in more complex case like: newtype Parser a = P (String -> [(a,String)]) parse :: Parser a -> String -> [(a,String)] parse (P p) inp = p inp item :: Parser Char item = P (\inp -> case inp of [] -> [] (x:xs) -> [(x,xs)]) instance Functor Parser where fmap g p = P (\inp -> case p inp of [] -> [] [(v, out)] -> [(g v, out)]) instance Applicative Parser where pure v = P (\inp -> [(v, inp)]) pg <*> px = P (\inp -> case parse pg inp of [] -> [] [(g, out)] -> parse (fmap g px) out) When I do: parse (pure (\x y -> (x,y)) <*> item <*> item) "abc" The answer is: [(('a','b'),"c")] But I don't understand what exactly happens. First: pure (\x y -> (x,y)) => P (\inp -> [(\x y -> (x,y), inp)]) Then: P (\inp -> [(\x y -> (x,y), inp)]) <*> item => ??? Can someone explain what's happens step by step please. Thank you.