# Pastebin BDYGOBhK
-- Tried to define the same all function this way using recursion but gives error
all''' :: (a -> Bool) -> [a] -> Bool
all''' f [] = True
all''' f (x:xs) = f x && all''' xs
{- this is the error
2-functions-higherorder.hs:10:26: error:
    • Couldn't match expected type ‘Bool’
                  with actual type ‘[a0] -> Bool’
    • Probable cause: ‘all''’ is applied to too few arguments
      In the second argument of ‘(&&)’, namely ‘all''' xs’
      In the expression: f x && all''' xs
      In an equation for ‘all'''’: all''' f (x : xs) = f x && all''' xs
   |
10 | all''' f (x:xs) = f x && all''' xs
   |                          ^^^^^^^^

2-functions-higherorder.hs:10:32: error:
    • Couldn't match expected type ‘a0 -> Bool’ with actual type ‘[a]’
    • In the first argument of ‘all''’, namely ‘xs’
      In the second argument of ‘(&&)’, namely ‘all'' xs’
      In the expression: f x && all'' xs
    • Relevant bindings include
        xs :: [a] (bound at 2-functions-higherorder.hs:10:13)
        x :: a (bound at 2-functions-higherorder.hs:10:11)
        f :: a -> Bool (bound at 2-functions-higherorder.hs:10:8)
        all''' :: (a -> Bool) -> [a] -> Bool
          (bound at 2-functions-higherorder.hs:9:1)
   |
10 | all''' f (x:xs) = f x && all'' xs

-}