【Haskell】アプリカティブむずかしい

sequenceA [(+3), (*4)]を計算してみようと思います。

sequenceA [(+3), (*4)]
= (:) <$> (+3) <*> sequenceA [(*4)]
= (:) <$> (+3) <*> ((:) <$> (*4) <*> pure [])
= (:) <$> (+3) <*> ((:) <$> (*4) <*> (\_ -> []))
= (:) <$> (+3) <*> ((\_ -> (:)) <*> (*4) <*> (\_ -> []))
= (:) <$> (+3) <*> ((\x -> (:) ((*4) x)) <*> (\_ -> []))
= (:) <$> (+3) <*> (\x -> (:) ((*4) x) [])
= (\_ -> (:)) <*> (+3) <*> (\x -> (:) ((*4) x) [])
= \x -> (:) ((+3) x) (:) ((*4) x) []
= \x -> [x+3, x*4]