Evaluation strategy of haskell

Question

If I have these functions:

add :: Int -> Int -> Int
add a b = a+b


map :: (a->b) -> [a] -> [b]
map f [] = []
map f (x:xs) = f x : map f xs

When I call map (add 1) [1,2,3] the output is [2,3,4] but how does the evaluation of Haskell work in this case?

My ideas:

It matches with the second pattern because map is a function and [1,2,3] a list.

Then we would have:

map (add 1) [1,2,3]
= add 1 : map add [2,3]
= add 1 : add 2 : map add [3]
= add 1 : add 2 : add 3 : map add []
= add 1 : add 2 : add 3 : []

However this cannot be true. Can someone help me please?

Answer 1

add 1 is the unary function f, not add. So you end up with the following.

map (add 1) [1,2,3]
    == add 1 1 : map (add 1) [2,3]
    == add 1 1 : add 1 2 : map (add 1) [3]
    == add 1 1 : add 1 2 : add 1 3 : map (add 1) []
    == 2 : 3 : 4 : []
    == [2,3,4]