Defining foldl in terms of foldr

Question

myFoldl :: (a -> b -> a) -> a -> [b] -> a

myFoldl f z xs = foldr step id xs z
    where step x g a = g (f a x)

I am currently readi

Answer 1

The thing will be become obvious when to expand the expression of foldr step id xs z:

As Adam Smith said in the comments:

foldr step id xs z = (foldr step id xs) z

Consider foldr step id xs firstly

foldr step id xs
= x1 `step` (foldr step id xs1)
= x1 `step` (x2 `step` (foldr step id xs2))
...
= x1 `step` (x2 `step` ... (xn `step` (foldr step id []))...)
= x1 `step` (x2 `step` ... (xn `step` id)...)

where

xs = (x1:xs1)
xs1 = (x2:xs2), xs = (x1:x2:xs2) 
....
xsn = (xn:[]), xs = (x1:x2...xsn) respectively 

Now, apply above function with argument z, i.e.

(x1 `step` (x2 `step` ... (xn `step` id)...)) z

and let

g = (x2 `step` ... (xn `step` id)...) 

gives

(x1 `step` g) z

i.e.

(step x1 g) z

and now apply the where part of foldl:

where step x g a = g (f a x)

gives

(step x1 g) z = step x1 g z = g (step z x1)

where

g (step z x1) = (x2 `step` (x3 step ... (xn `step` id)...) (step z x1)

let

g' = (x3 step ... (xn `step` id)...)

gives

(x2 `step` g') (step z x1)
= step x2 g' (step z x1)
= g' (step (step z x1) x2))
= (x3 step ... (xn `step` id)...) (step (step z x1) x2))

repeats the same steps, finally we have,

(xn `step` id) (step ....(step (step z x1) x2)....)
= step xn id (step ....(step (step z x1) x2)....)
= id (step (step ....(step (step z x1) x2)....) xn)
= (step (step ....(step (step z x1) x2)....) xn)
= foldl step z xs

and now, it is obvious that why use id function. finally, see why

foldl step z xs = (step (step ....(step (step z x1) x2)....) xn)

initial case:

foldl step z' [] = z'

recursive case:

foldl step z (x1:xs1) 
= foldl step (step z x1) xs1
= foldl step (step (step z x1) x2) xs2
...
= foldl step (step (step ....(step (step z x1) x2)....) xn) []
= (step (step ....(step (step z x1) x2)....) xn)

where

z' = (step (step ....(step (step z x1) x2)....) xn) in initial case

Just same as above.