In foldl definition possible wrong in SML/NJ 110.75, I found that the relation foldl (op -) 2 [1] = foldr (op -) 2 [1] holds. But when I tried the above in Haskell I found that the above relation rewritten in Haskell as foldl (-) 2 [1] == foldr (-) 2 [1] doesn't hold. Why is this? Does Haskell have different definition for fold than SML/NJ?
Thanks