mulAssignment.hs

(x, y, z) = (4, 5, 6)
(x:xs) = [1, 2, 3, 4]
(x, y) = (4, 5)

List unpacking (second line) should be (x:xs).
The cons in between x and xs are actually a function:

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

and part of the constructors of the list type. (The other being Nil / the empty list / [] )
So (x:xs) "undoes" the last construction step of the linked list to yield an element of the list and whatever the rest of the list constructed until there is.

Thanks for that! I've never used Haskell myself so this script was directly copied from the original poster. Will correct it!