Apologies for the maybe double post (no subject and hadn't confirmed the mailing list first time I sent this). Hi, Here's a simple expression double ( double [1,2,3,4] ) !! 0 The 1st element of a list is double ( 1*2 : double [2,3,4] ) !! 0 1*2*2 : double ( double [2,3,4 ] ) !! 0 1*2*2 4 calculated without having to traverse the rest of the list due to laziness. But how far does the laziness extend? For example, if I take the 2nd element double (double [1,2,3,4]) !! 1 Do we have 1*2*2 : double ( 2*2 : double [3,4] ) !! 1 1*2*2 : 2*2*2 : double ( double [3,4] ) !! 1 2*2*2 8 Resulting in 8. I assume that the 1st list entry 1*2*2 doesn't actually get calculated, but the expression does get formed? The second element (index 1) gets formed, isolated (by !!) and calculated as the return value. And then things can happily stop. What I'm checking is that the laziness doesn't some how magically extend to not having to form the preceding expressions that make up the resulting list? Or can Haskell recognize that this is like doing function composition i.e., (double . double) [ 1 2, 3, 4 ] !! 1 Thanks for any answer. Cheers, Matt. p.s, I kinda hope things are how I've assumed as the example finally shows me some concrete benefit of function composition which has never seemed important given it's so easily re-written as application.