10 Oct
2010
10 Oct
'10
7:44 p.m.
On Mon, Oct 11, 2010 at 00:51, Ozgur Akgun
My point was: you need to find/define two operators, not just one. That still holds :)
No it doesn't. f $ g $ h $ x == f (g (h x)) == f . g . h $ x == x $$ h $$ g $$ f if you have the correct associativity for ($$) --Max