Hi > The following is one of the laws. > > (x >>= f) >>= g == x >>= (\v -> f v >>= g) Or stated another way: (x >>= f) >>= g == x >>= (f >>= g) Now it should be easier to see that this is simply associativity. It's easy enough to violate, if you want to - but I don't have any nice simple examples to hand. Thanks Neil