Jason Dusek wrote:
Actually, the third law is:
(m >>= f) >>= g == m >>= (\x -> f x >>= g)
Yeah. I just saw that in the tutorial. And it actually makes more sense that way (esp after reading some of the tutorial): 1) A "computation" is something that may have a result. 2) In "m >>= f", m is a computation and f is a function that takes a value and returns a computation. >>= takes the output of m and makes it the input of f. 3) Therefore, the value of "m >>= f" is a computation. 4) Because "m >>= f" is a computation, it can be the left parameter of another >>=, as in "(m >>= f) >>= g". 5) If "x" is a value then "f x" is a computation. Thus, "f x" can be the left parameter of >>=, as in "f x >>= g", and the result of that is a computation. 6) Therefore, (\x -> f x >>= g) is a function that takes a value ("x") and returns a computation, so it can be the right parameter of >>=. I think I get it (at least this part). Cheers, Daniel.