Re: [Haskell-cafe] lazy evaluation is not complete

On 10 Feb 2009, at 07:57, Max Rabkin wrote:

On Mon, Feb 9, 2009 at 10:50 PM, Iavor Diatchki wrote:

...

I 0 * _ = I 0 I x * I y = I (x * y)

Note that (*) is now non-commutative (w.r.t. _|_). Of course, that's what we need here, but it means that the "obviously correct" transformation of

just to improve slightly: I 0 |* _ = I 0 I x |* I y = I (x * y) _ *| I 0 = I 0 I x *| I y = I (x * y) I x * | y = (I x |* I y) `unamb` (I x *| I y) Now it is commutative :) Bob