[Haskell-cafe] Monad laws in presence of bottoms

[Miguel Mitrofanov]

22.02.2012, 11:20, "wren ng thornton" <wren at freegeek.org>:
> On 2/22/12 1:45 AM, Miguel Mitrofanov wrote:
>
>>>  However, there is no free ordering on:
>>>
>>>        { (a0,b) | b<- B } \cup { (a,b0) | a<- A }
>>  What? By definition, since, a0<= a and b0<= b, we have (a0, b0)<= (a0, b) and (a0, b0)<= (a0, b0), so, (a0, b0) is clearly the bottom of A\times B.
>
> Sorry, the ordering relation on domain products is defined by:
>
>      (a1,b1) <=_(A,B) (a2,b2) if and only if a1 <=_A a2 and b1 <=_B b2

So, you're agreeing with me? a0 <=_A a, b0 <=_B b0, so (a0, b0) <=_(A,B) (a, b0). Right?