22 Feb
2012
22 Feb
'12
3:36 a.m.
On 2/22/12 2:20 AM, wren ng thornton wrote:
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
Sorry, I misread what you wrote. Yes, of course you're correct. I can't recall off-hand why it is that domain products break things; I'll have to look it up. -- Live well, ~wren