This was wrong: > (l1 F w1) > l1 F w2 > (l1 F w2) >: l2 F w1 > (l2 F w1) > l2 F w2 > (l2 F w2) >: l3 F w1 > (l3 F w1) >: l3 F w2
0: l1 F w1 1: l1 F w2 2: l2 F w1 3: l2 F w2 4: l3 F w1 5: l3 F w2 So the proper set of inequalities is: (l3 F w2) > l1 F w1 (l1 F w1) >: l3 F w1 (l3 F w1) > l1 F w2 (l1 F w2) >: l2 F w1 (l2 F w1) >: l2 F w2 This means: (l3 F w2) > l1 F w2 (l1 F w2) >: l2 F w2 and (l1 F w1) >: l3 F w1 (l3 F w1) > l2 F w1 If F=: * then neither w1 nor w2 can be zero. If w1 is not 0 then l1 is different from l3 and l2 if w2 is not 0 then l2 is different from l3 and l1 Any solution where w1 < 0 has a corresponding solution where all values have been negated, so we can ignore those cases for now. With w1 > 0 and F=:* l3 > l1 l1 > l2 if l3 > l1 then w2 < 0 so l3 < l1 l1 <: l2 This is a contradiction. That said, I have not fully thought through the case where these numbers are complex. I can define < and > in terms of /: for those cases but I have not studied the rules for complex inequalities to know whether the above postulates are all valid for those cases. -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
