Steve Luke wrote:
The two statements are the same, they are just stated in reverse order. The first one is positive, so it is easier to prove, which is why I used it. But the second statement also fails the same way. It says f R(a,b) is true, and a is not equal to b, the R(b,a) must be false. In my scenario, the 'order' relationship is true and the equality is false, so to be antisymmetric, the 'anti_order' relationship must be false. But it is not, so this rule fails, just like the positive rule.
yeah, thanks for correcting mine.
after you pointed out, i rechecked it carefully, i should notice it says "equivalently" which has the very enough implication.
and also look into it, if a binary relation R on a set X is antisymmetric, it's supposed to behave like this. if R(a,b) and R(b,a), which seems not antisymmetric, only resulted from no distinct elements of X, i.e. from the elements of the same value. just like the formal definition suggested:
if there is no pair of distinct elements of X each of which is related by R to the other.
thanks.