View Single Post laminatedevildoll is offline May4-05, 07:00 PM
P: 233 Quote Quote by honestrosewater I can't really follow what
you're saying. You have two order relations here: R and <. And
you've defined R to be <. So you really have nothing to prove-
just check your definition of < - is it an order relation? Did
you want to prove something else? You haven't said on what set
the relations are defined, but assume they're defined on {1, 2,
3}. Using the usual definition of <, "y precedes x if and only
if y less than or equal to x" means "if y precedes x, then y <
x" and "if y < x, then y precedes x". So R = {(1, 1), (2, 2),
(3, 3), (1, 2), (1, 3), (2, 3)}. Is this what you meant to say?
I think you only meant "For all x and y in R, if (y, x) is in R,
then y < x." That also means if y > x, then (y, x) is not in R.
Is that what you wanted? This is actually unrelated to the
previous problem. But, I think that I have to go through the
three properties of an order relation to prove this.  The
problem:  Let x,y is