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 an element in R, real number. We define a relation <= (precedes symbol) on R. Verify that <= (precedes symbol) defined by
x<= (precedes) y if and only if y<= x (less than or equal to)
Verify that <= (precedes symbol) is an order relation.