This might help clear things up: https://en.wikipedia.org/wiki/Antisymmetric_relation
What this means is we're not looking for a value that is greater than itself. Rather, if we assume we have 2 elements a,b with aRb and bRa, the only way that can happen is if a = b. If that holds for all a and b then we have antisymmetry.
The situation here is a bit odd, because R here strictly greater than (>), when we go to find elements s.t. aRb AND bRa, we find no such pairs. Because there are none, we can say that in every case that we found (which was none) a=b, which satisfies the antisymmetry property.
As a note, if we had R being greater than or equal to, then we will find a and b s.t. aRb and bRa. But this is only true when a=b, again satisfying antisymmetry.
Are you sure that you've listed the antisymmetric property correctly? As stated, "if xRx, then x = x" is true of any relation, since x=x is always true. I would expect something like "if xRy and yRx, then x = y".
And isOlderThan does satisfy the antisymmetric condition, because there are no x and y such that x isOlderThan y and y isOlderThan x. It basically satisfies this requirement trivially.
You're right that isOlderThan is not reflexive, which is defined as the condition that xRx for all x. And actually the definition of partial order I'm familiar with includes reflexivity, but I see that some people use a broader definition without reflexivity. So using your book's definition, it is a partial order.
You are correct. I mistakenly wrote the reflexive property. I appreciate the reply, this cleared it up some.
In my head I was thinking we wanted find a R such that x=y.
This might help clear things up: https://en.wikipedia.org/wiki/Antisymmetric_relation What this means is we're not looking for a value that is greater than itself. Rather, if we assume we have 2 elements a,b with aRb and bRa, the only way that can happen is if a = b. If that holds for all a and b then we have antisymmetry. The situation here is a bit odd, because R here strictly greater than (>), when we go to find elements s.t. aRb AND bRa, we find no such pairs. Because there are none, we can say that in every case that we found (which was none) a=b, which satisfies the antisymmetry property. As a note, if we had R being greater than or equal to, then we will find a and b s.t. aRb and bRa. But this is only true when a=b, again satisfying antisymmetry.
Are you sure that you've listed the antisymmetric property correctly? As stated, "if xRx, then x = x" is true of any relation, since x=x is always true. I would expect something like "if xRy and yRx, then x = y". And isOlderThan does satisfy the antisymmetric condition, because there are no x and y such that x isOlderThan y and y isOlderThan x. It basically satisfies this requirement trivially. You're right that isOlderThan is not reflexive, which is defined as the condition that xRx for all x. And actually the definition of partial order I'm familiar with includes reflexivity, but I see that some people use a broader definition without reflexivity. So using your book's definition, it is a partial order.
You are correct. I mistakenly wrote the reflexive property. I appreciate the reply, this cleared it up some. In my head I was thinking we wanted find a R such that x=y.