Let R be a relation on $\displaystyle \bold{N}^2 $ defined by $\displaystyle (x,y)R(u,v) $ if and only if $\displaystyle x \leq u $ or $\displaystyle y \leq v $. Is R a partial order? Prove your answer.

This is what I came up with so far.

So, just to be courteous to the people who read this, a relation on a set is called a partial ordering if the relation is reflexive, antisymmetric, and transitive.

Therefore, I tried to prove that R is reflexive, antisymmetric, and transitive.

Here's what I did.

Claim: R is a partial order.

Proof: For R to be a partial order, it must be reflexive, antisymmetric, and transitive.

In order to show R is reflexive, let $\displaystyle (x,y) \in \bold{N}^2 $. Notice that $\displaystyle x \leq x $. Therefore, $\displaystyle (x,y)R(x,y) $, and hence, R is reflexive.

In order to show R is antisymmetric, let $\displaystyle (x,y) \in \bold{N}^2 $ and $\displaystyle (u,v) \in \bold{N}^2$. Assume $\displaystyle (x,y)R(u,v) $ and $\displaystyle (u,v)R(x,y) $. Therefore, $\displaystyle x \leq u $ or $\displaystyle y \leq v $, and $\displaystyle u \leq x $ or $\displaystyle v \leq y $.

After this point, I am not sure how to continue. I am thinking that I would have to break this into cases, but I'm not sure how to break it into cases because I have two different statements that both involve "or". It's looking like the transitive proof will also have cases, but I'm not sure how to attempt those either. If I can clear up the antisymmetric one, the transitive one can probably follow pretty easily. I will post a continuation after I get some hints on how to go about continuing. Thanks.