Prove $(n,m)R(r,s) \equiv (n>r) \text{ or } (n=r \text{ and } m\geq s)$ is an order relation.

Question

Prove $(n,m)R(r,s) \equiv (n>r)\text{ or } (n=r\text{ and } m\geq s)$ is an order relation.

So I have to prove reflexivity, antysimmetry and transitivity.

I could prove reflexivity but I'm having lots of trouble proving antysimmetry, I tried proving it by the counter-reciprocal but I got a mess of logic operators.

Any tip is welcome.

Answer 1

Suppose $(n,m)\mathrel{R}(r,s)$ and $(r,s)\mathrel{R}(n,m)$

Can we have $n>r$? No: this would contradict $(r,s)\mathrel{R}(n,m)$, because both $r>n $ and $r=n$ are false. Therefore $n=r$. Now we know $m\ge s$ and $s\ge m$, so also $m=s$.

For transitivity it's similar.

Suppose $(n,m)\mathrel{R}(r,s)$ and $(r,s)\mathrel{R}(x,y)$.

If $n>r$, then it is obvious that $n>x$, because either $r>x$ or $r=x$.

Suppose $n=r$. Then we have two cases.

If $r>x$, then $n>x$.

If $r=x$, we have $m\ge s$ and $s\ge y$. Then $r=x$ and $m\ge y$.

In each case we have $(n,m)\mathrel{R}(x,y)$ as we wished to prove.

Answer 2

The "mess of logical operators" is more or less inevitable, since you're trying to prove an if-then statement containing phrases each of which has an "or" in it.

You're trying to prove: if $(n,m)R(r,s)$ and $(r,s)R(n,m)$, then $(n,m)=(r,s)$ (which is to say, $n=r$ and $m=s$). Try proving this directly by splitting into two cases: $n\ne r$ and $n=r$. In one case, the implication you're trying to prove will be vacuously true; in the other, it will depend upon $m$ and $s$.