Proof that $(\mathbb{N}\times\mathbb{N},\preceq), (a,b)\preceq(c,d)\iff a\leq c\wedge b\geq d$ is a poset I'm having a trouble accepting this anti-symmetric property as proof. It seems faulty to me because it's assuming that the relation is anti-symmetric from the outset and not proving that it is.

Anti-symmetric: 
If $(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$, $(a,b)\preceq(c,d)\wedge(c,d)\preceq(a,b)$, then $a\leq c$, $b\geq d$, $c\leq a$, $d,\geq b$. Therefore $a=c\wedge b=d$.

Is this fine? I would have thought you'd need to show something along the lines of


*

*Anti-symmetric: $\forall(a,b),(c,d)\in\mathbb{N}\times\mathbb{N}$:
\begin{align*}
  (a,b)\preceq(c,d)&\iff a\leq c\wedge b\geq d\\
  &\iff
\end{align*}


but now I'm stuck. So maybe the above method is right, but it still doesn't feel right.

Note that $\preceq$ is not $\leq$. The former is what you have to show to be anti-symmetric whereas the latter is given as anti-symmetric.
So if $(a,b)\preceq(c,d)\preceq(a,b)$ then $a\leq c\leq a$ and $b\geq d\geq b$. Now what can we deduce?
