Let $[[a,b]]_{\boxtimes}$ and $[[c,d]]_{\boxtimes}$ be two integers succeeding $[[0,0]]_{\boxtimes}$, where $[[x,y]]_{\boxtimes}\preceq [[z,w]]_{\boxtimes}\iff x+w\le z+y$.
How to prove that their product, $[[xz+yw,xw+yz]]_{\boxtimes}$ also succeeds $[[0,0]]_{\boxtimes}$?
Note that I'm not assuming ordered ring axioms - I'm trying to prove that the previously defined relation fulfills them.
Notice that $[[0,0]]_{\boxtimes}\preceq [[z,w]]_{\boxtimes}\iff w\leq z$.
Thus, if $[[0,0]]_{\boxtimes}\preceq [[z,w]]_{\boxtimes}$ and $[[0,0]]_{\boxtimes}\preceq [[x,y]]_{\boxtimes}$, then $z-w\geq0$ and $x-y\geq0$. Since $$[[0,0]]_{\boxtimes}\preceq [[xz+yw,xw+yz]]_{\boxtimes}\Leftrightarrow xz+yw\geq xw+yz\Leftrightarrow(z-w)(x-y)\geq0,$$ we can conclude that $[[0,0]]_{\boxtimes}\preceq [[z,w]]_{\boxtimes}$ and $[[0,0]]_{\boxtimes}\preceq [[x,y]]_{\boxtimes}$ implies $[[0,0]]_{\boxtimes}\preceq [[xz+yw,xw+yz]]_{\boxtimes}$
Probably you have noticed that the integer $m$ is intended to be the set $\{(z,w)\in\mathbb{N}^2:z-w=m\}$ (of course this is not a proper definition since it is circular, but will help you for the intuitive part of the calculations).