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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.

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    You should explain your notation, because the notation used here is non-standard. Presumably, you mean where $a,b,c,d$ are natural numbers? – 2017-01-01
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    Yes, I'm using the formal construction of integers here. $(a,b){\boxtimes}(c,d){\iff}a+d=b+c$. – 2017-01-01
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    @asdasdfsss I recommend you to read "Landau, Edmund. Foundations of Analysis: The Arithmetic of Whole, Rational, Irrational and Complex Numbers. A Suppl. to Text-books on the Differential and Integral Calculus. Transl. by F. Steinhardt. Chelsea Publ. Comp., 1966". This textbook shows the construction of integers, rationals, reals and complex numbers (and their order and algebraic properties) just starting with the natural numbers. – 2017-01-01

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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).

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    It's dangerous to use the minus symbol, $-$, when talking about natural numbers. In natural numbers, $w\leq z$ means there is a natural number $n$ such that $w+n=z$. While that seems like the same thing as $z-w\geq 0$, the problem is that $z-w$ is often left undefined in the natural numbers. But yes, this is the heart of the proof. – 2017-01-01
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    If we define $(z-w)$ and $(x-y)$ as natural numbers such that $(z-w)+w=z$ and $(x-y)+y=x$, without even considering $-$ as an operation, how to prove that $(z-w)(x-y)=((zx-wx)-yz)+wy$? – 2017-01-01
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    It's not much harder than proving that $\boxtimes$ is well-defined, @asdasdfsss – 2017-01-01
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    @asdasdfsss you can use the fact that for $w,x,y,z\in\mathbb{N}$ such that $z\geq w$, we have that $y(z−w)=yz−yw$. Then your equality follows from $((z−w)+w)((x−y)+y)=zx$. – 2017-01-01