Proof

Let $x,y,z\in \mathbb Z$ be any integers. We first show the trichotomy of the order relation for integers "$ < `$".

If $x < y$, then it follows from the definition of the order relation for integers that $x-y = [0,a]$ for some natural number $a\neq 0$.
Analogously, if $x > y$, it follows that $x-y = [b,0]$ for some natural number $b\neq 0$.
Finally, if $x=y$, it follows that $x-y = [0,0]$.
Since $a\neq 0\neq b$, we have, therefore the implications
- $x < y\Longrightarrow x \neq y$ and $x \not > 0$.
- $x > y\Longrightarrow x \neq y$ and $x \not < 0$.
- $x = y\Longrightarrow x \not > y$ and $x \not < 0$.

Now, we shot the transitivity.

Let $x < y$ and $y < z$.
Applying the definition of the order relation for integers we get $x-y = [0,a]$ and $y-z=[0,b]$ for some natural numbers $a,b\neq 0$.
Adding both equations results in $x - y + y - z = [0,a+b]$.
By the existence of inverse integers with respect to addition we get $x-z=[0,a+b]$.
Since $a+b\neq 0$ in the set of natural numbers $\mathbb N$, we get the relation $x < z$ in the set of integers $\mathbb Z$.

The transitivity of the relations "$>$", "$\ge$", and "$\le$" follows analogously.

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