I do not consider the following to prove all of the so-called trichotomy of order, since I take that to be a statement that exactly one of $<.=,>$ holds for any given pair of numbers.
The notation $x^\prime$ indicates the successor of $x$, as in Peano's axioms. The expression $a<b$ is defined to be $a<b\iff\exists_c a+c=b.$
The objective is to show that, for the natural numbers beginning with 1 we have
$$ \forall_{x,y}\left[x\ne y\implies x<y\lor x>y\right]. $$
It is easily shown that every number is either 1 or a successor. Furthermore, $\forall_{x}x^{\prime}>1$. We shall also accept $x^\prime +y=x +y^\prime$.
Let $\mathcal{M}$ be the set of numbers $x$ such that, for all numbers $y$ the implication $x\ne y\implies x<y\lor x>y$ holds. That is
$$ \mathcal{M}\equiv\left\{ x\backepsilon\forall_{y}\left[x\ne y\implies x<y\lor x>y\right]\right\} . $$
The principle of induction requires us to show that $1\in\mathcal{M}$ and that for all $x\in\mathcal{M}$ it follows that $x^{\prime}\in\mathcal{M}.$ For the case of $x=1$ we have
$$ x=1\implies x=y=1\lor\exists_{z}y=z^{\prime}>1=x. $$
Since $1\ne y=z^{\prime}$ we have $1\in\mathcal{M}.$
Now we assume $x\in\mathcal{M},$ which means we need to consider the following three cases:
In the first case we have $y$ such that $x<y.$ So
$$ x<y\iff\exists_{c}\left[y=x+c\land\left(c>1\lor c=1\right)\right]. $$
This has two sub-cases. In the first sub-case
$$ 1<c\land x<y=x+c $$
$$ \iff\exists_{d}\left[y=x+c=x+d^{\prime}\iff y=x^{\prime}+d\right] $$
$$ \iff x^{\prime}<y. $$
In the second sub-case
$$ 1=c\land x<y=x+c=x^{\prime}. $$
The second case is $y$ such that $x=y.$ So $x^{\prime}=y+1>y.$
The third case is $y$ such that $x>y.$ So we have
$$ x=y+c\implies x^{\prime}=y+c^{\prime}>y. $$
The induction hypothesis assumes that every $y$ is in one of these cases, so $x^{\prime}\ne y\implies x^{\prime}<y\lor y<x^{\prime}.$
We thereby conclude $x\in\mathcal{M}\implies x^{\prime}\in\mathcal{M}.$
Since I have two up-votes, and no disagreement, I sense that others approve of my proof. It makes sense to me, so I am accepting it as valid.
It is motivated by Thurston's method of proof, and, IMO, is more elegant and satisfying than the proof outlined in BBFSK; discussed here.