Ordering of Natural Numbers is Provable

Theorem
Let $x, y \in \N$.

Suppose $x < y$.

Then $\sqbrk x < \sqbrk y$ is a theorem of minimal arithmetic.

Proof
Fix $x$, and let $z > 0$ be arbitrary.

Proceed by induction on $z$.

Basis for the Induction
Suppose $z = 1$.

The following is a formal proof:
 * By Equality is Reflexive, $\sqbrk x = \sqbrk x$.
 * By Rule of Addition, $\sqbrk x < \sqbrk x \lor \sqbrk x = \sqbrk x$.
 * By Axiom $\text M 8$, $\sqbrk x < \map s {\sqbrk x} \iff \paren {\sqbrk x < \sqbrk x \lor \sqbrk x = \sqbrk x}$
 * By Biconditional Elimination and Modus Ponendo Ponens, $\sqbrk x < \map s {\sqbrk x}$

But:
 * $\map s x = x + 1 = x + z$

Therefore:
 * $\sqbrk {x + z} = \sqbrk {\map s x} = \map s {\sqbrk x}$

and the above proves:
 * $\sqbrk x < \sqbrk {x + z}$

Induction Hypothesis
Suppose that there is a formal proof of:
 * $\sqbrk x < \sqbrk {x + z}$

Induction Step
The following is a formal proof:
 * By the induction hypothesis, $\sqbrk x < \sqbrk {x + z}$
 * By Rule of Addition, $\sqbrk x < \sqbrk {x + z} \lor \sqbrk x = \sqbrk {x + z}$
 * By Axiom $\text M 8$, $\sqbrk x < \map s {\sqbrk {x + z} } \iff \sqbrk x < \sqbrk {x + z} \lor \sqbrk x = \sqbrk {x + z}$
 * By Biconditional Elimination and Modus Ponendo Ponens, $\sqbrk x < \map s {\sqbrk {x + z} }$

But $\map s {\sqbrk {x + z} } = \sqbrk {\map s {x + z} } = \sqbrk {x + \map s z}$.

Therefore, the above proves:
 * $\sqbrk x < \sqbrk {x + \map s z}$

which satisfies the induction step.

Thus, by the Principle of Mathematical Induction, there is such a proof for every $z > 0$.

But by definition of ordering:
 * $x < y \iff \exists z > 0: x + z = y$.

But then the formal proof above is also a proof for $\sqbrk x < \sqbrk y$.