Inequality of Natural Numbers is Provable

Theorem
Let $x, y \in \N$ be natural numbers.

Suppose $x \ne y$.

Let $\sqbrk a$ denote the unary representation of $a \in \N$.

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

Lemma
As $x \ne y$, by Ordering on Natural Numbers is Trichotomy, either $x < y$ or $y < x$.

If $y < x$, the result follows from the lemma.

If $x < y$, then:
 * $\sqbrk y \ne \sqbrk x$

is a theorem by the lemma.

Hence, the following formal proof:
 * By Rule of Assumption, suppose $\sqbrk x = \sqbrk y$
 * By Equality is Symmetric, $\sqbrk y = \sqbrk x$
 * By Principle of Non-Contradiction, this contradicts $\sqbrk y \ne \sqbrk x$
 * By Proof by Contradiction, $\sqbrk x \ne \sqbrk y$

demonstrates the result.

The overall result follows from Proof by Cases.