Lower Section of Natural Number is Provable

Theorem
Let $x \in \N$ be a natural number.

Then the following WFF:
 * $\forall y: y = 0 \lor y = \map s 0 \lor \dotso \lor y = \sqbrk {x - 1} \lor \neg \paren {y < \sqbrk x}$

is a theorem of minimal arithmetic.

Proof
Proceed by induction on $x$.

Basis for the Induction
Let $x = 0$.

Then the result follows immediately from Axiom $\text M 7$.

Induction Hypothesis
Suppose that there is a formal proof of:
 * $\forall y: y = 0 \lor y = \map s 0 \lor \dotso \lor y = \sqbrk {x - 1} \lor \neg \paren {y < \sqbrk x}$

Induction Step
The following is a formal proof:
 * By Universal Instantiation, let $y_0$ be arbitrary.
 * Suppose $y_0 = 0 \lor y_0 = \map s 0 \lor \dotso \lor y_0 = \sqbrk {x - 1}$.
 * Then, the result follows from Rule of Addition.
 * Otherwise, $\neg {\paren {y_0 < \sqbrk x} }$.
 * By Law of Excluded Middle, $y_0 = \sqbrk x \lor y_0 \ne \sqbrk x$.
 * Suppose $y_0 = \sqbrk x$.
 * Then, the result again follows from Rule of Addition.
 * Finally, $y_0 \ne \sqbrk x$.
 * By Rule of Conjunction and Conjunction of Negations, $\neg \paren {y_0 < \sqbrk x \lor y_0 = \sqbrk x}$.
 * By Axiom $\text M 8$ and Modus Tollendo Tollens: $\neg \paren {y_0 < \map s {\sqbrk x} }$.
 * The result follows from Rule of Addition.

The result in question is:
 * $y_0 = 0 \lor y_0 = \map s 0 \lor \dotso \lor y_0 = \sqbrk {x - 1} \lor y_0 = \sqbrk x \lor \neg \paren {y_0 < \map s {\sqbrk x} }$

But as $\map s {\sqbrk x}$ is the same as $\sqbrk {\map s x}$, the induction step is satisfied by Universal Generalisation.

The result follows from Principle of Mathematical Induction.