Unary Representation of Natural Number

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

Then:
 * $x = \map s {\map s {\dots \map s 0} }$

for some term consisting of finitely many applications of the successor mapping to the constant $0$.

Proof
We shall proceed by induction.

Base Case
When $x = 0$, we have:
 * $0 = 0$

which satisfies the theorem, and follows from Equality is Reflexive.

Induction Step
Let $x \in \N$ satisfy the theorem.

Then $x = \map s {\map s {\dots \map s 0} }$.

Denote the term as $\phi$.

By Leibniz's Law:
 * $\map s x = \map s x \iff \map s x = \map s \phi$

But by Equality is Reflexive,
 * $\map s x = \map s x$

is a tautology.

Therefore, by Biconditional Elimination and Modus Ponendo Ponens:
 * $\map s x = \map s \phi$

But $\map s \phi$ satisfies the theorem for $\map s x$.

Thus, by Principle of Mathematical Induction, the theorem holds for every $x \in \N$.