Unary Representation of Natural Number

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

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

where there are $x$ 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 follows from Equality is Reflexive.

But that is zero applications of the successor mapping to the constant $0$.

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

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

where there are $x$ applications of the successor mapping.

Denote the term as $\phi$.

But by Substitution Property of Equality:
 * $\map s x = \map s \phi$

But $\map s \phi$ consists of $x + 1 = \map s x$ applications of the successor mapping to the constant $0$.

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