# Successor to Natural Number

## Theorem

Let $\N_{> 0}$ be the 1-based natural numbers:

$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$

Let $<$ be the ordering on $\N_{> 0}$:

$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$

Let $a \in \N_{>0}$.

Then there exists no natural number $n$ such that $a < n < a + 1$.

## Proof

Using the following axioms:

 $(\text A)$ $:$ $\ds \exists_1 1 \in \N_{> 0}:$ $\ds a \times 1 = a = 1 \times a$ $(\text B)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds a \times \paren {b + 1} = \paren {a \times b} + a$ $(\text C)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds a + \paren {b + 1} = \paren {a + b} + 1$ $(\text D)$ $:$ $\ds \forall a \in \N_{> 0}, a \ne 1:$ $\ds \exists_1 b \in \N_{> 0}: a = b + 1$ $(\text E)$ $:$ $\ds \forall a, b \in \N_{> 0}:$ $\ds$Exactly one of these three holds: $\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y}$ $(\text F)$ $:$ $\ds \forall A \subseteq \N_{> 0}:$ $\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0}$

Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.

Then by the definition of ordering on natural numbers:

 $\ds a + x$ $=$ $\ds n$ Definition of Ordering on Natural Numbers: $a < n$ $\ds n + y$ $=$ $\ds a + 1$ Definition of Ordering on Natural Numbers: $a < n$ $\ds \implies \ \$ $\ds \left({a + x}\right) + y$ $=$ $\ds a + 1$ substitution for $n$ $\ds \implies \ \$ $\ds a + \left({x + y}\right)$ $=$ $\ds a + 1$ Natural Number Addition is Associative $\ds \implies \ \$ $\ds x + y$ $=$ $\ds 1$ Addition on $1$-Based Natural Numbers is Cancellable

By Axiom $D$, either:

$y = 1$

or:

$y = t + 1$ for some $t \in \N_{>0}$

Then either:

$x + 1 = 1$ when $y = 1$

or:

$x + \left({t + 1}\right) = \left({x + t}\right) + 1 = 1$ when $y = t + 1$

Both of these conclusions violate Natural Number is Not Equal to Successor.

Hence the result.

$\blacksquare$