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 (strict) ordering on $\N_{> 0}$ defined as Ordering on Natural Numbers:
 * $\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$

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

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

Proof
Using the following axioms:

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

Then by the definition of ordering on natural numbers:

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.