Successor to Natural Number

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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:

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


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

Then by the definition of ordering on natural numbers:

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

$\blacksquare$


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