Natural Number is Not Equal to Successor

Theorem
Let $\N_{> 0}$ be the 1-based natural numbers:
 * $\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$

Then:
 * $\forall n \in \N_{> 0}: n \ne n + 1$

Proof
Using the following axioms:

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $n \ne n + 1$

Basis for the Induction
$P \left({1}\right)$ is the case:
 * $1 \ne 1 + 1$

From axiom $E$:
 * $E: \quad \forall a, b \in \N_{> 0}: \text{ exactly 1 of these three holds: } a = b \lor \left({\exists x \in \N_{> 0}: a + x = b}\right) \lor \left({\exists y \in \N_{> 0}: a = b + y}\right)$

Putting $a = b = 1$ this means:


 * $1 = 1$

or
 * $1 = 1 + 1$

As $1 = 1$ it follows that $1 \ne 1 + 1$ and so $P \left({1}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $k \ne k + 1$

Then we need to show:
 * $k + 1 \ne \left({k + 1}\right) + 1$

Induction Step
This is our induction step:

As $k \ne k + 1$ from the induction hypothesis, it follows that:
 * $k + 1 \ne \left({k + 1}\right) + 1$

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{> 0}: n \ne n + 1$