Natural Number is Not Equal to Successor/Proof 2

Proof
Using the following axioms:

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

Hence, taking $a = n$ and $b = n + 1$, we see that since:


 * $a + 1 = b$

it follows that the middle condition of the three holds:


 * $\exists x \in \N_{>0}: a + x = b$

Therefore, since exactly one the three holds it must be that:


 * $n \ne n + 1$

as desired.