Natural Number is Not Equal to Successor/Proof 2
Jump to navigation
Jump to search
Theorem
Let $\N_{> 0}$ be the $1$-based natural numbers:
- $\N_{> 0} = \set {1, 2, 3, \ldots}$
Then:
- $\forall n \in \N_{> 0}: n \ne n + 1$
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:\( \) | ||||||
\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | |||||||||
\((\text F)\) | $:$ | \(\ds \forall A \subseteq \N_{> 0}:\) | \(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |
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.
$\blacksquare$