Natural Number is Not Equal to Successor

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 1

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:\( \)
					\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)
\((\text F)\)	$:$			\(\ds \forall A \subseteq \N_{> 0}:\)	\(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)

Proof by induction:

For all $n \in \N_{>0}$, 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:

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

$\blacksquare$

Proof 2

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.

$\blacksquare$