# Natural Number Addition is Commutative/Proof 3

## Theorem

The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative:

$\forall x, y \in \N_{> 0}: x + y = y + x$

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

Let $x \in \N_{> 0}$ be arbitrary.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$x + n = n + x$

### Basis for the Induction

From Natural Number Commutes with 1 under Addition, we have that:

$\forall x \in \N_{> 0}: x + 1 = 1 + x$

and so $\map P 1$ holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$x + k = k + x$

Then we need to show:

$x + \paren {k + 1} = \paren {k + 1} + x$

### Induction Step

This is our induction step:

 $\ds x + \paren {k + 1}$ $=$ $\ds \paren {x + k} + 1$ Natural Number Addition is Associative $\ds$ $=$ $\ds \paren {k + x} + 1$ Induction Hypothesis $\ds$ $=$ $\ds k + \paren {x + 1}$ Natural Number Addition is Associative $\ds$ $=$ $\ds k + \paren {1 + x}$ Basis for the Induction $\ds$ $=$ $\ds \paren {k + 1} + x$ Natural Number Addition is Associative

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

$\blacksquare$