# Natural Number Addition is Associative/Proof 2

## Theorem

The operation of addition on the set of natural numbers $\N$ is associative:

$\forall x, y, z \in \N: x + \paren {y + z} = \paren {x + y} + z$

## Proof

Consider the von Neumann construction of natural numbers $\N$, as elements of the minimally inductive set $\omega$.

We are to show that:

$\paren {x + y} + n = x + \paren {y + n}$

for all $x, y, n \in \N$.

From the definition of addition, we have that:

 $\ds \forall m, n \in \N: \,$ $\ds m + 0$ $=$ $\ds m$ $\ds m + n^+$ $=$ $\ds \paren {m + n}^+$

Let $x, y \in \N$ be arbitrary.

For all $n \in \N$, let $\map P n$ be the proposition:

$\paren {x + y} + n = x + \paren {y + n}$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds \paren {x + y} + 0$ $=$ $\ds x + y$ $\ds$ $=$ $\ds x + \paren {y + 0}$

and so $\map P 0$ holds.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

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

Then we need to show:

$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$

### Induction Step

This is our induction step:

 $\ds \paren {x + y} + k^+$ $=$ $\ds \paren {\paren {x + y} + k}^+$ Definition of Addition in Minimally Inductive Set $\ds$ $=$ $\ds \paren {x + \paren {y + k} }^+$ induction Hypothesis $\ds$ $=$ $\ds x + \paren {\paren {y + k}^+}$ Definition of Addition in Minimally Inductive Set $\ds$ $=$ $\ds x + \paren {y + k^+}$ Definition of Addition in Minimally Inductive Set

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

$\blacksquare$