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 1

Consider the natural numbers defined as a naturally ordered semigroup.

By definition, the operation in a semigroup is associative.

Hence the result.

$\blacksquare$

Proof 2

Consider the von Neumann construction of natural numbers $\N$, as elements of the minimal infinite successor 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 Minimal Infinite Successor Set $\ds$ $=$ $\ds \paren {x + \paren {y + k} }^+$ induction Hypothesis $\ds$ $=$ $\ds x + \paren {\paren {y + k}^+}$ Definition of Addition in Minimal Infinite Successor Set $\ds$ $=$ $\ds x + \paren {y + k^+}$ Definition of Addition in Minimal Infinite Successor Set

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

$\blacksquare$

Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

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

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

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

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

Basis for the Induction

From Axiom $\text C$, we have by definition that:

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

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:

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

Then we need to show:

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

Induction Step

This is our induction step:

 $\ds \paren {x + y} + \paren {k + 1}$ $=$ $\ds \paren {\paren {x + y} + k} + 1$ Basis for the Induction $\ds$ $=$ $\ds \paren {x + \paren {y + k} } + 1$ Induction Hypothesis $\ds$ $=$ $\ds x + \paren {\paren {y + k} + 1}$ Basis for the Induction $\ds$ $=$ $\ds x + \paren {y + \paren {k + 1} }$ Basis for the Induction

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

$\blacksquare$