# Natural Number Addition Commutativity with Successor

## Contents

## Theorem

Let $\N$ be the natural numbers.

Then:

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

## Proof 1

Proof by induction:

From the definition of addition, we have that:

\(\displaystyle \forall m, n \in \N: \ \ \) | \(\displaystyle m + 0\) | \(=\) | \(\displaystyle m\) | ||||||||||

\(\displaystyle m + n^+\) | \(=\) | \(\displaystyle \left({m + n}\right)^+\) |

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

- $\forall m \in \N: m^+ + n = \left({m + n}\right)^+$

### Basis for the Induction

By definition, we have:

\(\displaystyle \forall m \in \N: \ \ \) | \(\displaystyle m^+ + 0\) | \(=\) | \(\displaystyle m^+\) | by definition | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + 0}\right)^+\) | by definition: $m = m + 0$ |

Thus $P \left({0}\right)$ is seen to be true.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.

So this is our induction hypothesis $P \left({k}\right)$:

- $\forall m \in \N: m^+ + k = \left({m + k}\right)^+$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:

- $\forall m \in \N: m^+ + k^+ = \left({m + k^+}\right)^+$

### Induction Step

This is our induction step:

\(\displaystyle m^+ + k^+\) | \(=\) | \(\displaystyle \left({m^+ + k}\right)^+\) | by definition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({m + k}\right)^+}\right)^+\) | from the induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + k^+}\right)^+\) | by definition |

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

Therefore:

- $\forall m, n \in \N: m^+ + n = \left({m + n}\right)^+$

$\blacksquare$

## Proof 2

Using the following axioms:

\((A)\) | $:$ | \(\displaystyle \exists_1 1 \in \N_{> 0}:\) | \(\displaystyle a \times 1 = a = 1 \times a \) | |||||

\((B)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a \) | |||||

\((C)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle a + \paren {b + 1} = \paren {a + b} + 1 \) | |||||

\((D)\) | $:$ | \(\displaystyle \forall a \in \N_{> 0}, a \ne 1:\) | \(\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1 \) | |||||

\((E)\) | $:$ | \(\displaystyle \forall a, b \in \N_{> 0}:\) | \(\displaystyle \)Exactly one of these three holds:\( \) | |||||

\(\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \) | ||||||||

\((F)\) | $:$ | \(\displaystyle \forall A \subseteq \N_{> 0}:\) | \(\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \) |

Proof by induction:

From Axiom:Axiomatization of 1-Based Natural Numbers, we have by definition that:

\(\displaystyle \forall m, n \in \N: \ \ \) | \(\displaystyle m + 0\) | \(=\) | \(\displaystyle m\) | ||||||||||

\(\displaystyle \left({m + n}\right)^+\) | \(=\) | \(\displaystyle m + n^+\) |

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

- $\forall m \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$

### Basis for the Induction

When $n = 1$, we have:

- $\left({m + 1}\right) + 1 = \left({m + 1}\right) + 1$

which holds trivially.

Thus $P \left({1}\right)$ is seen to be true.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis $P \left({k}\right)$:

- $\forall m \in \N: \left({m + 1}\right) + k = \left({m + k}\right) + 1$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:

- $\forall m \in \N: \left({m + 1}\right) + \left({k + 1}\right) = \left({m + \left({k + 1}\right)}\right) + 1$

### Induction Step

This is our induction step:

\(\displaystyle \left({m + \left({k + 1}\right)}\right) + 1\) | \(=\) | \(\displaystyle \left({\left({m + k}\right) + 1}\right) + 1\) | Natural Number Addition is Associative/Proof 3 | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({m + 1}\right) + k}\right) + 1\) | from the induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({m + 1}\right) + \left({k + 1}\right)\) | Natural Number Addition is Associative/Proof 3 |

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

Therefore:

- $\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$

$\blacksquare$

## Also defined as

Some treatments of Peano's axioms define the non-successor element (or **primal element**) to be $1$ and not $0$.

The treatments are similar, but the $1$-based system results in an algebraic structure which has no identity element for addition, and so no zero for multiplication.

Thus, in the context of 1-based natural numbers, this result can be written:

- $\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$