# Natural Number Addition Commutativity with Successor

## 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:

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