# Natural Number Addition is Cancellable

## Contents

## Theorem

Let $\N$ be the natural numbers.

Let $+$ be addition on $\N$.

Then:

- $\forall a, b, c \in \N: a + c = b + c \implies a = b$
- $\forall a, b, c \in \N: a + b = a + c \implies b = c$

That is, $+$ is cancellable on $\N$.

## Proof 1

Consider the natural numbers $\N$ defined as a naturally ordered semigroup $\left({\N, +, \le}\right)$.

From Naturally Ordered Semigroup: $NO 2$, every element of $\left ({\N, +}\right)$ is cancellable.

$\blacksquare$

## Proof 2

By Natural Number Addition is Commutative, we only need to prove the first statement.

Proof by induction.

Consider the natural numbers $\N$ defined in terms of Peano's Axioms.

From the definition of addition in Peano structure, we have that:

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

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

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

- $\forall a, b \in \N: a + n = b + n \implies a = b$

### Basis for the Induction

$P \left({0}\right)$ is the proposition:

- $\forall a, b \in \N: a + 0 = b + 0 \implies a = b$

which holds because of the definition of $+$.

This is our basis for the induction.

### Induction Hypothesis

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

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

- $\forall a, b \in \N: a + k = b + k \implies a = b$

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

- $\forall a, b \in \N: a + s \left({k}\right) = b + s \left({k}\right) \implies a = b$

### Induction Step

This is our induction step:

\(\displaystyle a + s \left({k}\right)\) | \(=\) | \(\displaystyle b + s \left({k}\right)\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle s \left({a + k}\right)\) | \(=\) | \(\displaystyle s \left({b + k}\right)\) | from the definition of $+$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle a + k\) | \(=\) | \(\displaystyle b + k\) | Peano's Axiom $P3$: $s$ is injective | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle a\) | \(=\) | \(\displaystyle b\) | from the induction hypothesis |

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

Therefore:

- $\forall a, b \in \N: a + n = b + n \implies a = b$

$\blacksquare$