Natural Number Addition is Cancellable

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Let $\N$ be the natural numbers.

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


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

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.


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.


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