Natural Number Addition is Cancellable

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.

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:

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:

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

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