Natural Number Addition is Cancellable/Proof 2

Proof
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 $\map P n$ be the proposition:
 * $\forall a, b \in \N: a + n = b + n \implies a = b$

Basis for the Induction
$\map P 0$ 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 $\map P k$ is true, where $k \in \N$, then it logically follows that $\map P {\map s k}$ is true.

So this is our induction hypothesis $\map P k$:
 * $\forall a, b \in \N: a + k = b + k \implies a = b$

Then we need to show that $\map P {\map s k}$ follows directly from $\map P k$:
 * $\forall a, b \in \N: a + \map s k = b + \map s k \implies a = b$

Induction Step
This is our induction step:

So $\map P k \implies \map P {\map s k}$ and the result follows by the Principle of Mathematical Induction for Peano Structure.

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