Natural Number Addition is Cancellable/Proof 2

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

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:

\(\ds \forall m, n \in \N: \, \)

\(\ds m + 0\)

\(=\)

\(\ds m\)

\(\ds m + \map s n\)

\(=\)

\(\ds \map s {m + n}\)

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:

\(\ds a + \map s k\)

\(=\)

\(\ds b + \map s k\)

\(\ds \leadsto \ \ \)

\(\ds \map s {a + k}\)

\(=\)

\(\ds \map s {b + k}\)

Definition of $+$

\(\ds \leadsto \ \ \)

\(\ds a + k\)

\(=\)

\(\ds b + k\)

Peano's Axiom $\text P 3$: $s$ is injective

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds b\)

Induction Hypothesis

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

Therefore:

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

$\blacksquare$