# Proof: By Induction

The proof is in two steps:

### Step $$(1)$$ Showing by induction of $$n$$ that $$n+0=0+n$$ for all $$n\in\mathbb N$$.

It follows from the definition of addition that $0+0=0+0\text{ (Base case)}.$ Now, let assume that $$n_0+0=0+n_0$$ has been proven for all $$n_0\le n$$, where we use "$$\le$$" as the order relation of natural numbers. Then it follows again from the definition of addition that $n^+ + 0=n^+$ Here, $$n^+$$ denotes the unique successor of $$n$$. Because by assumption $$n=0+n$$, we have that $n^+=(0+n)^+=0+n^+.$ This proves the commutativity law $$n+0=0+n$$ for all $$n\in\mathbb N$$.

### Step $$(2)$$ Showing by induction of $$m$$ that $$n+m=m+n$$ for all $$m\in\mathbb N$$.

It follows from the step $$(1)$$ that $n+0=0+n\text{ (Base case)}.$ Now, let assume that $$n+m_0=m_0+n$$ has been proven for all $$m_0\le m$$. Then it follows from the definition of addition that $n+m^+=(n+m)^+$ Because by assumption $$n+m=m+n$$, we have that $(n+m)^+=(m+n)^+=m+n^+.$ These two steps prove the commutativity law $$n+m=m+n$$ for all $$n,m\in\mathbb N$$.

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013