(related to Proposition: Addition of Natural Numbers Is Commutative)
The proof is in two steps:
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 \(n\in\mathbb N\).
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 \(n,m\in\mathbb N\).