(related to Proposition: Multiplication of Natural Numbers is Commutative)
The proof is in two steps:
It follows from the definition of multiplication that \[0\cdot 0=0\cdot 0\text{ (Base case)}.\] Now, let assume that \(n_0\cdot 0=0\cdot n_0\) 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 multiplication that \[n^+ \cdot 0=0.\quad\quad ( * )\] Here, \(n^+\) denotes the unique successor of \(n\). On the other side, we have by definition of multiplication and by assumption \[0\cdot n^+=(0\cdot n)+0=(n\cdot 0)+0=0+0=0.\quad\quad ( * * )\] In the last step (\(0 + 0=0\)) we have used the definition of addition of natural numbers. The comparison of \(( * )\) and \(( * * )\) proves the commutativity law \(n\cdot 0=0\cdot n\) for all \(n\in\mathbb N\).
It follows from the step \((1)\) that \[n\cdot 0=0\cdot n\text{ (Base case)}.\] Now, let assume that \(n\cdot m_0=m_0\cdot n\) has been proven for all \(m_0\le m\). Then it follows from the definition of multiplication and by assumption that \[n\cdot m^+=(n\cdot m) + n=(m\cdot n) + n.\quad\quad ( \ast ) \] From the definition of multiplication and the definition of addition, we have \(n\cdot 1=(n\cdot 0)+n=0+n=n\). Using that result, applying \(\ast\) and also the right-distributivity law for natural numbers we get \[(m\cdot n)+n=(m\cdot n)+(n\cdot 1)=(m+1)\cdot n=m^+\cdot n\quad\quad ( \ast \ast ) \] The comparison of \(( \ast )\) and \(( \ast\ast )\) proves the commutativity law \(n\cdot m=m\cdot n\) for all \(n,m\in\mathbb N\).
