(related to Proposition: Multiplication of Natural Numbers Is Associative)
We shall show the associativity of "\( \cdot \)" by induction. For arbitrary \(n,m\in\mathbb N\), it follows from the definition of multiplication that
\[n\cdot (m\cdot 0)=n\cdot 0=0=(n\cdot m)\cdot 0.\]
Now, let assume that \(n\cdot (m\cdot p_0)=(n\cdot m)\cdot p_0\) has been proven for all \(p_0\le p\). Then it follows \[\begin{array}{rcl} n\cdot (m\cdot p^+)&=&n((m\cdot p)+m)\quad\quad(i)\\ &=&n\cdot (m\cdot p)+ n\cdot m\quad\quad(ii)\\ &=&(n\cdot m)\cdot p + n\cdot m\quad\quad(iii)\\ &=&(n\cdot m)\cdot (p + 1)\quad\quad(iv)\\ &=&(n\cdot m)\cdot p^+. \end{array}\] Here, we have used in step \((i)\) the definition of multiplication, in steps \((ii)\) and \(iii)\) the left-distributivity property of multiplication of natural numbs \(p^+\) denotes the unique successor of \(p\).
This proves the associativity of "\( \cdot \)" for all \(n,m,p\in\mathbb N\).
