Prove Proposition 2.3.5.
Proposition 2.3.5 (multiplication is associative).
For any natural numbers $a$, $b$, $c$, we have $(a\times b)\times c=a\times(b\times c)$.
We will prove this by induction on $a$, holding $b$ and $c$ fixed. Let $S(a)$ be the statement that $(a\times b)\times c=a\times(b\times c)$ for any natural numbers $a$, $b$, $c$. We need to prove the base
case $S(0)$ and the implication step $S(a)\implies S(a++)$.
Base Case
The base case $S(0)$ states that $(0\times b)\times c=0\times(b\times c)$. By Definition 2.3.1, the expression $0\times(b\times c)$ is identically zero. Definition 2.3.1 also gives $(0\times b)=0$, which means $(0\times b)\times c=0\times c=0$. And so the base case is true.
Induction Step
We need to show $S(a++)$, that is, $((a++)\times b)\times c=(a++)\times(b\times c)$. We assume $S(a),$ that is, $(a\times b)\times c=a\times(b\times c)$. We proceed as follows
$$\begin{alignat*}{3} ((a++)\times b)\times c & =((a\times b)+b)\times c & \hspace{4cm} & & \text{Definition 2.3.1}\\ \\ & =((a\times b)\times c)+(b\times c) & \hspace{4cm} & & \text{Proposition 2.3.4}\\ \\ & =(a\times(b\times c))+(b\times c) & \hspace{4cm} & & \text{induction hypothesis}\\ \\ & =(a++)\times(b\times c) & \hspace{4cm} & & \text{Definition 2.3.1} \end{alignat*}$$
By showing the base case and the induction step, we have proved that $(a\times b)\times c=a\times(b\times c)$ for any natural numbers.
