Prove that when $c=e$, for any $a, b \in \boldsymbol{N}$, from equations (1) and (2) we get
$$(a+b)+e=(a+b)^{+}=a+b^{+}=a+(b+e),$$

so equation (8) holds. Assume that equation (8) holds for some $c=n$ and any $a, b \in \boldsymbol{N}$. When $c=n^{+}$, for any $a, b \in \boldsymbol{N}$, we have (using equation (2) and the assumption)
$$\begin{aligned} (a+b)+n^{+} & =((a+b)+n)^{+}=(a+(b+n))^{+} \\ & =a+(b+n)^{+}=a+\left(b+n^{+}\right) \end{aligned}$$

Thus, equation (8) holds for $c=n^{+}$ and any $a, b \in \boldsymbol{N}$. By the principle of mathematical induction, the conclusion is proved.