Proof. We prove the statement for $c$ using the principle of mathematical induction. When $c=e$, by the definition of multiplication, we have
$$(a+b) \cdot e=a+b=(a \cdot e)+(b \cdot e),$$

so the conclusion holds. Assume that the conclusion holds for $c=n$. Then, for $c=n^{+}$, by the definition of multiplication, the inductive hypothesis, and the commutative and associative laws of addition, we get
$$\begin{aligned} (a+b) \cdot n^{+} & =((a+b) \cdot n)+(a+b) \\ & =((a \cdot n)+(b \cdot n))+(a+b) \\ & =((a \cdot n)+a)+((b \cdot n)+b) \\ & =\left(a \cdot n^{+}\right)+\left(b \cdot n^{+}\right) \end{aligned}$$

Thus, the conclusion also holds for $c=n^{+}$. The proof is complete.