On page 51 of Naive Set Theory, Halmos states the following:
Multiplication is associative and commutative. The proofs are straightforward adaptations of the ones that worked for addition. The distributive law (i.e., the assertion that $k \cdot (m + n) = k \cdot m + k \cdot n$ whenever $k$, $m$, and $n$ are natural numbers) is another easy consequence of the principle of mathematical induction. (Use induction on $n$)
However, with the tools developed up to this point in the book, it is not apparent to me how this is so. Any attempt to define any one of the above properties seems to depend on another property being usable.
Consider that multiplication is here defined as a function $p_m$ such that $p_m(0) = 0$ and $p_m(n^+) = p_m(n) + m$ for all $n$ in the set of natural numbers. Let us attempt to prove commutativity.
In the trivial case, we can first establish that using zero in multiplication always results in a product of zero, like so:
$$p_0(0) = 0 \cdot 0 = 0 = 0 \cdot 0$$ Induction hypothesis: $$0 \cdot n = 0 = n \cdot 0$$ To be demonstrated: $$\forall n \in \omega(0 \cdot n^+ = n^+ \cdot 0)$$ This proof is simple enough: $$0 \cdot n^+ = (0 \cdot n) + 0 = 0 + 0 = 0 = n^+ \cdot 0$$
This allows us to say that for $0$, the commutativity of multiplication is established. We may now take as our induction hypothesis the following:
$$m \cdot n = n \cdot m$$ To be demonstrated: $$\forall n,m \in \omega(m \cdot n^+ = n^+ \cdot m)$$ Between our definitions and hypothesis, we can do the following: $$m \cdot n^+ = (m \cdot n) + m$$ $$= (n \cdot m) + m$$ . . . but this is where I get stuck. This would appear to require distributivity as an assumption to move forward. However, any proof of distributivity that would actually be useful would, in turn, appear to depend on commutativity already being proven.
I am aware that there are proofs independent of the way Halmos constructs set theory, but since he so casually states that these proofs should be straightforwardly adapted from earlier proofs, I am curious how he imagines we may prove any of these properties using just what's on offer in the book to that point.
I believe you can prove distributivity without needing commutativity of multiplication. We show that
$$k \cdot (m+n) = k \cdot m + k \cdot n$$
by induction on $\,n.\,$
Base case $\,n=0:\,$
$$ \begin{align} k \cdot (m+0) &= k \cdot m\\ &= k \cdot m + 0\\ &= k \cdot m + k \cdot 0 \end{align} $$
Induction step: Suppose the statement holds for $\,n.\,$ Then
$$ \begin{align} k \cdot (m+n^+) &= k \cdot (m+n)^+\\ &= k \cdot (n+m)^+\\ &= k \cdot (n+m^+)\\ &= k \cdot (m^++n)\\ &= k \cdot m^+ + k \cdot n\\ &= (k \cdot m + k) + k \cdot n\\ &= k \cdot m + (k + k \cdot n)\\ &= k \cdot m + (k \cdot n + k)\\ &= k \cdot m + k \cdot n^+\\ \end{align} $$
In the above we have made use of both commutativity and associativity, but only for addition.