= Solution
Fix $m,n$ and induct on $k$. For $k=0$,
$$
m\cdot(n+0)=m\cdot n=m\cdot n+0=m\cdot n+m\cdot0.
$$
For the induction step, use the <multiplication> <recursion>, the <induction hypothesis>, and <addition> <associativity> from (i):
$$
\begin{aligned}
m\cdot(n+k^+)&=m\cdot(n+k)^+\\
&=m\cdot(n+k)+m\\
&=(m\cdot n+m\cdot k)+m\\
&=m\cdot n+(m\cdot k+m)\\
&=m\cdot n+m\cdot k^+.
\end{aligned}
$$
Thus
$$
\boxed{m\cdot(n+k)=m\cdot n+m\cdot k.}
$$
This establishes the indicated distributive law without assuming <multiplication> is commutative.
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