= Solution
Fix $m$ and induct on $n$. The initial equality $m+0=0+m$ follows from the defining <zero> clause and (ii). Assuming $m+n=n+m$, the successor clause and (iii) give
$$
m+n^+=(m+n)^+=(n+m)^+=n+m^+=n^++m.
$$
This proves
$$
\boxed{m+n=n+m\quad(m,n\in\omega).}
$$
The extension to arbitrary <ordinals> is false. By the <limit> clause for <ordinal addition>,
$$
1+\omega=\sup_{n<\omega}(1+n)=\omega,
$$
whereas $\omega+1$ is the <successor ordinal> of $\omega$ and is strictly greater than it. Therefore
$$
\boxed{1+\omega\ne\omega+1.}
$$
Finite <addition> is commutative; <ordinal addition> records the order of concatenated well-orders and need not be commutative.
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