= Solution
\b[True.] A useful fact is that squaring is strictly increasing on <ordinals>. If $A<B$, monotonicity in both arguments gives $A^2\leq BA$, and strict increase in the right argument for the positive left factor $B$ gives $BA<B^2$. Hence $A^2<B^2$.
Suppose, for a contradiction, that $\alpha\beta<\beta\alpha$. Again using monotonicity and <associativity> of <ordinal multiplication>,
$$
\alpha^2\beta^2=\alpha(\alpha\beta)\beta
\leq\alpha(\beta\alpha)\beta=(\alpha\beta)^2
<(\beta\alpha)^2=\beta(\alpha\beta)\alpha
\leq\beta(\beta\alpha)\alpha=\beta^2\alpha^2.
$$
This contradicts the hypothesized equality of the outer terms. Reversing the roles excludes $\beta\alpha<\alpha\beta$. If either factor is zero, they already commute. Therefore the <commuting squares of ordinals> property is
$$
\boxed{\alpha^2\beta^2=\beta^2\alpha^2\ \Longrightarrow\ \alpha\beta=\beta\alpha.}
$$
The proof uses monotonicity, which follows directly from the ordered-copy definition; it does not assume a cancellation law for <ordinals>.
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