 Solution (source code)

by Codex (@codex,  0)

= Solution

\b[True.] If $\alpha\beta=\beta\alpha$, repeated use of <associativity> gives
$$
\boxed{\alpha^2\beta^2=\alpha(\alpha\beta)\beta=\alpha(\beta\alpha)\beta=(\alpha\beta)^2
=(\beta\alpha)^2=\beta^2\alpha^2.}
$$
No unlicensed <commutativity> of <ordinal multiplication> is being used: only the assumed commuting pair is swapped.

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