= Solution
\b[True: <ordinal multiplication> on the left distributes over <ordinal addition>.] If $\alpha=0$ both sides are zero. Otherwise fix $\alpha>0$ and $\beta$, and induct on $\gamma$. At zero the identity follows from $\alpha0=0$. At a successor, use the recursive <ordinal multiplication> rule and the <associativity> just proved:
$$
\alpha(\beta+(\gamma+1))
=\alpha((\beta+\gamma)+1)
=\alpha(\beta+\gamma)+\alpha
=(\alpha\beta+\alpha\gamma)+\alpha
=\alpha\beta+\alpha(\gamma+1).
$$
For a nonzero <limit ordinal> $\lambda$, the <ordinals> $\beta+\xi$ are <cofinal> in $\beta+\lambda$. <Continuity> of <ordinal multiplication> in its right argument gives
$$
\alpha(\beta+\lambda)
=\sup_{\xi<\lambda}\alpha(\beta+\xi)
=\sup_{\xi<\lambda}(\alpha\beta+\alpha\xi)
=\alpha\beta+\alpha\lambda.
$$
For the last <equality>, $\alpha\xi$ is increasing and <cofinal> in $\alpha\lambda$, which is a <limit ordinal> because $\alpha>0$, and <ordinal addition> is continuous in its right argument. This establishes the limit step and the identity for all <ordinals>.
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