= Solution
First give the recursive definitions, keeping the right argument as the recursion <variable>. For <ordinals> $\alpha,\beta$ and nonzero <limit ordinals> $\lambda$,
$$
\alpha+0=\alpha,\qquad \alpha+(\beta+1)=(\alpha+\beta)+1,\qquad
\alpha+\lambda=\sup_{\xi<\lambda}(\alpha+\xi),
$$
$$
\alpha\cdot0=0,\qquad \alpha\cdot(\beta+1)=\alpha\beta+\alpha,\qquad
\alpha\lambda=\sup_{\xi<\lambda}\alpha\xi.
$$
Here $\xi+1$ is the <successor ordinal>, and the <supremum> of a <set> of <ordinals> is its <union>. These are <ordinal arithmetic>, defined by <transfinite recursion>, and differ from <cardinal sum> and <ordinal multiplication>.
\b[<Associativity> of <ordinal addition> is true.] Fix $\alpha,\beta$ and use <transfinite induction> on $\gamma$. The case $\gamma=0$ is immediate. In the successor step the recursive definitions and the induction hypothesis give
$$
(\alpha+\beta)+(\gamma+1)=((\alpha+\beta)+\gamma)+1
=(\alpha+(\beta+\gamma))+1=\alpha+(\beta+(\gamma+1)).
$$
For a nonzero <limit ordinal> $\lambda$, the <ordinals> $\beta+\xi$ for $\xi<\lambda$ form an increasing <cofinal subset> of the <limit ordinal> $\beta+\lambda$. The limit clause for <ordinal addition>, restricted to this <cofinal subset>, gives
$$
(\alpha+\beta)+\lambda
=\sup_{\xi<\lambda}((\alpha+\beta)+\xi)
=\sup_{\xi<\lambda}(\alpha+(\beta+\xi))
=\alpha+(\beta+\lambda).
$$
<Cofinal> restriction is valid because <ordinal addition> by a fixed <ordinal> on the left is increasing and continuous in its right argument, directly from its recursive definition. This completes the induction.
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