[proofplan] We fix the left factors $\alpha$ and $\beta$ and prove the identity by transfinite induction on $\gamma$. The zero and successor cases follow directly from the recursive definition of ordinal multiplication, except that the successor case also uses the right-distributive law $\alpha \cdot (\xi+\beta)=\alpha\cdot\xi+\alpha\cdot\beta$, which we prove as an internal claim. At a limit ordinal $\lambda$, both sides are determined by suprema over earlier stages; the induction hypothesis identifies the approximating ordinals, and continuity of ordinal multiplication in its right argument gives the desired equality. [/proofplan] [step:Prove right-distributivity of multiplication over ordinal addition] We first record the auxiliary identity needed at successor stages. We shall use the following standard recursive facts about ordinal arithmetic: ordinal addition is associative, ordinal addition is continuous in its right argument at nonzero limits, ordinal multiplication is defined by $\theta\cdot 0=0$, $\theta\cdot(\eta+1)=\theta\cdot\eta+\theta$, and $\theta\cdot\lambda=\sup_{\delta<\lambda}\theta\cdot\delta$ for nonzero limit ordinals $\lambda$. These are exactly the recursion clauses for addition and multiplication, together with associativity of addition. For every ordinal $\xi$, \begin{align*} \alpha \cdot (\xi+\beta)=\alpha\cdot\xi+\alpha\cdot\beta. \end{align*} [claim:Right-distributivity in the second factor] For all ordinals $\alpha$, $\xi$, and $\beta$, \begin{align*} \alpha \cdot (\xi+\beta)=\alpha\cdot\xi+\alpha\cdot\beta. \end{align*} [/claim] [proof] Fix ordinals $\alpha$ and $\xi$. We prove the asserted identity by transfinite induction on $\beta$. If $\beta=0$, then $\xi+0=\xi$ and $\alpha\cdot 0=0$. Hence \begin{align*} \alpha\cdot(\xi+0)=\alpha\cdot\xi=\alpha\cdot\xi+0=\alpha\cdot\xi+\alpha\cdot 0. \end{align*} Suppose the identity holds for an ordinal $\beta$. Using the successor clauses for ordinal addition and ordinal multiplication, and using [associativity of ordinal addition](/theorems/1473) in the middle equality, we obtain \begin{align*} \alpha\cdot(\xi+(\beta+1)) &=\alpha\cdot((\xi+\beta)+1) \\ &=\alpha\cdot(\xi+\beta)+\alpha \\ &=(\alpha\cdot\xi+\alpha\cdot\beta)+\alpha \\ &=\alpha\cdot\xi+(\alpha\cdot\beta+\alpha) \\ &=\alpha\cdot\xi+\alpha\cdot(\beta+1). \end{align*} This proves the successor step. Now let $\lambda$ be a nonzero limit ordinal, and suppose the identity holds for every $\delta<\lambda$. By the limit clause for ordinal addition, \begin{align*} \xi+\lambda=\sup_{\delta<\lambda}(\xi+\delta). \end{align*} Using the limit clause for ordinal multiplication in its right argument and the induction hypothesis, we get \begin{align*} \alpha\cdot(\xi+\lambda) &=\sup_{\delta<\lambda}\alpha\cdot(\xi+\delta) \\ &=\sup_{\delta<\lambda}(\alpha\cdot\xi+\alpha\cdot\delta). \end{align*} By the limit clause for ordinal multiplication, $\alpha\cdot\lambda=\sup_{\delta<\lambda}\alpha\cdot\delta$. Therefore the preceding supremum is \begin{align*} \sup_{\delta<\lambda}(\alpha\cdot\xi+\alpha\cdot\delta) =\alpha\cdot\xi+\sup_{\delta<\lambda}\alpha\cdot\delta =\alpha\cdot\xi+\alpha\cdot\lambda, \end{align*} where the first equality is the limit clause for ordinal addition applied to the ordinal $\alpha\cdot\xi$ and to the increasing net of right summands whose supremum is $\alpha\cdot\lambda$. Thus the identity holds for $\lambda$. By transfinite induction on $\beta$, the claim follows. [/proof] [guided] The successor step in the main proof will require rewriting \begin{align*} \alpha\cdot(\beta\cdot\gamma)+\alpha\cdot\beta \end{align*} as \begin{align*} \alpha\cdot(\beta\cdot\gamma+\beta). \end{align*} This is not commutativity; ordinal multiplication is not commutative. The correct tool is right-distributivity of multiplication over ordinal addition in the second factor. We prove it directly from the recursive definitions. Fix ordinals $\alpha$ and $\xi$, and induct on the ordinal $\beta$. For $\beta=0$, the recursive definitions give $\xi+0=\xi$ and $\alpha\cdot 0=0$, so \begin{align*} \alpha\cdot(\xi+0)=\alpha\cdot\xi=\alpha\cdot\xi+0=\alpha\cdot\xi+\alpha\cdot 0. \end{align*} Assume the identity holds for $\beta$. The successor clause for addition gives \begin{align*} \xi+(\beta+1)=(\xi+\beta)+1. \end{align*} Then the successor clause for multiplication gives \begin{align*} \alpha\cdot(\xi+(\beta+1)) &=\alpha\cdot((\xi+\beta)+1) \\ &=\alpha\cdot(\xi+\beta)+\alpha. \end{align*} Substituting the induction hypothesis and reassociating ordinal addition, \begin{align*} \alpha\cdot(\xi+\beta)+\alpha &=(\alpha\cdot\xi+\alpha\cdot\beta)+\alpha \\ &=\alpha\cdot\xi+(\alpha\cdot\beta+\alpha) \\ &=\alpha\cdot\xi+\alpha\cdot(\beta+1). \end{align*} The last equality is exactly the successor clause for multiplication. Finally let $\lambda$ be a nonzero limit ordinal. The limit clause for addition says \begin{align*} \xi+\lambda=\sup_{\delta<\lambda}(\xi+\delta). \end{align*} Since ordinal multiplication is defined at a nonzero limit right factor by taking the supremum of the earlier products, we have \begin{align*} \alpha\cdot(\xi+\lambda) &=\sup_{\delta<\lambda}\alpha\cdot(\xi+\delta). \end{align*} For each $\delta<\lambda$, the induction hypothesis identifies the approximant: \begin{align*} \alpha\cdot(\xi+\delta)=\alpha\cdot\xi+\alpha\cdot\delta. \end{align*} Therefore \begin{align*} \alpha\cdot(\xi+\lambda) &=\sup_{\delta<\lambda}(\alpha\cdot\xi+\alpha\cdot\delta). \end{align*} The right summands have supremum $\alpha\cdot\lambda$, because the limit clause for ordinal multiplication gives \begin{align*} \alpha\cdot\lambda=\sup_{\delta<\lambda}\alpha\cdot\delta. \end{align*} Applying the limit clause for ordinal addition to the fixed left summand $\alpha\cdot\xi$ and this increasing family of right summands gives \begin{align*} \sup_{\delta<\lambda}(\alpha\cdot\xi+\alpha\cdot\delta) &=\alpha\cdot\xi+\sup_{\delta<\lambda}\alpha\cdot\delta \\ &=\alpha\cdot\xi+\alpha\cdot\lambda. \end{align*} This completes the proof of right-distributivity. [/guided] [/step] [step:Prove the zero case for the induction on $\gamma$] Fix ordinals $\alpha$ and $\beta$. For $\gamma=0$, the recursive definition of ordinal multiplication gives \begin{align*} (\alpha\cdot\beta)\cdot 0=0 \end{align*} and \begin{align*} \beta\cdot 0=0, \qquad \alpha\cdot(\beta\cdot 0)=\alpha\cdot 0=0. \end{align*} Hence \begin{align*} (\alpha\cdot\beta)\cdot 0=\alpha\cdot(\beta\cdot 0). \end{align*} [/step] [step:Use the induction hypothesis and right-distributivity at successor ordinals] Assume that $\gamma$ is an ordinal such that \begin{align*} (\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma). \end{align*} Then, using the successor clause for ordinal multiplication, the induction hypothesis, the right-distributivity claim with $\xi=\beta\cdot\gamma$, and the successor clause again, we compute \begin{align*} (\alpha\cdot\beta)\cdot(\gamma+1) &=((\alpha\cdot\beta)\cdot\gamma)+(\alpha\cdot\beta) \\ &=\alpha\cdot(\beta\cdot\gamma)+\alpha\cdot\beta \\ &=\alpha\cdot((\beta\cdot\gamma)+\beta) \\ &=\alpha\cdot(\beta\cdot(\gamma+1)). \end{align*} Thus the desired identity holds at $\gamma+1$. [guided] At a successor ordinal, the definition of ordinal multiplication expands the right factor by adding one copy of the left factor. Applying this to the left-hand side gives \begin{align*} (\alpha\cdot\beta)\cdot(\gamma+1) =((\alpha\cdot\beta)\cdot\gamma)+(\alpha\cdot\beta). \end{align*} The induction hypothesis applies to the first term: \begin{align*} (\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma). \end{align*} Substituting this equality yields \begin{align*} (\alpha\cdot\beta)\cdot(\gamma+1) =\alpha\cdot(\beta\cdot\gamma)+\alpha\cdot\beta. \end{align*} Now the auxiliary right-distributivity identity applies with $\xi=\beta\cdot\gamma$. It gives \begin{align*} \alpha\cdot(\beta\cdot\gamma)+\alpha\cdot\beta =\alpha\cdot((\beta\cdot\gamma)+\beta). \end{align*} Finally, the successor clause for ordinal multiplication says \begin{align*} \beta\cdot(\gamma+1)=(\beta\cdot\gamma)+\beta. \end{align*} Therefore \begin{align*} (\alpha\cdot\beta)\cdot(\gamma+1) =\alpha\cdot(\beta\cdot(\gamma+1)). \end{align*} This proves the successor step. [/guided] [/step] [step:Pass to limit ordinals by comparing the defining suprema] Let $\lambda$ be a nonzero limit ordinal, and assume that for every ordinal $\delta<\lambda$, \begin{align*} (\alpha\cdot\beta)\cdot\delta=\alpha\cdot(\beta\cdot\delta). \end{align*} By the limit clause for ordinal multiplication, \begin{align*} (\alpha\cdot\beta)\cdot\lambda =\sup_{\delta<\lambda}((\alpha\cdot\beta)\cdot\delta). \end{align*} Using the induction hypothesis for each $\delta<\lambda$, this becomes \begin{align*} (\alpha\cdot\beta)\cdot\lambda =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta). \end{align*} If $\beta=0$, then $\beta\cdot\lambda=0$ and both sides are $0$. Assume henceforth that $\beta>0$. Since $\lambda$ is a nonzero limit ordinal, the limit clause for ordinal multiplication gives \begin{align*} \beta\cdot\lambda=\sup_{\delta<\lambda}\beta\cdot\delta. \end{align*} Moreover the family $(\beta\cdot\delta)_{\delta<\lambda}$ is cofinal in $\beta\cdot\lambda$ by this equality, and $\beta\cdot\lambda$ is a nonzero limit ordinal: for every $\delta<\lambda$ we have $\delta+1<\lambda$ and \begin{align*} \beta\cdot\delta<\beta\cdot\delta+eta=\beta\cdot(\delta+1)<\beta\cdot\lambda. \end{align*} The limit clause for multiplication by $\alpha$ therefore yields \begin{align*} \alpha\cdot(\beta\cdot\lambda) =\sup_{\rho<\beta\cdot\lambda}\alpha\cdot\rho. \end{align*} Because ordinal multiplication is monotone in its right argument, taking the supremum over the cofinal subfamily $\rho=\beta\cdot\delta$ gives \begin{align*} \sup_{\rho<\beta\cdot\lambda}\alpha\cdot\rho =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta). \end{align*} Hence \begin{align*} \alpha\cdot(\beta\cdot\lambda) =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta), \end{align*} and comparison with the expression for $(\alpha\cdot\beta)\cdot\lambda$ gives \begin{align*} (\alpha\cdot\beta)\cdot\lambda=\alpha\cdot(\beta\cdot\lambda). \end{align*} [guided] At a limit ordinal, ordinal multiplication is defined by taking the supremum of all earlier stages. Thus the left-hand side expands as \begin{align*} (\alpha\cdot\beta)\cdot\lambda =\sup_{\delta<\lambda}((\alpha\cdot\beta)\cdot\delta). \end{align*} The induction hypothesis is available for every $\delta<\lambda$, so every approximant in this supremum can be rewritten: \begin{align*} (\alpha\cdot\beta)\cdot\delta=\alpha\cdot(\beta\cdot\delta). \end{align*} Hence \begin{align*} (\alpha\cdot\beta)\cdot\lambda =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta). \end{align*} We now compare this with the right-hand side. First separate the degenerate case. If $\beta=0$, then $\beta\cdot\lambda=0$, so \begin{align*} (\alpha\cdot 0)\cdot\lambda=0 \qquad\text{and}\qquad \alpha\cdot(0\cdot\lambda)=\alpha\cdot 0=0. \end{align*} Now assume $\beta>0$. Since $\lambda$ is a nonzero limit ordinal, multiplication by $\beta$ in the right argument is defined by \begin{align*} \beta\cdot\lambda=\sup_{\delta<\lambda}\beta\cdot\delta. \end{align*} This says that the ordinals $\beta\cdot\delta$ form a cofinal family in $\beta\cdot\lambda$. We also need to know that $\beta\cdot\lambda$ is itself a limit ordinal before applying the limit clause to multiplication by $\alpha$. For every $\delta<\lambda$, the successor $\delta+1$ is still below $\lambda$, and because $\beta>0$, \begin{align*} \beta\cdot\delta<\beta\cdot\delta+eta=\beta\cdot(\delta+1)<\beta\cdot\lambda. \end{align*} Thus no approximant is final, so $\beta\cdot\lambda$ is a nonzero limit ordinal. Now the limit clause for multiplication by $\alpha$ applies to the limit right factor $\beta\cdot\lambda$: \begin{align*} \alpha\cdot(\beta\cdot\lambda) =\sup_{\rho<\beta\cdot\lambda}\alpha\cdot\rho. \end{align*} Because ordinal multiplication is monotone in its right argument, replacing the full indexing set $\{\rho: \rho<\beta\cdot\lambda\}$ by the cofinal subfamily $\{\beta\cdot\delta: \delta<\lambda\}$ does not change the supremum. Hence \begin{align*} \alpha\cdot(\beta\cdot\lambda) =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta). \end{align*} Both sides are therefore the same supremum: \begin{align*} (\alpha\cdot\beta)\cdot\lambda =\sup_{\delta<\lambda}\alpha\cdot(\beta\cdot\delta) =\alpha\cdot(\beta\cdot\lambda). \end{align*} This proves the limit step. [/guided] [/step] [step:Conclude by transfinite induction] The zero case, successor step, and limit step establish by transfinite induction on $\gamma$ that for the fixed ordinals $\alpha$ and $\beta$, \begin{align*} (\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma) \end{align*} for every ordinal $\gamma$. Since $\alpha$ and $\beta$ were arbitrary, the identity holds for all ordinals $\alpha$, $\beta$, and $\gamma$. [/step]