[proofplan] We prove the three monotonicity statements separately by transfinite induction on the right endpoint $\gamma$. Addition is handled directly from the recursive definition of ordinal addition. Multiplication then follows from the recursive definition of ordinal multiplication together with the already proved monotonicity of addition. Exponentiation is proved last, using the multiplication result and the fact that positive ordinal bases have nonzero powers. [/proofplan] [step:Prove monotonicity of ordinal addition in the right argument] Fix an ordinal $\alpha$. We prove by transfinite induction on the ordinal $\gamma$ that for every ordinal $\beta$ with $\beta \leq \gamma$, \begin{align*} \alpha + \beta \leq \alpha + \gamma. \end{align*} If $\gamma = 0$, then $\beta \leq 0$ implies $\beta = 0$, so the desired inequality is equality. Suppose $\gamma = \delta + 1$ and assume the assertion holds for $\delta$. Let $\beta \leq \delta + 1$. If $\beta = \delta + 1$, then the desired inequality is equality. If $\beta < \delta + 1$, then $\beta \leq \delta$, so the induction hypothesis gives \begin{align*} \alpha + \beta \leq \alpha + \delta. \end{align*} By the recursive definition of ordinal addition, \begin{align*} \alpha + (\delta + 1) = (\alpha + \delta) + 1, \end{align*} and every ordinal is less than or equal to its successor. Hence \begin{align*} \alpha + \beta \leq \alpha + \delta \leq (\alpha + \delta) + 1 = \alpha + (\delta + 1). \end{align*} Now suppose $\gamma = \lambda$ is a limit ordinal and assume the assertion holds for every ordinal $\delta < \lambda$. Let $\beta \leq \lambda$. If $\beta = \lambda$, then the desired inequality is equality. If $\beta < \lambda$, then by the recursive definition of ordinal addition at a limit ordinal, \begin{align*} \alpha + \lambda = \sup_{\delta < \lambda}(\alpha + \delta). \end{align*} Since $\beta < \lambda$, the ordinal $\alpha + \beta$ is one of the ordinals in this supremum, and therefore \begin{align*} \alpha + \beta \leq \sup_{\delta < \lambda}(\alpha + \delta) = \alpha + \lambda. \end{align*} By transfinite induction, $\alpha + \beta \leq \alpha + \gamma$ whenever $\beta \leq \gamma$. [/step] [step:Prove monotonicity of ordinal multiplication in the right argument] Fix an ordinal $\alpha$. We prove by transfinite induction on the ordinal $\gamma$ that for every ordinal $\beta$ with $\beta \leq \gamma$, \begin{align*} \alpha \cdot \beta \leq \alpha \cdot \gamma. \end{align*} If $\gamma = 0$, then $\beta \leq 0$ implies $\beta = 0$, so the desired inequality is equality. Suppose $\gamma = \delta + 1$ and assume the assertion holds for $\delta$. Let $\beta \leq \delta + 1$. If $\beta = \delta + 1$, then the desired inequality is equality. If $\beta < \delta + 1$, then $\beta \leq \delta$, and the induction hypothesis gives \begin{align*} \alpha \cdot \beta \leq \alpha \cdot \delta. \end{align*} By the recursive definition of ordinal multiplication, \begin{align*} \alpha \cdot (\delta + 1) = \alpha \cdot \delta + \alpha. \end{align*} Applying the addition monotonicity already proved, with left argument $\alpha \cdot \delta$ and $0 \leq \alpha$, gives \begin{align*} \alpha \cdot \delta = \alpha \cdot \delta + 0 \leq \alpha \cdot \delta + \alpha = \alpha \cdot (\delta + 1). \end{align*} Thus \begin{align*} \alpha \cdot \beta \leq \alpha \cdot \delta \leq \alpha \cdot (\delta + 1). \end{align*} Now suppose $\gamma = \lambda$ is a limit ordinal and assume the assertion holds for every ordinal $\delta < \lambda$. Let $\beta \leq \lambda$. If $\beta = \lambda$, then the desired inequality is equality. If $\beta < \lambda$, then by the recursive definition of ordinal multiplication at a limit ordinal, \begin{align*} \alpha \cdot \lambda = \sup_{\delta < \lambda}(\alpha \cdot \delta). \end{align*} Since $\beta < \lambda$, the ordinal $\alpha \cdot \beta$ is one of the ordinals in this supremum, and therefore \begin{align*} \alpha \cdot \beta \leq \sup_{\delta < \lambda}(\alpha \cdot \delta) = \alpha \cdot \lambda. \end{align*} By transfinite induction, $\alpha \cdot \beta \leq \alpha \cdot \gamma$ whenever $\beta \leq \gamma$. [/step] [step:Show that positive ordinal powers are nonzero] Assume $\alpha > 0$. We prove by transfinite induction on the ordinal $\delta$ that \begin{align*} \alpha^\delta > 0. \end{align*} For $\delta = 0$, the recursive definition of ordinal exponentiation gives $\alpha^0 = 1$, so $\alpha^0 > 0$. Suppose $\delta = \eta + 1$ and assume $\alpha^\eta > 0$. By the recursive definition of ordinal exponentiation, \begin{align*} \alpha^{\eta + 1} = \alpha^\eta \cdot \alpha. \end{align*} Since $\alpha^\eta > 0$ and $\alpha > 0$, the ordinal product $\alpha^\eta \cdot \alpha$ is nonzero. Hence $\alpha^{\eta + 1} > 0$. Suppose $\delta = \lambda$ is a limit ordinal and assume $\alpha^\eta > 0$ for every ordinal $\eta < \lambda$. By the recursive definition of ordinal exponentiation at a limit ordinal, \begin{align*} \alpha^\lambda = \sup_{\eta < \lambda} \alpha^\eta. \end{align*} Because $\lambda$ is a limit ordinal, $0 < \lambda$, so $0$ is an ordinal with $0 < \lambda$. Thus $\alpha^0 = 1$ appears among the ordinals whose supremum defines $\alpha^\lambda$, and therefore \begin{align*} 1 = \alpha^0 \leq \sup_{\eta < \lambda} \alpha^\eta = \alpha^\lambda. \end{align*} Hence $\alpha^\lambda > 0$. Thus $\alpha^\delta > 0$ for every ordinal $\delta$. [/step] [step:Prove monotonicity of ordinal exponentiation in the right argument for positive bases] Assume $\alpha > 0$. We prove by transfinite induction on the ordinal $\gamma$ that for every ordinal $\beta$ with $\beta \leq \gamma$, \begin{align*} \alpha^\beta \leq \alpha^\gamma. \end{align*} If $\gamma = 0$, then $\beta \leq 0$ implies $\beta = 0$, so the desired inequality is equality. Suppose $\gamma = \delta + 1$ and assume the assertion holds for $\delta$. Let $\beta \leq \delta + 1$. If $\beta = \delta + 1$, then the desired inequality is equality. If $\beta < \delta + 1$, then $\beta \leq \delta$, and the induction hypothesis gives \begin{align*} \alpha^\beta \leq \alpha^\delta. \end{align*} By the recursive definition of ordinal exponentiation, \begin{align*} \alpha^{\delta + 1} = \alpha^\delta \cdot \alpha. \end{align*} From the previous step, $\alpha^\delta > 0$. Since $\alpha > 0$, we have $1 \leq \alpha$. Applying multiplication monotonicity in the right argument, with left argument $\alpha^\delta$ and $1 \leq \alpha$, gives \begin{align*} \alpha^\delta = \alpha^\delta \cdot 1 \leq \alpha^\delta \cdot \alpha = \alpha^{\delta + 1}. \end{align*} Therefore \begin{align*} \alpha^\beta \leq \alpha^\delta \leq \alpha^{\delta + 1}. \end{align*} Now suppose $\gamma = \lambda$ is a limit ordinal and assume the assertion holds for every ordinal $\delta < \lambda$. Let $\beta \leq \lambda$. If $\beta = \lambda$, then the desired inequality is equality. If $\beta < \lambda$, then by the recursive definition of ordinal exponentiation at a limit ordinal, \begin{align*} \alpha^\lambda = \sup_{\delta < \lambda}\alpha^\delta. \end{align*} Since $\beta < \lambda$, the ordinal $\alpha^\beta$ is one of the ordinals in this supremum, and therefore \begin{align*} \alpha^\beta \leq \sup_{\delta < \lambda}\alpha^\delta = \alpha^\lambda. \end{align*} By transfinite induction, $\alpha^\beta \leq \alpha^\gamma$ whenever $\beta \leq \gamma$ and $\alpha > 0$. [/step] [step:Combine the three monotonicity statements] The first step proves \begin{align*} \alpha + \beta \leq \alpha + \gamma \end{align*} for all ordinals $\alpha$, $\beta$, and $\gamma$ with $\beta \leq \gamma$. The second step proves \begin{align*} \alpha \cdot \beta \leq \alpha \cdot \gamma \end{align*} under the same hypotheses. The fourth step proves, under the additional hypothesis $\alpha > 0$, that \begin{align*} \alpha^\beta \leq \alpha^\gamma. \end{align*} These are exactly the three asserted inequalities. [/step]