[proofplan] Let $\mu=\max\{\kappa,\lambda\}$. The standard monotonicity inequalities reduce the theorem to proving the two absorption identities $\mu+\mu=\mu$ and $\mu\cdot\mu=\mu$ for each infinite cardinal $\mu$. Using Choice, we represent $\mu$ by its initial ordinal and prove $\mu\cdot\mu=\mu$ by transfinite induction over infinite initial ordinals: order $\mu\times\mu$ by increasing maximum coordinate, then lexicographically inside each level, and use the induction hypothesis to show every initial segment has size $<\mu$. The additive identity then follows by injecting the disjoint union of two copies of $\mu$ into $\mu\times\mu$. [/proofplan]

[step:Reduce the theorem to the square of one infinite cardinal] Let $\kappa$ and $\lambda$ be cardinals, and assume at least one of them is infinite. Define the cardinal \begin{align*} \mu := \max\{\kappa,\lambda\}. \end{align*} Then $\mu$ is infinite, $\kappa \leq \mu$, and $\lambda \leq \mu$. For cardinal addition, monotonicity gives \begin{align*} \mu \leq \kappa+\lambda \leq \mu+\mu. \end{align*} The lower bound holds because one of $\kappa,\lambda$ is equal to $\mu$, and the corresponding summand embeds into the disjoint union. Thus, once $\mu+\mu=\mu$ is known, antisymmetry of cardinal comparison gives \begin{align*} \kappa+\lambda=\mu. \end{align*} For cardinal multiplication, assume additionally that $\kappa\neq 0$ and $\lambda\neq 0$. Since one of $\kappa,\lambda$ is equal to $\mu$ and the other is nonzero, fixing an element in the nonzero factor gives an injection from $\mu$ into the product. Monotonicity gives \begin{align*} \mu \leq \kappa\cdot\lambda \leq \mu\cdot\mu. \end{align*} Thus, once $\mu\cdot\mu=\mu$ is known, antisymmetry gives \begin{align*} \kappa\cdot\lambda=\mu. \end{align*} It remains to prove $\mu+\mu=\mu$ and $\mu\cdot\mu=\mu$ for every infinite cardinal $\mu$. [/step]

[step:Prove by induction that every infinite initial ordinal absorbs its square] By the [Axiom of Choice](/page/Axiom%20of%20Choice), every cardinal is represented by a unique initial ordinal. We prove, by transfinite induction over infinite initial ordinals $\mu$, that \begin{align*} |\mu\times\mu|=\mu. \end{align*} Fix an infinite initial ordinal $\mu$, and assume that for every infinite initial ordinal $\nu<\mu$, \begin{align*} |\nu\times\nu|=\nu. \end{align*} Define a strict well-order $\prec$ on $\mu\times\mu$ as follows. For $(\alpha,\beta),(\gamma,\delta)\in\mu\times\mu$, set \begin{align*} (\alpha,\beta)\prec(\gamma,\delta) \end{align*} if, in lexicographic order, the triple \begin{align*} \bigl(\max\{\alpha,\beta\},\alpha,\beta\bigr) \end{align*} is less than \begin{align*} \bigl(\max\{\gamma,\delta\},\gamma,\delta\bigr). \end{align*} This is a well-order because it is the pullback of the lexicographic well-order on ordinal triples. Fix $(\gamma,\delta)\in\mu\times\mu$, and define \begin{align*} \rho := \max\{\gamma,\delta\}+1. \end{align*} Then $\rho<\mu$, since $\mu$ is a limit ordinal: every infinite initial ordinal is a limit ordinal. The initial segment below $(\gamma,\delta)$ in the order $\prec$ is contained in $\rho\times\rho$. Let $\nu:=|\rho|$ be the cardinality of $\rho$. Since $\rho<\mu$ and $\mu$ is initial, we have $\nu<\mu$. If $\nu$ is finite, then $|\rho\times\rho|=\nu^2<\mu$ because $\mu$ is infinite. If $\nu$ is infinite, let $\nu$ also denote its initial ordinal representative; by the induction hypothesis, \begin{align*} |\rho\times\rho|=\nu\cdot\nu=\nu<\mu. \end{align*} Thus every proper initial segment of $(\mu\times\mu,\prec)$ has cardinality $<\mu$. Let $\theta$ be the order type of $(\mu\times\mu,\prec)$, and let \begin{align*} e:\theta\to\mu\times\mu \end{align*} be the unique order isomorphism. For each $\xi<\theta$, the initial segment $\{\eta<\theta:\eta<\xi\}$ has cardinality $<\mu$, by the preceding paragraph applied to $e(\xi)$. Hence no ordinal $\xi<\theta$ has cardinality at least $\mu$. Since $\mu$ is the least ordinal of cardinality $\mu$, this implies $\theta\leq\mu$. Therefore there is an injection \begin{align*} \mu\times\mu \hookrightarrow \mu. \end{align*} Conversely, the map \begin{align*} i:\mu &\to \mu\times\mu\\ \alpha &\mapsto (\alpha,0) \end{align*} is injective. By the Cantor--Schroeder--Bernstein theorem, the two injections imply \begin{align*} |\mu\times\mu|=\mu. \end{align*} This completes the induction. [/step]

[step:Deduce additive absorption from multiplicative absorption] Let $\mu$ be an infinite cardinal, represented by its initial ordinal. Define the disjoint union \begin{align*} \mu\sqcup\mu := (\mu\times\{0\})\cup(\mu\times\{1\}). \end{align*} Since $\mu$ is infinite, $2\leq\mu$. Hence there is an injection \begin{align*} j:\mu\sqcup\mu &\to \mu\times\mu\\ (\alpha,\varepsilon) &\mapsto (\alpha,\varepsilon). \end{align*} Using the already proved identity $|\mu\times\mu|=\mu$, we obtain \begin{align*} \mu+\mu = |\mu\sqcup\mu| \leq |\mu\times\mu|=\mu. \end{align*} The reverse inequality $\mu\leq\mu+\mu$ is given by the injection \begin{align*} \alpha\mapsto(\alpha,0). \end{align*} Therefore, by antisymmetry of cardinal comparison, \begin{align*} \mu+\mu=\mu. \end{align*} [/step]

[step:Conclude both absorption identities for the original cardinals] Returning to the cardinals $\kappa$ and $\lambda$, set \begin{align*} \mu=\max\{\kappa,\lambda\}. \end{align*} The preceding steps proved \begin{align*} \mu+\mu=\mu \qquad\text{and}\qquad \mu\cdot\mu=\mu \end{align*} for every infinite cardinal $\mu$. The inequalities from the first step give \begin{align*} \mu \leq \kappa+\lambda \leq \mu+\mu=\mu, \end{align*} and hence \begin{align*} \kappa+\lambda=\mu=\max\{\kappa,\lambda\}. \end{align*} If $\kappa\neq0$ and $\lambda\neq0$, the product inequalities from the first step give \begin{align*} \mu \leq \kappa\cdot\lambda \leq \mu\cdot\mu=\mu, \end{align*} and hence \begin{align*} \kappa\cdot\lambda=\mu=\max\{\kappa,\lambda\}. \end{align*} This proves both asserted absorption identities. [/step]