FALSE: $\kappa \oplus \mu = \lambda \oplus \mu$ implies $\kappa = \lambda$

FALSE. Cardinal addition is cancellative: for all cardinals $\kappa, \lambda, \mu$ (Cardinal (initial ordinal) and cardinality, Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations),

$\kappa \oplus \mu = \lambda \oplus \mu \quad \Longrightarrow \quad \kappa = \lambda .$

The claim is plausible because it is true for finite cardinals, where $\oplus$ is the ordinary addition of natural numbers (Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense) and cancellation is a Peano fact. It fails at the first infinite cardinal, and it fails for the same reason that infinite arithmetic is easy: absorption (Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda = \kappa$, and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$) makes $\oplus$ throw away the smaller argument, and an operation that forgets one of its inputs cannot be cancelled.

The finite case really is cancellative, and nothing above contradicts it. For $m, n, k \in \omega$ read as cardinals, $m \oplus k = n \oplus k$ is $m +_{\mathbb{N}} k = n +_{\mathbb{N}} k$ by Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\lvert A \rvert$ in the finite sense equal to $\lvert A \rvert$ in the cardinal sense, and Peano addition is cancellative. The witness above is forced to use an infinite $\mu$, and once $\mu$ is infinite every $\kappa \le \mu$ gives the same sum.

Multiplication fails in the same way, and for the same reason. Absorption: for cardinals $\kappa, \lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda = \kappa$, and $\kappa \otimes \lambda = \kappa$ when $\lambda \ne 0$ also gives $\aleph_0 \otimes \aleph_0 = \aleph_0 = 1 \otimes \aleph_0$ with $1 \ne \aleph_0$, so $\otimes$ is not cancellative either, even away from the trivial obstruction at $\mu = 0$.

What survives. Monotonicity survives: $\kappa \le \lambda$ still gives $\kappa \oplus \mu \le \lambda \oplus \mu$ (Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$, the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$). It is the strict form that fails, and cancellation is exactly the strict form in disguise. Exponentiation is the one place on this page where a strict increase survives at every cardinal, and that is Assuming the Axiom of Choice, $2^{\kappa} = \lvert \mathcal{P}(\kappa) \rvert$, and Cantor's theorem in cardinal form: $\kappa < 2^{\kappa}$.