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),

$\kappa \oplus \mu =\lambda \oplus \mu ⟹\kappa =\lambda \mathrm{.}$

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 $\mid A\mid$ in the finite sense equal to $\mid A\mid$ 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 $\lambda \mathrm{\ne }0$) 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 $\mid A\mid$ in the finite sense equal to $\mid A\mid$ 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 $\lambda \mathrm{\ne }0$ also gives ${\mathrm{\aleph }}_0\otimes {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0=1\otimes {\mathrm{\aleph }}_0$ with $1\mathrm{\ne }{\mathrm{\aleph }}_0$, 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, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$.