 Nimber multiplication

For finite nimbers, multiplication is recursively defined from smaller arguments bywhere $\oplus$ is Nim-sum and mex selects the smallest missing nonnegative integer. Thus zero times anything is zero and one times $b$ is $b$. For $a=b=2$, the displayed set is $\{0,2,3\}$, so $2\otimes 2=1$.

Together with Nim-sum, this is field arithmetic on finite nimbers, rather than ordinary integer multiplication. In particular $2\otimes 2=1$ cannot be computed by a usual integer product or bitwise exclusive or.^[ref]