Solution
= Solution
The promise is <Simon's problem>. If $z=0^n$, then $f(x)=f(y)$ implies $x=y$, so $f$ is an <injective function>. If $z\ne0^n$, then each value has exactly the two preimages $x$ and $x\mathbin\oplus z$; thus $f$ is <two-to-one function>[two-to-one] and is constant on the two-element cosets of $\{0^n,z\}\leq(\mathbb Z_2)^n$.