Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 12F / e / Solution

Since the empty language does not lie inRice's construction gives $K{\le }_m\mathbf{Two}$. Conversely, $\mathbf{Two}$ is computably enumerable: dovetail the enumerator for $W_w$ until two distinct words appear, then accept. The diagonal halting set $K$ is many-one complete for computably enumerable sets, so $\mathbf{Two}{\le }_{m}K$. Therefore