It is easy to see that \(A \leq_m B\) implies \(A \leq_T B\). (For if \(f(x)\) is a \(m\)-reduction of \(A\) to \(B\), then consider the program which first computes \(f(n)\) and then, using \(B\) an as oracle, checks if \(f(n) \in B\), outputting 1 if so and 0 if not.) It thus follows that \(K\) is also Turing complete—i.e., it embodies the maximum “degree of unsolvability” among the the c.e. sets when this notion is understood in terms of Turing reducibility as well as many-one reducibility.