 This illustrates a sense in which deciding membership in \(K\) can also be understood as universal for computably enumerable sets or, alternatively, that there is no c.e. set which is any “harder” to solve than \(K\). Nonetheless, there are problems that are harder to solve than \(K\) in the sense that they could not be solved even if we possessed a decision algorithm for \(K\). For instance, it will follow from results given below that while \(K\) is \(m\)-reducible to \(\TOT\), \(\TOT\) is not \(m\)-reducible to \(K\). This illustrates how \(m\)-reducibility can be used to study the relative difficulty of solving computational problems.