 The set of combinators \(\{\textsf{S},\textsf{K}\}\) is called a combinatory base, that is, these two combinators are the undefined constants of \(\text{CL}_\triangleright\). To give an idea of a proof in this calculus, the following steps may be pieced together to prove \(\textsf{SKK} (\textsf{KSK})\triangleright \textsf{S}\).  \(\textsf{KSK}\triangleright \textsf{S}\) is an instance of an axiom. Then \(\textsf{SKK}(\textsf{KSK}) \triangleright\textsf{SKKS}\) is obtained by right monotonicity, and further, \(\textsf{SKK}(\textsf{KSK})\triangleright\textsf{S}\) results by instances of the \(\textsf{S}\) and \(\textsf{K}\) axioms together with applications of the transitivity rule.