 We could ask a slightly different question: If we start with the base \(\{\textsf{S},\textsf{K}\}\) and we omit \(\textsf{S}\), then we get the base \(\{\textsf{K}\}\) and all the combinators strongly normalize, but what if we omit \(\textsf{K}\)? Do the combinators over \(\{\textsf{S}\}\) strongly normalize or at least normalize? The answer is “no.” A term (discovered by Henk Barendregt in the early 1970s) that shows the lack of strong normalization is \(\textsf{SSS}(\textsf{SSS})(\textsf{SSS})\). The first \(\textsf{S}\) is the head of a (indeed, the only) redex, and the head reduction sequence of this term is infinite. Since \(\{\textsf{S}\}\) does not contain any combinator with a cancellative effect, the existence of an infinite reduction sequence for a term means that the term has no nf. There are shorter combinators over the base \(\{\textsf{S}\}\) without an nf, for example, \(\textsf{S}(\textsf{SS})\textsf{SSSSS}\) comprises only eight occurrences of \(\textsf{S}\).