“…,n, in the previous equation contains a leading word of a relation in S ∪ R, as a subword. Proving by contradiction, let us assume, say, that u(1) f (2)v(1) contains such leading word w. If w has no letters x, y then w is a subword in f (2), which is impossible. Otherwise, w = u f i v, for some i, where the "wings" u and v are in the set of non-overlapping words M. As just above, we cannot have w completely inside f (2) and so either u has overlapping with u(1) (hence, equal to u(1)) or v has overlapping v(1) (hence, equal to u(1)), or both.…”