A new proof of a result of Higman

Given two strings x, y ∈ Σ∗, say that x is a subsequence of y (denoted x y) if x results from removing zero or more characters from y. For a language L ⊆ Σ∗, define SUBSEQ(L) to be the set of all subsequences of strings in L. We give a new proof of a result of Higman, which states, If L is any language over a finite alphabet, then SUBSEQ(L) is regular. Higman’s original proof makes use of the theory of well quasi-orders. The current proof is completely different and makes no mention of well quasi-orders. It also provides a different insight into the relationships between L and SUBSEQ(L) for various L. Fix an alphabet Σ. For x,w ∈ Σ∗ we let x w denote the condition that x is a subsequence of w, that is, x = x1x2 · · ·xn for x1, x2, . . . , xn ∈ Σ, and y ∈ L(Σx1Σx2Σ · · ·ΣxnΣ). For a language L ⊆ Σ∗, define SUBSEQ(L) := {x ∈ Σ∗ | (∃w ∈ L) x w }. The following theorem was essentially proved by Higman [1] using well quasi-order theory. Theorem 1 (Higman [1]). SUBSEQ(L) is regular for any L ⊆ Σ∗. Clearly, SUBSEQ(SUBSEQ(L)) = SUBSEQ(L) for any L, since is transitive. We’ll say that L is -closed if L = SUBSEQ(L). So Theorem 1 is equivalent to the statement that a language L is regular if L is -closed. The remainder of this note is to prove Theorem 1. We do so directly, without recourse to well quasi-orders. ∗Computer Science and Engineering Department, Columbia, SC 29208. [email protected]