proposition: Lemma eqseq_pivot2r x s1 s2 s3 s4 : x \notin s2 -> x \notin s4 -> (s1 ++ x :: s2 == s3 ++ x :: s4) = (s1 == s3) && (s2 == s4).
proof: Proof. move=> xNs2 xNs4; rewrite -(can_eq revK) !rev_pivot. by rewrite eqseq_pivot2l ?mem_rev // !(can_eq revK) andbC. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat)