proposition: Lemma lseq_pos xs p h : p != null -> h \In lseq p xs -> exists x, exists r, exists h', [/\ xs = x :: behead xs, p :-> x :+ (p .+ 1 :-> r :+ h') = h & h' \In lseq r (behead xs)].
proof: Proof. by apply: lseg_neq. Qed.
imports: mathcomp(ssreflect ssrbool eqtype seq ssrfun) LemmaOverloading(heaps rels hprop stmod stsep stlogR)