proposition: Lemma subset_nil xs ys : {subset xs <= ys} -> ys = [::] -> xs = [::].
proof: Proof. by move=>X E; move: E X=>->; apply: subset_nilR. Qed.
imports: Coq(ssreflect ssrbool ssrfun) mathcomp(ssrnat seq path eqtype choice bigop) pcm(options prelude pred)