proposition: Lemma subset_respects_set_eq : forall {T : Type} {s1 s2 s1' s2' : set T}, s1 <--> s1' -> s2 <--> s2' -> s1' \subset s2' -> s1 \subset s2.
proof: Proof. firstorder. Qed.
imports: mathcomp.ssreflect.ssreflect Classes.RelationClasses Classes.Morphisms List Tactics mathcomp(ssrfun ssrbool ssrnat seq)