proposition: Lemma monoidfunassoc {X Y Z W : monoid} (f : monoidfun X Y) (g : monoidfun Y Z) (h : monoidfun Z W) : monoidfuncomp f (monoidfuncomp g h) = monoidfuncomp (monoidfuncomp f g) h.
proof: Proof. use monoidfun_paths. use idpath. Qed.
imports: UniMath.Algebra.BinaryOperations UniMath.MoreFoundations.Subtypes UniMath.MoreFoundations.Sets UniMath.MoreFoundations.Orders