reserve X,Y,z,s for set, L,L1,L2,A,B for List of X, x for Element of X,
  O,O1,O2,O3 for Operation of X, a,b,y for Element of X, n,m for Nat;

theorem
  L1 c= L2 & O1 c= O2 implies L1|O1 c= L2|O2
  proof
    assume L1 c= L2 & O1 c= O2; then
    L1|O1 c= L2|O1 & L2|O1 c= L2|O2 by RELAT_1:123,124;
    hence thesis;
  end;
