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 <> {} & L1 c= L2 implies L2 \& O c= L1 \& O
  proof
    assume A1: L1 <> {};
    assume A2: L1 c= L2;
    let z be object;
    assume
A3: z in L2 \& O;
    now
      set c = the Element of L1;
      c in L1 by A1; then
      reconsider c as Element of X;
      c.O in {x.O: x in L1} by A1;
      hence {x.O: x in L1} <> {};
      let Y; assume Y in {x.O: x in L1}; then
      consider x such that
A4:   Y = x.O & x in L1;
      Y in {a.O: a in L2} by A4,A2;
      hence z in Y by A3,SETFAM_1:def 1;
    end;
    hence z in L1 \& O by SETFAM_1:def 1;
  end;
