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 Th26:
  O1 c= O2 implies L\&O1 c= L\&O2
  proof
    assume
A1: O1 c= O2;
    let z be object; assume
A2: z in L\&O1; then
    {x.O1: x in L} <> {} by SETFAM_1:def 1; then
    consider Y being object such that
A3: Y in {x.O1: x in L} by XBOOLE_0:def 1;
    consider y such that
A4: Y = y.O1 & y in L by A3;
A5: y.O2 in {x.O2: x in L} by A4;
    now
      let Y; assume Y in {x.O2: x in L}; then
      consider a such that
A6:   Y = a.O2 & a in L;
      a.O1 in {x.O1: x in L} by A6; then
      z in a.O1 & a.O1 c= Y by A1,A2,A6,Th1,SETFAM_1:def 1;
      hence z in Y;
    end;
    hence thesis by A5,SETFAM_1:def 1;
  end;
