theorem Th16:
  <%>E in B implies A c= A ^^ B & A c= B ^^ A
proof
  assume
A1: <%>E in B;
  thus A c= A ^^ B
  proof
    let x be object;
    assume
A2: x in A;
    then reconsider xa = x as Element of E^omega;
    xa ^ {} in A ^^ B by A1,A2,Def1;
    hence thesis;
  end;
  thus A c= B ^^ A
  proof
    let x be object;
    assume
A3: x in A;
    then reconsider xa = x as Element of E^omega;
    {} ^ xa in B ^^ A by A1,A3,Def1;
    hence thesis;
  end;
end;
