reserve
  X for non empty set,
  FX for Filter of X,
  SFX for Subset-Family of X;

theorem Th29:
  for X be set, A be Subset of X holds
  {B where B is Element of BoolePoset X: A c= B} =
  {B where B is Subset of X: A c= B}
  proof
    let X be set, A be Subset of X;
    set C = {B where B is Element of BoolePoset X: A c= B};
    set D = {B where B is Subset of X: A c= B};
    now
      hereby
        let x be object;
        assume x in C;
        then consider b0 be Element of BoolePoset X such that
A1:     x=b0 and
A2:     A c= b0;
        x is Subset of X by A1,WAYBEL_8:26;
        hence x in D by A1,A2;
      end;
      let x be object;
      assume x in D;
      then consider b0 be Subset of X such that
A3:   x=b0 and
A4:   A c= b0;
      x is Element of BoolePoset X by A3,WAYBEL_8:26;
      hence x in C by A3,A4;
    end;
    then C c= D & D c= C;
    hence thesis;
  end;
