reserve A,A1,A2,B,C,D for Ordinal,
  X,Y for set,
  x,y,a,b,c for object,
  L,L1,L2,L3 for Sequence,
  f for Function;

theorem
  A in sup X implies ex B st B in X & A c= B
proof
  assume that
A1: A in sup X and
A2: for B st B in X holds not A c= B;
  for B st B in X holds B in A by A2,ORDINAL1:16;
  then sup X c= A by Th20;
  hence contradiction by A1,ORDINAL1:5;
end;
