
theorem th1: ::: isn't it just not X in X?
for X being set ex o being object st not o in X
proof
let X be set;
now assume for u being object st u in bool X holds u in X;
  then bool X c= X;
  then B: card bool X c= card X by CARD_1:11;
  card X in card bool X by CARD_1:14;
  hence contradiction by B,CARD_1:3;
  end;
  hence thesis;
end;
