theorem
  not x in X & not y in X implies X = X \/ {x,y} \ {x,y}
proof
A1: (X \/ {x,y}) \ {x,y} = X \ {x,y} by XBOOLE_1:40;
  assume ( not x in X)& not y in X;
  hence thesis by A1,Th119;
end;
