reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th21:
  for x, y being Element of L holds (x | y) | (x | x) = x
proof
  let x, y be Element of L;
  set X = x | y;
  x | ((y | x) | x) = X by Th19;
  hence thesis by Th16;
end;
