reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th14:
  for x, y being Element of L holds x | ((y | x) | x) = y | x
proof
  let x, y be Element of L;
  ((y | x) | (y | x)) | x = y | x by Th13;
  hence thesis by Th10;
end;
