reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th10:
  for x, y being Element of L holds x | ((((y | x) | (y | x)) | x
  ) | x) = y | x
proof
  let x, y be Element of L;
  ((y | x) | (((y | x) | (y | x)) | (y | x))) |((y | x) | (y | x)) = y | x
  by Th6;
  hence thesis by Th9;
end;
