reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th8:
  for x, y being Element of L holds (x | (((y | y) | x) | x)) | y = y | y
proof
  let x, y be Element of L;
  set Y = y | y;
  set X = x | y;
  Y | (y | (X | y)) = y by Th7;
  hence thesis by Th3;
end;
