reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th9:
  for x, y being Element of L holds ((x | y) | (((x | y) | (x | y))
  | (x | y))) | ((x | y) | (x | y)) = y | ((((x | y) | (x | y )) | y) | y)
proof
  let x, y be Element of L;
  y | ((((x | y) | (x | y)) | y) | y) | (x | y) = (x | y) | (x | y) by Th8;
  hence thesis by Th2;
end;
