reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th5:
  for x being Element of L holds x | ((x | x) | x) = x | x
proof
  let x be Element of L;
  (x | ((x | x) | x)) | (x | (x | ((x | x) | x))) = x by Th3;
  hence thesis by Th4;
end;
