reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th4:
  for x, y being Element of L holds x | ((x | ((x | x) | x)) | (y |
  (x | ((x | 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 Th1;
end;
