reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th52:
  for x, y, z being Element of L holds (x | ((y | z) | x)) | (y | y) = y
proof
  let x, y, z be Element of L;
  (y | z) | (y | y) = y by Th21;
  hence thesis by Th51;
end;
