reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th39:
  for x, y, z being Element of L holds ((x | y) | y) | (y | (z | x
  )) = (y | (z | x)) | (y | (z | x))
proof
  let x, y, z be Element of L;
  set X = y | (z | x);
  X | (x | y) = y by Th23;
  hence thesis by Th36;
end;
