reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th30:
  for x, y, z being Element of L holds (x | (y | (x | z))) | y = y | (z | x)
proof
  let x, y, z be Element of L;
  z | x = x | z by Th20;
  hence thesis by Th28;
end;
