reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th18:
  for x, y, z being Element of L holds ((x | (y | z)) | z) | x = x | (y | z)
proof
  let x, y, z be Element of L;
  set X = y | z;
  (x | X) | (x | z) = x by Th16;
  hence thesis by Th15;
end;
