reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th17:
  for x, y, z being Element of L holds x | ((x | y) | (z | y)) = x | y
proof
  let x, y, z be Element of L;
  (x | y) | (x | (z | y)) = x by Th15;
  hence thesis by Th16;
end;
