reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th15:
  for x, y, z being Element of L holds (x | y) | (x | (z | y)) = x
proof
  let x, y, z be Element of L;
  y | ((x | y) | y) = x | y by Th14;
  hence thesis by Def1;
end;
