reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th65:
  for x, y, z being Element of L holds (x | (y | y)) | (x | (z | z
  )) = (x | (z | y)) | (x | (z | y))
proof
  let x, y, z be Element of L;
  set X = x | (y | y);
  (x | (y | y)) | (x | (z | z)) = (x | (y | y)) | (x | (z | X)) by Th61;
  hence thesis by Th64;
end;
