reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th32:
  for x, y being Element of L holds x | (y | (x | y)) = x | x
proof
  let x, y be Element of L;
  (x | y) | (x | ((x | y) | y)) = x by Th15;
  hence thesis by Th30;
end;
