reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th36:
  for x, y being Element of L holds (x | (y | x)) | y = y | y
proof
  let x, y be Element of L;
  set X = x;
  set Y = y;
  Y | (X | (Y | X)) = (X | (Y | X)) | Y by Th20;
  hence thesis by Th32;
end;
