reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th49:
  for x, y, z being Element of L holds x | ((y | y) | (z | (x | (x
  | y)))) = x | y
proof
  let x, y, z be Element of L;
  (y | y) | (y | y) = y by Th21;
  hence thesis by Th47;
end;
