reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th41:
  for x, y, z being Element of L holds x | (y | ((y | x) | z)) = x | (y | y)
proof
  let x, y, z be Element of L;
  x | (y | (z | (x | y))) = x | (y | ((y | x) | z)) by Th37;
  hence thesis by Th38;
end;
