reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th40:
  for x, y, z, u being Element of L holds (x | y) | (z | u) = (u | z) | (y | x)
proof
  let x, y, z, u be Element of L;
  (x | y) | (z | u) = (x | y) | (u | z) by Th33;
  hence thesis by Th37;
end;
