reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th37:
  for x, y, z being Element of L holds (x | y) | z = z | (y | x)
proof
  let x, y, z be Element of L;
  set X = x | y;
  X | z = z | X by Th20;
  hence thesis by Th33;
end;
