reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th22:
  for x, y, z being Element of L holds (x | y) | (y | (z | x)) = y
proof
  let x, y, z be Element of L;
  set Y = x | ((y | x) | x);
  Y = x | y by Th19;
  hence thesis by Def1;
end;
