reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th12:
  for x, y being Element of L holds x | (y | (x | x)) = x | x
proof
  let x, y be Element of L;
  set Y = (x | x);
  Y | (x | x) = x by Th11;
  hence thesis by Th11;
end;
