reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th11:
  for x, y being Element of L holds (x | x) | (y | x) = x
proof
  let x, y be Element of L;
  set Y = y | x;
  x | (((Y | Y) | x) | x) = Y by Th10;
  hence thesis by Th7;
end;
