reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th19:
  for x, y being Element of L holds x | ((y | x) | x) = x | y
proof
  let x, y be Element of L;
  (x | ((y | x) | x)) | (y | ((y | x) | x)) = y by Def1;
  hence thesis by Th17;
end;
