reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th51:
  for x, y, z being Element of L holds (x | ((y | z) | x)) | (y |
  y) = (y | z) | (y | y)
proof
  let x, y, z be Element of L;
  set Y = y | z;
  set X = x | ((y | z) | x);
  (Y | Y) | (X | Y) = Y by Th11;
  hence thesis by Th50;
end;
