reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th48:
  for x, y, z being Element of L holds ((x | (y | z)) | (x | (y |
  z))) | (y | y) = x | (y | y)
proof
  let x, y, z be Element of L;
  set Z = (y | y);
  set Y = (y | z);
  (x | Y) | (x | Y) = (x | Y) | (x | Z) by Th46;
  hence thesis by Th26;
end;
