reserve L for satisfying_Sh_1 non empty ShefferStr;

theorem Th26:
  for x, y, z being Element of L holds ((x | y) | (x | z)) | z = x | z
proof
  let x, y, z be Element of L;
  (x | z) | (z | (x | y)) = z by Th24;
  hence thesis by Th22;
end;
