theorem Th100:
  for q,z,x holds (q | (x | z)) = ((z | ((x | x) | q)) | ((q | q)
  | ((x | x) | q)))
proof
  let q,z,x;
  (z | z) | (z | z) = z by SHEFFER1:def 13;
  hence thesis by Th99;
end;
