theorem Th129:
  for x,z,y holds (((y | y) | z) | (x | z)) | (((y | y) | z) | (x
  | z)) = ((x | x) | ((y | y) | z)) | z
proof
  let x,z,y;
  ((z | ((z | z) | y)) | (z | ((z | z) | y))) = z by Th128;
  hence thesis by Th117;
end;
