theorem Th130:
  for x,z,y holds ((((y | y) | z) | (x | z)) | (((y | y) | z) | (
  x | z))) = z | (y | (x | x))
proof
  let x,z,y;
  (((x | x) | ((y | y) | z)) | z) = z | (y | (x | x)) by Th125;
  hence thesis by Th129;
end;
