theorem
  Lang(w, id (E^omega)) = {w}
proof
  {}(E^omega, E^omega) \/ id (E^omega) = {} \/ id (E^omega) by PARTIT_2:def 1
    .= id (E^omega);
  hence Lang(w, id (E^omega)) = Lang(w, {}(E^omega, E^omega)) by Th49
    .= {w} by Th50;
end;
