theorem Th8: for P be complete PNPair st untn(A,B) in rng P
  holds A in rng P & B in rng P & A 'U' B in rng P
  proof
    let P be complete PNPair;
    assume
A1: untn(A,B) in rng P;
    tau rng P = rng P by LTLAXIO3:def 11;
    hence A in rng P & B in rng P & A 'U' B in rng P by A1,LTLAXIO3:22;
  end;
