theorem Th16:
  Ant(f) is_Subsequence_of Ant(g) & Suc(f) = Suc(g) & Ant(f) |=
  Suc(f) implies Ant(g) |= Suc(g)
proof
  assume that
A1: Ant(f) is_Subsequence_of Ant(g) and
A2: Suc(f) = Suc(g) & Ant(f) |= Suc(f);
  let A,J,v such that
A3: J,v |= rng(Ant(g));
  now
    let p;
    assume
A4: p in rng(Ant(f));
    rng(Ant(f)) c= rng(Ant(g)) by A1,Th1;
    hence J,v |= p by A3,A4;
  end;
  then J,v |= rng(Ant(f));
  then J,v |= Ant(f);
  hence thesis by A2;
end;
