theorem Th3:
  ex BSeq st for n holds BSeq.n = ASeq.n /\ B
proof
  deffunc F(Element of NAT) = ASeq.$1 /\ B;
  consider f being Function such that
A1: dom f = NAT &
   for n being Element of NAT holds f.n = F(n) from FUNCT_1:sch 4;
  now
   let m;
   reconsider mm=m as Element of NAT by ORDINAL1:def 12;
    ASeq.m /\ B in bool Omega;
    then f.mm in bool Omega by A1;
   hence f.m in bool Omega;
  end;
  then for x being object st x in NAT holds f.x in bool Omega;
  then reconsider f as SetSequence of Omega by A1,FUNCT_2:3;
  now
    let m be Nat;
    reconsider m1 = m as Element of NAT by ORDINAL1:def 12;
    ASeq.m1 /\ B in Sigma;
    hence f.m in Sigma by A1;
  end;
  then rng f c= Sigma by NAT_1:52;
  then reconsider f as SetSequence of Sigma by RELAT_1:def 19;
  take f;
  let n;
   reconsider n as Element of NAT by ORDINAL1:def 12;
   f.n = F(n) by A1;
  hence thesis;
end;
