theorem Th5:
  (for n holds BSeq.n = ASeq.n /\ B) implies (Intersection ASeq)
  /\ B = Intersection BSeq
proof
  assume
A1: for n holds BSeq.n = ASeq.n /\ B;
A2: now
    let x be object;
    assume
A3: x in (Intersection BSeq);
A4: for n holds x in ASeq.n /\ B
    proof
      let n;
      x in BSeq.n by A3,PROB_1:13;
      hence thesis by A1;
    end;
A5: for n holds x in ASeq.n & x in B
    proof
      let n;
      x in ASeq.n /\ B by A4;
      hence thesis by XBOOLE_0:def 4;
    end;
    then x in (Intersection ASeq) by PROB_1:13;
    hence x in (Intersection ASeq) /\ B by A5,XBOOLE_0:def 4;
  end;
  now
    let x be object;
    assume
A6: x in (Intersection ASeq) /\ B;
    then
A7: x in (Intersection ASeq) by XBOOLE_0:def 4;
A8: for n holds x in ASeq.n /\ B
    proof
      let n;
      x in ASeq.n & x in B by A6,A7,PROB_1:13,XBOOLE_0:def 4;
      hence thesis by XBOOLE_0:def 4;
    end;
    for n holds x in BSeq.n
    proof
      let n;
      x in ASeq.n /\ B by A8;
      hence thesis by A1;
    end;
    hence x in (Intersection BSeq) by PROB_1:13;
  end;
  hence thesis by A2,TARSKI:2;
end;
