theorem
  card PX <= card X
proof
  assume card PX > card X;
  then card Segm card X in card Segm card PX by NAT_1:41;
  then consider Pi being object such that
A1: Pi in PX and
A2: for x being object st x in X holds (proj PX).x <> Pi by Th66;
  reconsider Pi as Element of PX by A1;
  consider q being Element of X such that
A3: q in Pi by Th85;
  reconsider Pq = (proj PX).q as Element of PX;
A4: Pq = Pi or Pq misses Pi by EQREL_1:def 4;
  q in Pq by EQREL_1:def 9;
  hence contradiction by A2,A3,A4,XBOOLE_0:3;
end;
