reserve f for Function;
reserve p,q for FinSequence;
reserve A,B,C for set,x,x1,x2,y,z for object;
reserve k,l,m,n for Nat;
reserve a for Nat;
reserve D for non empty set;
reserve d,d1,d2,d3 for Element of D;
reserve L,M for Element of NAT;
reserve f for Function of A,B;
reserve f for Function;
reserve x1,x2,x3,x4,x5 for object;
reserve p for FinSequence;
reserve ND for non empty set;
reserve y1,y2,y3,y4,y5 for Element of ND;
reserve X, A for non empty finite set,
  PX for a_partition of X,
  PA1, PA2 for a_partition of A;

theorem
  PA1 is_finer_than PA2 implies card PA2 <= card PA1
proof
  defpred P[object,object] means
     ex A,B being set st A = $1 & B = $2 & A c= B;
  assume
A1:  PA1 is_finer_than PA2;
A2: for e being object st e in PA1 ex u being object st u in PA2 & P[e,u]
   proof let e be object;
      reconsider ee=e as set by TARSKI:1;
     assume e in PA1;
     then ee in PA1;
     then ex u being set st u in PA2 & ee c= u by SETFAM_1:def 2,A1;
     hence thesis;
   end;
  consider f being Function of PA1, PA2 such that
A3: for e being object st e in PA1 holds P[e,f.e] from FUNCT_2:sch 1 (A2);
  assume card PA1 < card PA2;
  then card Segm card PA1 in card Segm card PA2 by NAT_1:41;
  then consider p2i being object such that
A4: p2i in PA2 and
A5: for x being object st x in PA1 holds f.x <> p2i by Th66;
  reconsider p2i as Element of PA2 by A4;
  consider q being Element of A such that
A6: q in p2i by Th85;
  reconsider p2q = f.((proj PA1).q) as Element of PA2;
A7: p2q=p2i or p2q misses p2i by EQREL_1:def 4;
  P[((proj PA1).q),f.((proj PA1).q)] by A3;
  then q in (proj PA1).q & (proj PA1).q c= p2q by EQREL_1:def 9;
  hence contradiction by A5,A6,A7,XBOOLE_0:3;
end;
