
theorem Th7: :: Partcard:
for X being set, P being a_partition of X holds card P c= card X
proof
 let X be set, P be a_partition of X;
 defpred P[object,object] means
  ex D1 being set st D1 = $1 & $2 = the Element of D1;
 :: !!! ChoiceFunc
 A1: for e being object st e in P ex u being object st P[e,u]
  proof let e be object;
   assume e in P;
    reconsider ee = e as set by TARSKI:1;
   take u = the Element of ee;
   thus P[e,u];
  end;
  consider f being Function such that
A2: dom f = P and
A3: for e being object st e in P holds P[e,f.e] from CLASSES1:sch 1(A1);
A4: f is one-to-one proof
    let x1, x2 be object such that
   A5: x1 in dom f and
   A6: x2 in dom f and
   A7: f.x1 = f.x2;
   A8: x1 <> {} by A2,A5;
   A9: x2 <> {} by A2,A6;
     reconsider xx1=x1, xx2=x2 as set by TARSKI:1;
     P[x1,f.x1] & P[x2,f.x2] by A5,A6,A2,A3;
     then f.x1 = the Element of xx1 & f.x2 = the Element of xx2;
       then xx1 meets xx2 by A7,A8,A9,XBOOLE_0:3;
    hence x1 = x2 by A5,A6,A2,EQREL_1:def 4;
   end;
  rng f c= X proof
    let y be object;
    assume y in rng f;
     then consider x being object such that
  A10: x in dom f and
  A11: y = f.x by FUNCT_1:def 3;
     reconsider x as set by TARSKI:1;
     P[x,f.x] by A2,A3,A10;
     then
  A12: f.x = the Element of x;
      x <> {} by A2,A10;
      then f.x in x by A12;
    hence y in X by A10,A2,A11;
  end;
 hence card P c= card X by A2,A4,CARD_1:10;
end;
