
theorem Th10:
for X being set holds
 card { {X,[x,X]} where x is Element of X : x in X } = card X
proof
 let X be set;
 set uG = X;  set A = { {uG,[x,uG]} where x is Element of uG : x in uG };
 deffunc F(object) = {uG,[$1,uG]};
 consider f being Function such that
A1: dom f = uG and
A2: for x being object st x in uG holds f.x = F(x) from FUNCT_1:sch 3;
   now
     let x1,x2 be object;
     assume that A3: x1 in dom f and A4: x2 in dom f and A5: f.x1 = f.x2;
       F(x1) = f.x1 & F(x2) = f.x2 by A3,A4,A1,A2;
      then [x1,uG] = uG or [x1,uG] = [x2,uG] by A5,ZFMISC_1:6;
     hence x1 = x2 by Th2,XTUPLE_0:1;
   end;
then A6: f is one-to-one;
A7: rng f = A proof
     thus rng f c= A proof
       let y be object;
       assume y in rng f;
       then consider a being object such that
     A8: a in dom f and
     A9: f.a = y by FUNCT_1:def 3;
       y = {uG,[a,uG]} by A8,A9,A1,A2;
      hence thesis by A8,A1;
     end;
     thus A c= rng f proof
       let a be object;
       assume a in A;
       then consider x being Element of uG such that
     A10: a = {uG,[x,uG]} and
     A11: x in uG;
       f.x = a by A10,A11,A2;
      hence thesis by A1,A11,FUNCT_1:def 3;
     end;
   end;
   A, uG are_equipotent by A1,A6,A7,WELLORD2:def 4;
 hence card A = card uG by CARD_1:5;
end;
