reserve i for Nat,
  j for Element of NAT,
  X,Y,x,y,z for set;

theorem Th23:
  Vars c= Rank omega proof consider V being ManySortedSet of NAT such that
A1: Vars = Union V and
A2: V.0 = the set of all [{}, i] where i is Element of NAT and
A3: for n being Nat holds V.(n+1) =
  {[varcl a, j] where a is Subset of V.n, j is Element of NAT: a is finite}
  by Def2;
  let x be object;
  assume x in Vars;
  then consider i being object such that
A4: i in dom V and
A5: x in V.i by A1,CARD_5:2;
  reconsider i as Element of NAT by A4;
  defpred P[Nat] means V.$1 c= Rank omega;
A6: P[ 0]
  proof
    let x be object;
    assume x in V.0;
    then consider i being Element of NAT such that
A7: x = [{}, i] by A2;
A8: Segm(i+1) = succ Segm i by NAT_1:38;
A9: {} c= i;
A10: i in i+1 by A8,ORDINAL1:6;
A11: {} in i+1 by A8,A9,ORDINAL1:6,12;
A12: the_rank_of {} = {} by CLASSES1:73;
A13: the_rank_of i = i by CLASSES1:73;
A14: {} in Rank (i+1) by A11,A12,CLASSES1:66;
    i in Rank (i+1) by A10,A13,CLASSES1:66;
    then
A15: x in Rank succ succ (i+1) by A7,A14,CLASSES1:45;
    succ succ (i+1) c= omega;
    then Rank succ succ (i+1) c= Rank omega by CLASSES1:37;
    hence thesis by A15;
  end;
A16: now
    let n be Nat such that
A17: P[n];
A18: V.(n+1) = {[varcl a, j] where a is Subset of V.n, j is Element of NAT:
    a is finite} by A3;
    thus P[n+1]
    proof
      let x be object;
      assume x in V.(n+1);
      then consider a being Subset of V.n, j being Element of NAT such that
A19:  x = [varcl a, j] and
A20:  a is finite by A18;
      a c= Rank omega by A17,XBOOLE_1:1;
      then a in Rank omega by A20,Th22;
      then reconsider i = the_rank_of a as Element of NAT by CLASSES1:66;
      reconsider k = j \/ i as Element of NAT by ORDINAL3:12;
A21:  the_rank_of varcl a = i by Th21;
A22:  the_rank_of j = j by CLASSES1:73;
A23:  k in succ k by ORDINAL1:6;
      then
A24:  i in succ k by ORDINAL1:12,XBOOLE_1:7;
A25:  j in succ k by A23,ORDINAL1:12,XBOOLE_1:7;
A26:  succ Segm k = Segm(k+1) by NAT_1:38;
      then
A27:  varcl a in Rank (k+1) by A21,A24,CLASSES1:66;
      j in Rank (k+1) by A22,A25,A26,CLASSES1:66;
      then
A28:  x in Rank succ succ (k+1) by A19,A27,CLASSES1:45;
      succ succ (k+1) c= omega;
      then Rank succ succ (k+1) c= Rank omega by CLASSES1:37;
      hence thesis by A28;
    end;
  end;
  for n being Nat holds P[n] from NAT_1:sch 2(A6,A16);
  then V.i c= Rank omega;
  hence thesis by A5;
end;
