reserve Al for QC-alphabet;
reserve a,a1,a2,b,c,d for set,
  X,Y,Z for Subset of CQC-WFF(Al),
  i,k,m,n for Nat,
  p,q for Element of CQC-WFF(Al),
  P for QC-pred_symbol of k,Al,
  ll for CQC-variable_list of k,Al,
  f,f1,f2,g for FinSequence of CQC-WFF(Al);
reserve A for non empty finite Subset of NAT;
reserve C for non empty set;

theorem Th3:
  for f being sequence of C, X being finite set st (for n,m st
m in dom f & n in dom f & n < m holds f.n c= f.m) & X c= union rng f holds ex k
  st X c= f.k
proof
  let f be sequence of C, X be finite set such that
A1: for n,m st m in dom f & n in dom f & n < m holds f.n c= f.m and
A2: X c= union rng f;
A3: now
    deffunc F(object) = min* {i : $1 in f.i};
    consider g being Function such that
A4: dom g = X &
for a being object st a in X holds g.a = F(a) from FUNCT_1:sch 3;
    reconsider A = rng g as finite set by A4,FINSET_1:8;
A5: now
      let b be object;
      assume b in A;
      then consider a being object such that
A6:   a in dom g & g.a = b by FUNCT_1:def 3;
      b = min* {i : a in f.i} by A4,A6;
      hence b in NAT;
    end;
    assume not X is empty;
    then ex c being object st c in dom g by A4;
    then reconsider A as non empty finite Subset of NAT by A5,FUNCT_1:3
,TARSKI:def 3;
    union A in A by Th2;
    then reconsider a = union A as Nat;
    take a;
    thus X c= f.a
    proof
      let b be object;
      assume
A7:   b in X;
      then consider c such that
A8:   b in c and
A9:   c in rng f by A2,TARSKI:def 4;
      consider d being object such that
A10:  d in dom f and
A11:  f.d = c by A9,FUNCT_1:def 3;
      reconsider d as Nat by A10;
      d in {i : b in f.i} by A8,A11;
      then reconsider Y = {i : b in f.i} as non empty set;
      now
        let a be object;
        assume a in Y;
        then ex i st i = a & b in f.i;
        hence a in NAT by ORDINAL1:def 12;
      end;
      then reconsider Y as non empty Subset of NAT by TARSKI:def 3;
A12:  g.b = min* Y by A4,A7;
      then reconsider Y9 = g.b as Nat;
      Y9 in Y by A12,NAT_1:def 1;
      then
A13:  ex i st i = g.b & b in f.i;
A14:  dom f = NAT by FUNCT_2:def 1;
A15:  now
        assume Y9 in a;
        then Y9 in {k where k is Nat : k < a} by AXIOMS:4;
        then consider k being Nat such that
A16:       k = Y9 & k < a;
A17:     a in NAT by ORDINAL1:def 12;
        Y9 in NAT by ORDINAL1:def 12;
        then f.Y9 c= f.a by A1,A14,A16,A17;
        hence thesis by A13;
      end;
      Y9 in A by A4,A7,FUNCT_1:3;
      hence thesis by A13,A15,Th2;
    end;
  end;
  now
    assume
A18: X is empty;
    take k = 0;
    {} c= f.k;
    hence thesis by A18;
  end;
  hence thesis by A3;
end;
