reserve x, x1, x2, y, X, D for set,
  i, j, k, l, m, n, N for Nat,
  p, q for XFinSequence of NAT,
  q9 for XFinSequence,
  pd, qd for XFinSequence of D;
reserve pN, qN for Element of NAT^omega;

theorem
  for n,k ex p st Sum p = card Domin_0(2*n+2+k,n+1) & dom p = k+1 & for
  i st i <= k holds p.i = card Domin_0(2*n+1+i,n)
proof
  let n,k;
  defpred P[set,set] means for j st $1=j holds $2=card Domin_0(2*n+1+j,n);
A1: for i st i in Segm(k+1) ex x be Element of NAT st P[i,x]
  proof
    let i such that
    i in Segm(k+1);
    P[i,card Domin_0(2*n+1+i,n)];
    hence thesis;
  end;
  consider p such that
A2: dom p=Segm(k+1) and
A3: for i be Nat st i in Segm(k+1) holds P[i,p.i] from STIRL2_1:sch 5(A1);
  take p;
A4: for i st i <= k holds p.i = card Domin_0(2*n+1+i,n)
  proof
    let i;
    assume i <= k;
    then i<k+1 by NAT_1:13;
    then i in Segm(k+1) by NAT_1:44;
    hence thesis by A3;
  end;
  now
    per cases;
    suppose
A5:   n=0;
      for x being object st x in dom p holds p.x = 1
      proof
        let x be object such that
A6:     x in dom p;
        reconsider i=x as Nat by A6;
        p.i=card Domin_0(2*n+1+i,n) by A2,A3,A6;
        hence thesis by A5,Th24;
      end;
      then p=(k+1)-->1 by A2,FUNCOP_1:11;
      then Sum p=(k+1)*1 by AFINSQ_2:58;
      hence Sum p = card Domin_0(2*n+2+k,n+1) by A5,Th30;
    end;
    suppose
      n>0;
      then reconsider n1=n-1 as Nat by NAT_1:20;
      defpred Q[Nat           ] means for q st dom q = $1+1 & for i st i <= $1
holds q.i = card Domin_0(2*n+1+i,n) holds Sum q = card Domin_0(2*n+2+$1,n+1);
A7:   for j st Q[j] holds Q[j+1]
      proof
        let j such that
A8:     Q[j];
        set CH2=(2*n+2+j) choose (n1+1);
        set CH1=(2*n+2+j) choose (n+1);
        set j1=j+1;
        let q such that
A9:     dom q=j1+1 and
A10:    for i st i<=j1 holds q.i=card Domin_0(2*n+1+i,n);
A11:    2*(n+1)<=2*(n+1)+j1 by NAT_1:11;
        j1 <=j1+1 by NAT_1:11;
        then Segm j1 c= Segm(j1+1) by NAT_1:39;
        then
A12:    dom (q|j1)=j1 by A9,RELAT_1:62;
A13:    for i st i<=j holds (q|j1).i = card Domin_0(2*n+1+i,n)
        proof
          let i;
          assume i<=j;
          then i<j1 by NAT_1:13;
          then i < len(q|j1) by A12;
          then i in dom (q|j1) & q.i= card Domin_0(2*n+1+i,n) by A10,A12,
AFINSQ_1:86;
          hence thesis by FUNCT_1:47;
        end;
        set CH4=(2*n+1+j1) choose n1;
        set CH3=(2*n+1+j1) choose n;
A14:    2*n<=2*n+(1+j1) & n1+1=n by NAT_1:11;
        q.j1=card Domin_0(2*n+1+j1,n) by A10;
        then
A15:    q.j1=CH3-CH4 by A14,Th28;
        j1 <j1+1 by NAT_1:13;
        then j1 < len q by A9;
        then j1 in dom q by AFINSQ_1:86;
        then
A16:    Sum (q|(j1+1))=Sum (q|j1) + q.j1 by AFINSQ_2:65;
        2*(n+1)<=2*n+2+j by NAT_1:11;
        then card Domin_0(2*n+2+j,n+1)=CH1-CH2 by Th28;
        then Sum(q|j1)=CH1-CH2 by A8,A12,A13;
        then Sum(q|j1)+q.j1=CH1+CH2-(CH3+CH4) by A15
          .=((2*n+2+j+1) choose (n+1))-(CH3+CH4) by NEWTON:22
          .=((2*n+2+j1) choose (n+1))-((2*n+2+j1) choose (n1+1)) by NEWTON:22
          .=card Domin_0(2*n+2+j1,n+1) by A11,Th28;
        hence thesis by A9,A16;
      end;
A17:  Q[0]
      proof
reconsider 2n1=2*n+1 as Nat;
        set 2CHn=(2*n+2) choose n;
        set 2CHn91=(2*n+2) choose (n+1);
        set CHn91=2n1 choose (n+1);
        set CHn1=2n1 choose n1;
        set CHn=2n1 choose n;
        let q;
        assume
        dom q= (0 qua Nat)+1 &
           for i st i <= 0 holds q.i = card Domin_0(2*n+1+i ,n);
        then
A18:    q.0=card Domin_0(2*n+1+(0 qua Nat),n) & len q = 1;
A19:    2*n+2=(2*n+1)+1;
        then
A20:    2CHn91 = CHn91 + CHn by NEWTON:22;
        n1+1=n;
        then
A21:    2CHn = CHn + CHn1 by A19,NEWTON:22;
        n<=n+(n+1) & (2*n+1)-n=n+1 by NAT_1:11;
        then
A22:    CHn = CHn91 by NEWTON:20;
        2*(n+1)=2*n+2;
        then
A23:    card Domin_0(2*n+2,n+1)=2CHn91-2CHn by Th28;
        card Domin_0(2n1,n1+1)=CHn-CHn1 & Sum <%q.0%>=q.0
            by Th28,AFINSQ_2:53,NAT_1:11;
        hence thesis by A20,A21,A22,A23,A18,AFINSQ_1:34;
      end;
      for j holds Q[j] from NAT_1:sch 2(A17,A7);
      hence Sum p = card Domin_0(2*n+2+k,n+1) by A2,A4;
    end;
  end;
  hence thesis by A2,A4;
end;
