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;

theorem Th16:
  for p, q, k st p|(2*k+len q) = (k-->0)^q^(k-->1) & q is
dominated_by_0 & 2*Sum q = len q & k > 0 holds min*{N:2*Sum(p|N) = N & N > 0} =
  2*k+len q
proof
  let p,q,k such that
A1: p|(2*k+len q)=(k-->0)^q^(k-->1) and
A2: q is dominated_by_0 and
A3: 2*Sum q=len q and
A4: k>0;
  set k0=k-->0;
A5: Sum k0=k*0 by AFINSQ_2:58;
  then
A6: 2*k > 0 by A4,XREAL_1:68;
  reconsider k1=k-->1 as XFinSequence of NAT;
  set M={N:2*Sum(p|N)=N & N>0};
  set kqk=k0^q^k1;
  Sum kqk = Sum (k0^q) + Sum k1 by AFINSQ_2:55;
  then
A7: Sum kqk = Sum k0+Sum q+Sum k1 by AFINSQ_2:55;
  Sum k1=k*1 by AFINSQ_2:58;
  then 2*Sum (p|(2*k+len q))=len q+2*k by A1,A3,A7,A5;
  then
A8: 2*k+len q in M by A6;
  M c= NAT
  proof
    let y be object;
    assume y in M;
    then ex i be Nat st i=y & 2*Sum(p|i)=i & i>0;
    hence thesis by ORDINAL1:def 12;
  end;
  then reconsider M as non empty Subset of NAT by A8;
  min*M=2*k+len q
  proof
    kqk = k0^(q^k1) by AFINSQ_1:27;
    then
A9: len kqk = len k0+len (q^k1) by AFINSQ_1:17;
A10: len kqk = k+(len q+len k1) by A9,AFINSQ_1:17;
    assume
A11: min*M<>2*k+len q;
    min*M in M by NAT_1:def 1;
    then
A12: ex i be Nat st i=min*M & 2*Sum(p|i)=i & i>0;
A14: 2*k+len q >= min*M by A8,NAT_1:def 1;
    then
A15: Segm(min*M) c= Segm(2*k+len q) by NAT_1:39;
    then
A16: p|min*M= kqk|min*M by A1,RELAT_1:74;
    now
      per cases;
      suppose
A17:    min*M <= k;
        k=dom k0 & kqk=k0^(q^k1) by AFINSQ_1:27;
        then
A18:    kqk|min*M=k0|min*M by A17,AFINSQ_1:58;
A19:    Sum (min*M-->0)=min*M * 0 by AFINSQ_2:58;
        k0|min*M =min*M-->0 by A17,Lm1;
        then Sum (p|min*M)=Sum (min*M-->0) by A1,A15,A18,RELAT_1:74;
        hence contradiction by A12,A19;
      end;
      suppose
        min*M > k;
        then reconsider mk=min*M-k as Nat by NAT_1:21;
        now
          per cases;
          suppose
A20:        min*M <= k+len q;
A21:        dom k0 = k;
            min*M=mk + k;
            then
A22:        (k0^q)|min*M=k0^(q|mk) by A21,AFINSQ_1:59;
            dom (k0^q)=len k0+len q by AFINSQ_1:def 3;
            then kqk|min*M=(k0^q)|min*M by A20,AFINSQ_1:58;
            then
A23:        Sum(p|min*M)=Sum(k0)+Sum(q|mk) by A16,A22,AFINSQ_2:55;
A24:        1 <= k by A4,NAT_1:14;
            Sum(k0)=k*0 by AFINSQ_2:58;
            then mk+k <= mk by A2,A12,A23,Th2;
            hence contradiction by A24,NAT_1:19;
          end;
          suppose
            min*M > k+len q;
            then reconsider mkL=min*M-(k+len q) as Nat by NAT_1:21;
A25:        2*Sum(p|min*M)=min*M by A12;
            dom (k0^q)=len k0 + len q & dom k0=k by AFINSQ_1:def 3;
            then min*M=dom(k0^q)+mkL;
            then kqk|min*M=(k0^q)^(k1|mkL) by AFINSQ_1:59;
            then
A26:        Sum(p|min*M)=Sum(k0^q)+Sum (k1|mkL) by A16,AFINSQ_2:55;
            min*M < len kqk by A11,A10,A14,XXREAL_0:1;
            then mkL < 2*k+len q-(k+len q) by A10,XREAL_1:9;
            then k1|mkL=mkL-->1 by Lm1;
            then
A27:        Sum(k1|mkL)=mkL*1 by AFINSQ_2:58;
            Sum(k0^q)=Sum(k0)+Sum q & Sum k0=k*0 by AFINSQ_2:55,58;
            hence contradiction by A3,A11,A26,A27,A25;
          end;
        end;
        hence contradiction;
      end;
    end;
    hence contradiction;
  end;
  hence thesis;
end;
