reserve a,b,i,j,k,l,m,n for Nat;

theorem DIS:
  for a,b be Integer holds
    a*b divides ((a,b)In_Power n).i - ((a,b) Subnomial n).i
  proof
    let a,b be Integer;
    per cases;
    suppose
      A1: i in dom ((a,b)In_Power (n+1-1)); then
      A1a: i in dom ((a,b)Subnomial (n+1-1)) by DOMN; then
      A1b: 1 <= i <= len ((a,b)Subnomial (n+1-1)) by FINSEQ_3:25; then
      A1c: 1-1 <= i-1 <= n by XREAL_1:9;
      reconsider m = i - 1 as Nat by A1b;
      n - m >= m - m by A1c,XREAL_1:9; then
      reconsider l = n - m as Nat by INT_1:3;
      A2: ((a,b)In_Power n).i = (n choose m)*a|^l*b|^m by A1,NEWTON:def 4;
      A3: ((a,b)In_Power n).i - ((a,b) Subnomial n).i =
        (n choose m)*a|^l*b|^m - a|^l*b|^m by A1a,Def2,A2
      .= ((n choose m)-1)*(a|^l*b|^m);
      per cases;
      suppose
        m = 0 or m = n; then
        (n choose m) = 1 by NEWTON:19,21;
        hence thesis by A3,INT_2:12;
      end;
      suppose
        m > 0 & m <> n; then
        0 < m < n by A1c,XXREAL_0:1; then
        m > 0 & n-m > m-m by XREAL_1:9; then
        a divides a|^l & b divides b|^m by NEWTON02:14; then
        a*b divides a|^l*b|^m by NEWTON02:2;
        hence thesis by A3,INT_2:2;
      end;
    end;
    suppose
      A1: not i in dom ((a,b)In_Power (n+1-1)); then
      not i in dom ((a,b)Subnomial (n+1-1)) by DOMN; then
      ((a,b)In_Power n).i = 0 & ((a,b) Subnomial n).i = 0 by A1,FUNCT_1:def 2;
      hence thesis by INT_2:12;
    end;
  end;
