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

theorem S2:
  for f be complex-valued FinSequence, n be Nat holds Sum ((f/^n)| 1) = f.(n+1)
  proof
    let f be complex-valued FinSequence, n be Nat;
    per cases;
    suppose
      A1: 1 in dom (f/^n);
A0:   f is FinSequence of COMPLEX by FINSEQ_1:107;
      thus Sum ((f/^n)| 1) = (f/^n).1 by S1
      .= f.(n+1) by A1,A0,FINSEQ_7:4;
    end;
    suppose
      not 1 in dom (f/^n); then
      A1: f/^n is empty by FINSEQ_5:6; then
      0 = len (f/^n)
      .= (len f) -'n by RFINSEQ:29; then
      len f = n or 0 + n >= ((len f)-n) +n by XREAL_0:def 2,XREAL_1:6; then
      len f + 0  < n + 1 by XREAL_1:8; then
      A2: not (n+1) in dom f by FINSEQ_3:25;
      thus Sum ((f/^n)| 1) = 0 by A1
      .= f.(n+1) by A2,FUNCT_1:def 2;
    end;
  end;
