
theorem Th2:
  for d being XFinSequence of NAT, n being Nat st for i being Nat
  st i in dom d holds n divides d.i holds n divides Sum d
proof
  let d be XFinSequence of NAT, n be Nat such that
A1: for i being Nat st i in dom d holds n divides d.i;
  per cases;
  suppose
    len d=0;
    then d={};
    then Sum d = 0;
    hence thesis by NAT_D:6;
  end;
  suppose
A2: len d > 0;
    then consider f being sequence of NAT such that
A3: f.0 = d.0 and
A4: for n being Nat st n+1 < len d holds f.(n + 1) = addnat
    . (f.n,d.(n + 1)) and
A5: addnat"**" d = f.(len d-1) by AFINSQ_2:def 8;
    defpred P[Nat] means $1 < len d implies n divides f.$1;
A6: now
      let k be Nat;
      assume
A7:   P[k];
      thus P[k+1]
      proof
        assume
A8:     k+1 < len d;
        then k+1 in Segm len d by NAT_1:44;
        then
A9:     n divides d.(k+1) by A1;
        f.(k+1) = addnat.(f.k,d.(k+1)) by A4,A8
          .= f.k + d.(k+1) by BINOP_2:def 23;
        hence thesis by A7,A8,A9,NAT_1:13,NAT_D:8;
      end;
    end;
    reconsider ld=len d-1 as Element of NAT by A2,NAT_1:20;
A10: ld < len d by XREAL_1:147;
    0 in Segm len d by A2,NAT_1:44;
    then
A11: P[0] by A1,A3;
A12:     addnat"**" d  = Sum d by AFINSQ_2:51;
    for k being Nat holds P[k] from NAT_1:sch 2(A11,A6);
    hence thesis by A5,A10,A12;
  end;
end;
