
theorem Th13:
for D be non empty set, Y be with_non-empty_element FinSequenceSet of D,
    s be non-empty sequence of Y, k,m be Nat st m in dom(s.k)
 ex n be Nat st n = (Partial_Sums(Length s)).k - len(s.k) + m - 1
       & (joined_seq s).n = (s.k).m
proof
   let D be non empty set, Y be with_non-empty_element FinSequenceSet of D,
       s be non-empty sequence of Y, k,m be Nat;
   assume A0: m in dom(s.k); then
A1:1 <= m & m <= len(s.k) by FINSEQ_3:25;
   now per cases;
    suppose A2: k = 0; then
    (Partial_Sums(Length s)).k = (Length s).0 by SERIES_1:def 1
      .= len(s.0) by Def3;
    hence (Partial_Sums(Length s)).k - len(s.k) + m - 1 is Nat
      by A1,A2,NAT_1:21;
    end;
    suppose k <> 0; then
    reconsider k1 = k-1 as Element of NAT by NAT_1:14,21;
    k = k1+1; then
    (Partial_Sums(Length s)).k = (Partial_Sums(Length s)).k1 + (Length s).k
      by SERIES_1:def 1
     .= (Partial_Sums(Length s)).k1 + len(s.k) by Def3; then
    reconsider n1= (Partial_Sums(Length s)).k - len(s.k) as Nat;
    n1+m >= m by NAT_1:11;
    hence (Partial_Sums(Length s)).k - len(s.k) + m - 1 is Nat
      by A1,XXREAL_0:2,NAT_1:21;
    end;
   end; then
   reconsider n = (Partial_Sums(Length s)).k - len(s.k) + m - 1 as Nat;
   take n;
   thus n = (Partial_Sums(Length s)).k - len(s.k) + m - 1;
   consider k2,m2 be Nat such that
A4: m2 in dom(s.k2) & (Partial_Sums(Length s)).k2 - len(s.k2) + m2 - 1 = n
  & (joined_seq s).n = (s.k2).m2 by Def4;
   m = m2 & k = k2 by A0,A4,Th10;
   hence (joined_seq s).n = (s.k).m by A4;
end;
