
theorem Th9:
for D be non empty set,
    Y be non empty with_non-empty_element FinSequenceSet of D,
    s be non-empty sequence of Y
 holds Partial_Sums(Length s) is increasing
proof
   let D be non empty set,
       Y be non empty with_non-empty_element FinSequenceSet of D,
       s be non-empty sequence of Y;
   now let n,m be Nat;
    assume
A1:  n in dom(Partial_Sums(Length s)) & m in dom(Partial_Sums(Length s))
         & n < m;
    defpred P[Nat] means
      (Partial_Sums(Length s)).n < (Partial_Sums(Length s)).(n+1+$1);
    (Partial_Sums(Length s)).(n+1+0)
     = (Partial_Sums(Length s)).n + (Length s).(n+1) by SERIES_1:def 1
    .= (Partial_Sums(Length s)).n + len(s.(n+1)) by Def3; then
A3: P[0] by XREAL_1:29;
A4: for k be Nat st P[k] holds P[k+1]
    proof
     let k be Nat;
     assume A5: P[k];
     (Partial_Sums(Length s)).(n+1+(k+1))
      = (Partial_Sums(Length s)).(n+1+k) + (Length s).(n+1+k+1)
           by SERIES_1:def 1
     .= (Partial_Sums(Length s)).(n+1+k) + len(s.(n+1+k+1)) by Def3; then
     (Partial_Sums(Length s)).(n+1+(k+1)) > (Partial_Sums(Length s)).(n+1+k)
       by XREAL_1:29;
     hence P[k+1] by A5,XXREAL_0:2;
    end;
A7: for k be Nat holds P[k] from NAT_1:sch 2(A3,A4);
    n+1 <= m by A1,NAT_1:13; then
    reconsider k = m - (n+1) as Nat by NAT_1:21;
    m = n+1+k;
    hence (Partial_Sums(Length s)).n < (Partial_Sums(Length s)).m by A7;
   end;
   hence Partial_Sums(Length s) is increasing by SEQM_3:def 1;
end;
