
theorem Th11:
  for s1, s2 be Real_Sequence st
  (for n be Nat holds 0 <= s1.n & s1.n <= s2.n) &
  (ex n be Nat st 1 <= n & s1.n < s2.n) & s2 is summable holds
  s1 is summable & Sum(s1) < Sum(s2)
  proof
    let s1, s2 be Real_Sequence;
    assume that
A1: for n be Nat holds 0 <= s1.n & s1.n <= s2.n and
A2: ex n be Nat st 1 <= n & s1.n < s2.n and
A3: s2 is summable;
    consider N be Nat such that
A4: 1 <= N & s1.N < s2.N by A2;
A5: for n be Nat st 0 <= n holds s1.n <= s2.n by A1;
    hence s1 is summable by A1,A3,SERIES_1:19;
    N - 1 in NAT by A4,INT_1:5; then
    reconsider N1 = N - 1 as Nat;
A6: N1 + 1 = N; then
A7: Partial_Sums(s1).N = Partial_Sums(s1).N1 + s1.N by SERIES_1:def 1;
A8: Sum(s1) = Partial_Sums(s1).N + Sum(s1^\(N+1)) by A1,A3,A5,SERIES_1:15,19;
A9: Partial_Sums(s2).N = Partial_Sums(s2).N1 + s2.N by A6,SERIES_1:def 1;
A10: Sum(s2) = Partial_Sums(s2).N + Sum(s2^\(N+1)) by SERIES_1:15,A3;
A11: for n be Nat holds 0 <= (s1 ^\ (N + 1)).n
    proof
      let n be Nat;
      (s1 ^\ (N + 1)).n = s1.(n + (N + 1)) by NAT_1:def 3;
      hence thesis by A1;
    end;
A12: s2 ^\ (N + 1) is summable by A3,SERIES_1:12;
    for n be Nat holds (s1 ^\ (N + 1)).n <= (s2 ^\ (N + 1)).n
    proof
      let n be Nat;
A13:  (s1 ^\ (N + 1)).n = s1.(n + (N + 1)) by NAT_1:def 3;
      (s2 ^\ (N + 1)).n = s2.(n + (N + 1)) by NAT_1:def 3;
      hence thesis by A1,A13;
    end; then
A14: Sum(s1 ^\ (N + 1)) <= Sum(s2 ^\ (N + 1)) by A11,A12,SERIES_1:20;
    Partial_Sums(s1).N1 + s1.N < Partial_Sums(s2).N1 + s2.N
    by XREAL_1:8,A4,A1,SERIES_1:14;
    hence thesis by A7,A8,A9,A10,A14,XREAL_1:8;
  end;
