
theorem Th9:
  for k being Nat holds for F being sequence of
ExtREAL holds ((for n being Element of NAT st n <> k holds F.n = 0.) implies (
for n being Element of NAT st n < k holds Ser(F).n = 0.) & for n being Element
  of NAT st k <= n holds Ser(F).n = F.k )
proof
  let k be Nat;
  let F be sequence of ExtREAL;
  defpred P[Nat] means $1 < k implies Ser(F).$1 = 0.;
  assume
A1: for n being Element of NAT st n <> k holds F.n = 0.;
A2: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    assume
A3: n < k implies Ser(F).n = 0.;
    assume
A4: n+1 < k;
    then
A5: n <= n+1 & F.(n+1) = 0. by A1,NAT_1:11;
    Ser(F).(n+1) = Ser(F).n + F.(n+1) by SUPINF_2:def 11
      .= 0. by A3,A4,A5,XXREAL_0:2;
    hence thesis;
  end;
A6: P[0]
  proof
    assume 0 < k;
    then F.0 = 0. by A1;
    hence thesis by SUPINF_2:def 11;
  end;
A7: for n being Nat holds P[n] from NAT_1:sch 2(A6,A2);
  defpred P[Nat] means k <= $1 implies Ser(F).$1 = F.k;
A8: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    assume
A9: k <= n implies Ser(F).n = F.k;
    assume
A10: k <= n+1;
    per cases by A10,NAT_1:8;
    suppose
A11:  k <= n;
      then k <> n+1 by NAT_1:13;
      then
A12:  F.(n+1) = 0. by A1;
      Ser(F).(n+1) = Ser(F).n + F.(n+1) by SUPINF_2:def 11
        .= F.k by A9,A11,A12,XXREAL_3:4;
      hence thesis;
    end;
    suppose
A13:  k = n + 1;
      then n < k by NAT_1:13;
      then
A14:  Ser(F).n = 0. by A7;
      Ser(F).(n+1) = Ser(F).n + F.(n+1) by SUPINF_2:def 11
        .= F.k by A13,A14,XXREAL_3:4;
      hence thesis;
    end;
  end;
A15: P[0]
  proof
A16: 0 <= k by NAT_1:2;
    assume k <= 0;
    then F.0 = F.k by A16,XXREAL_0:1;
    hence thesis by SUPINF_2:def 11;
  end;
  for n being Nat holds P[n] from NAT_1:sch 2(A15,A8);
  hence thesis by A7;
end;
