reserve a, b, k, n, m for Nat,
  i for Integer,
  r for Real,
  p for Rational,
  c for Complex,
  x for object,
  f for Function;
reserve l, n1, n2 for Nat;
reserve s1, s2 for Real_Sequence;

theorem
  (for n holds scf(r).n<>0) implies for n holds c_d(r).(n+1) >= tau |^ n
proof
  set s2=c_d(r);
  set s=scf(r);
  defpred P[Nat] means s2.($1+1) >= tau |^$1;
  sqrt 5 < sqrt 3^2 by SQUARE_1:27;
  then sqrt 5 < 3 by SQUARE_1:22;
  then 1+sqrt 5 < 1+3 by XREAL_1:8;
  then
A1: ((1+sqrt 5)/2)|^1 = (1+sqrt 5)/2 & (1+sqrt 5)/2 < 4/2 by XREAL_1:74;
  assume
A2: for n holds scf(r).n <>0;
  then
A3: s.1>=1 by Th40;
  then s2.(0+1)>=1 by Def6;
  then
A4: P[0] by NEWTON:4;
  let n;
A5: for n being Nat st P[n] & P[n+1] holds P[n+2]
  proof
    let n be Nat;
    assume that
A6: s2.(n+1) >= tau|^n and
A7: s2.(n+1+1) >= tau|^(n+1);
A8: tau|^(n+1)+tau|^n =((1+sqrt 5)/2)|^n * ((1+sqrt 5)/2) +((1+sqrt 5)/2)
    |^n by FIB_NUM:def 1,NEWTON:6
      .=((1+sqrt 5)/2)|^n * ((6+2*sqrt 5)/4);
    sqrt 5 >=0 by SQUARE_1:def 2;
    then (1+sqrt 5)/2 > 0 by XREAL_1:139;
    then
A9: tau|^(n+1) > 0 by FIB_NUM:def 1,PREPOWER:6;
A10: tau|^(n+2) =((1+sqrt 5)/2)|^n * ((1+sqrt 5)/2)|^2 by FIB_NUM:def 1
,NEWTON:8
      .=((1+sqrt 5)/2)|^n * ((1+sqrt 5)/2)^2 by WSIERP_1:1
      .=((1+sqrt 5)/2)|^n * ((1^2+2*1*sqrt 5+(sqrt 5)^2)/ 4)
      .=((1+sqrt 5)/2)|^n * ((1+2*sqrt 5+5)/ 4) by SQUARE_1:def 2
      .=((1+sqrt 5)/2)|^n * ((6+2*sqrt 5)/ 4);
A11: s2.(n+2+1) = s.(n+1+2)*s2.(n+1+1)+s2.(n+1) by Def6
      .=s.(n+3)*s2.(n+1+1)+s2.(n+1);
    n+3 >= 0+1 by XREAL_1:7;
    then s.(n+3) >= 1 by A2,Th40;
    then s.(n+3)*s2.(n+1+1) >= 1*(tau|^(n+1)) by A7,A9,XREAL_1:66;
    hence thesis by A6,A11,A8,A10,XREAL_1:7;
  end;
  s.2>=1 by A2,Th40;
  then s.2*s.1 >= 1 by A3,XREAL_1:159;
  then
A12: s.2*s.1+1 >= 1+1 by XREAL_1:6;
  s2.(1+1) = s.(0+2) * s2.(0+1) + s2.0 by Def6
    .=s.2 * s2.1 + 1 by Def6
    .=s.2 * s.1 + 1 by Def6;
  then
A13: P[1] by A12,A1,FIB_NUM:def 1,XXREAL_0:2;
  for n being Nat holds P[n] from FIB_NUM:sch 1(A4,A13,A5);
  hence thesis;
end;
