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
  scf(r).0 > 0 & (for n holds scf(r).n <> 0) implies for n holds c_n(r).
  n >= tau |^ n
proof
  assume that
A1: scf(r).0 > 0 and
A2: for n holds scf(r).n <>0;
  set s=scf(r);
A3: s.0 >= 0+1 by A1,INT_1:7;
  set s1=c_n(r);
  defpred P[Nat] means s1.$1 >= tau |^$1;
A4: for n being Nat st P[n] & P[n+1] holds P[n+2]
  proof
    let n be Nat;
    assume that
A5: s1.n >= tau|^n and
A6: s1.(n+1) >= tau|^(n+1);
A7: 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
A8: tau|^(n+1) > 0 by FIB_NUM:def 1,PREPOWER:6;
A9: 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);
A10: s1.(n+1+1) = s.(n+2)*s1.(n+1)+s1.n by Def5;
    n+2 >= 0+1 by XREAL_1:7;
    then s.(n+2) >= 1 by A2,Th40;
    then s.(n+2)*s1.(n+1) >= 1*(tau|^(n+1)) by A6,A8,XREAL_1:66;
    hence thesis by A5,A10,A7,A9,XREAL_1:7;
  end;
  s1.0 = s.0 by Def5;
  then
A11: P[0] by A3,NEWTON:4;
  s.1>=1 by A2,Th40;
  then s.1*s.0 >= 1 by A3,XREAL_1:159;
  then
A12: s.1*s.0+1 >= 1+1 by XREAL_1:6;
  let n;
  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
A13: (1+sqrt 5)/2 < 4/2 by XREAL_1:74;
  s1.1 = s.1*s.0+1 & ((1+sqrt 5)/2)|^1 = (1+sqrt 5)/2 by Def5;
  then
A14: P[1] by A12,A13,FIB_NUM:def 1,XXREAL_0:2;
  for n being Nat holds P[n] from FIB_NUM:sch 1(A11,A14,A4);
  hence thesis;
end;
