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;

theorem Th38:
  n >= 1 implies scf(r).n >= 0
proof
  defpred P[Nat] means scf(r).$1 >= 0;
  [\ r /] <= r by INT_1:def 6;
  then frac(r) >=0 by XREAL_1:48;
  then
A1: 1/frac(r) -1 >= 0 -1 by XREAL_1:9;
A2: for n being Nat st n>=1 holds P[n] implies P[n+1]
  proof
    let n be Nat;
    assume n>=1;
    [\ rfs(r).n /] <= rfs(r).n by INT_1:def 6;
    then frac(rfs(r).n) >=0 by XREAL_1:48;
    then
A3: 1/frac(rfs(r).n) -1 >= 0 -1 by XREAL_1:9;
    scf(r).(n+1) = [\ rfs(r).(n+1) /] by Def4
      .= [\ 1/frac(rfs(r).n) /] by Def3;
    then scf(r).(n+1) > 1/(rfs(r).n - [\ rfs(r).n /]) -1 by INT_1:def 6;
    then
    scf(r).(n+1) -(-1) > (1/frac(rfs(r).n) -1) - (1/frac(rfs(r).n) -1) by A3,
XREAL_1:14;
    hence thesis by INT_1:7;
  end;
  scf(r).1 = [\ rfs(r).(0+1) /] by Def4
    .= [\ 1/frac(rfs(r).0) /] by Def3
    .= [\ 1/frac(r) /] by Def3;
  then scf(r).1 > 1/frac(r) -1 by INT_1:def 6;
  then scf(r).1 -(-1) > (1/frac(r) -1) - (1/frac(r) -1) by A1,XREAL_1:14;
  then
A4: P[1] by INT_1:7;
  for n being Nat st n>=1 holds P[n] from NAT_1:sch 8(A4,A2);
  hence thesis;
end;
