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 Th33:
  (for n holds scf(r).n = 0) implies rfs(r).n = 0
proof
  assume that
A1: for n holds scf(r).n = 0 and
A2: rfs(r).n <> 0;
A3: scf(r).n = 0 by A1;
  set x = rfs(r).n;
A4: scf(r).n = [\x/] by Def4;
  per cases by A2;
  suppose
    x < 0;
    hence thesis by A3,A4,INT_1:def 6;
  end;
  suppose
    x >= 1;
    hence thesis by A3,A4,INT_1:54;
  end;
  suppose
    0 < x & x < 1;
    then
A5: 1/x > 1 by Th1;
A6: scf(r).(n+1) = 0 & scf(r).(n+1) = [\ rfs(r).(n+1) /] by A1,Def4;
    rfs(r).(n+1) = 1 / frac(x) by Def3
      .= 1/(x-scf(r).n) by Def4
      .= 1/(x-0) by A1
      .= 1/x;
    hence thesis by A5,A6,INT_1:54;
  end;
end;
