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 Th30:
  scf(i).0 = i & scf(i).(n+1) = 0
proof
  defpred P[Nat] means rfs(i).($1+1) = 0 & scf(i).($1+1) = 0;
  thus scf(i).0 = [\ rfs(i).0 /] by Def4
    .= [\ i /] by Def3
    .= i;
A1: for n st P[n] holds P[n+1]
  proof
    let n such that
A2: P[n];
    thus rfs(i).(n+1+1) = 1 / frac(rfs(i).(n+1)) by Def3
      .= 1 / (0 - 0) by A2
      .= 0;
    thus scf(i).(n+1+1) = [\ rfs(i).(n+1+1) /] by Def4
      .= [\ 0 /] by Th29
      .= 0;
  end;
  scf(i).(0+1) = [\ rfs(i).(0+1) /] by Def4
    .= [\ 0 /] by Th29
    .= 0;
  then
A3: P[0] by Th29;
  for n holds P[n] from NAT_1:sch 2(A3,A1);
  hence thesis;
end;
