reserve n, k, r, m, i, j for Nat;

theorem Th42:
  for k being non zero Element of NAT holds k divides n implies
  Fib (k) divides Fib (n)
proof
  let k be non zero Element of NAT;
  assume k divides n;
  then ex m being Nat st n = k * m by NAT_D:def 3;
  hence thesis by Th41;
end;
