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

theorem Th68:
  for n being non zero Nat, m being Nat st m <> 1 holds m divides
  Fib (n) implies not m divides Fib (n-'1)
proof
  let n be non zero Nat;
  let m be Nat;
  assume
A1: m <> 1;
  assume
A2: m divides Fib (n);
  n >= 1 by NAT_2:19;
  then n = n -' 1 + 1 by XREAL_1:235;
  then Fib (n-'1), Fib (n) are_coprime by Th67;
  then
A3: Fib (n-'1) gcd Fib (n) = 1 by INT_2:def 3;
  assume m divides Fib (n-'1);
  then m divides 1 by A2,A3,NAT_D:def 5;
  hence contradiction by A1,WSIERP_1:15;
end;
