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

theorem
  for n being Element of NAT st n > 1 & n <> 4 holds Fib (n) is prime
  implies n is prime
proof
  let n be Element of NAT such that
A1: n > 1 and
A2: n <> 4;
  assume
A3: Fib (n) is prime;
  assume not n is prime;
  then consider k being non zero Element of NAT such that
A4: k <> 1 and
A5: k <> 2 and
A6: k <> n and
A7: k divides n by A1,A2,Th49;
A8: Fib (k) <> Fib (n) by A1,A4,A6,Th48;
  Fib (k) <> 1 & Fib (k) divides Fib (n) by A4,A5,A7,Th42,Th47;
  hence contradiction by A3,A8,INT_2:def 4;
end;
