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

theorem Th48:
  for k,n being Element of NAT st n > 1 & k <> 0 & k <> 1 holds
  Fib (k) = Fib (n) iff k = n
proof
  let k, n be Element of NAT such that
A1: n > 1 and
A2: k <> 0 & k <> 1;
  k is non trivial by A2,NAT_2:def 1;
  then k >= 1 + 1 by NAT_2:29;
  then
A3: k > 1 by NAT_1:13;
  Fib (k) = Fib (n) implies k = n
  proof
    assume
A4: Fib (k) = Fib (n);
    assume
A5: k <> n;
    per cases by A5,XXREAL_0:1;
    suppose
      k > n;
      hence contradiction by A1,A4,Th46;
    end;
    suppose
      k < n;
      hence contradiction by A3,A4,Th46;
    end;
  end;
  hence thesis;
end;
