
theorem
  for n being Nat holds
    Fib (n+1) = (Fib n + sqrt(5*(Fib n)^2 + 4*(-1) to_power n))/2
   proof
     let n be Nat;
A1:  n - 0 is Element of NAT by NAT_1:21;
A2:  2 * Fib (n+1) = Fib n + Lucas n by FIB_NUM3:24;
A3:  (Lucas n)^2 - 5*(Fib n)^2 = (Lucas n) to_power 2 - 5*(Fib n)^2 by POWER:46
     .= (Lucas n) to_power 2 - 5 * (Fib n) to_power 2 by POWER:46
     .= - (5 * (Fib n) |^2 - (Lucas n) |^2)
     .= - (4 * (-1) to_power (n+1)) by A1,FIB_NUM3:30
     .= (-1) * ((-1) to_power (n+1) * 4)
     .= (-1) to_power 1 * ((-1) to_power (n+1) * 4)
     .= ((-1) to_power 1 * (-1) to_power (n+1)) * 4
     .= (-1) to_power (n + 1 + 1) * 4 by Th2
     .= (-1) to_power (n + 2) * 4
     .= (-1) to_power n * (-1) to_power 2 * 4 by Th2
     .= (-1) to_power n * 1 * 4 by FIB_NUM2:3,POLYFORM:5;
     Fib (n+1) = (Fib n + Lucas n) / 2 by A2;
     hence thesis by A3,SQUARE_1:def 2;
  end;
