
theorem
  for n being Nat st n >= 2 holds
    Fib (n+1) = [/ (Fib n + sqrt 5 * Fib n - 1)/2 \]
  proof
    let n be Nat;
    assume A1: n >= 2;
    then Fib (n+1) = [\ (Fib n + sqrt 5 * Fib n + 1) / 2 /] by Th24; then
A2: (Fib n + sqrt 5 * Fib n + 1)/2 - 1 < Fib (n+1) by INT_1:def 6;
    (Fib n + sqrt 5 * Fib n - 1)/2 + 1 > Fib (n+1)
    proof
      Fib (n+1) = [\ (Fib n + sqrt 5 * Fib n + 1) / 2 /] by Th24,A1; then
A3:   (Fib n + sqrt 5 * Fib n - 1)/2 + 1 >= Fib (n+1) by INT_1:def 6;
      (Fib n + sqrt 5 * Fib n - 1)/2 + 1 <> Fib (n+1)
      proof
        set tn = tau to_power n;
        set tbn = tau_bar to_power n;
        set t1 = tau to_power (n+1);
        set t2 = tau_bar to_power (n+1);
        set s5 = sqrt 5;
A4:     s5 > 0 by SQUARE_1:25;
        assume (Fib n + s5 * Fib n - 1)/2 + 1 = Fib (n+1); then
        Fib n * (1 + s5) + 1 = 2 * Fib (n+1); then
        ((tn - tbn)/s5) * (1+s5) + 1 = 2 * Fib (n+1) by FIB_NUM:7; then
        ((((tn - tbn)/s5) * (1+s5)) + 1) * s5 =
        (2 * ((t1-t2)/s5))*s5 by FIB_NUM:7; then
        ((tn - tbn) / s5 * (1+s5)) * s5 + 1 * s5 = 2*(((t1-t2)/s5)*s5); then
        ((tn - tbn) / s5 * (1+s5)) * s5 + s5 =
        2 * (t1-t2) by A4,XCMPLX_1:87; then
        (((tn-tbn)*(1+s5))/s5)*s5+s5 = 2 * t1 - 2 * t2 by XCMPLX_1:74; then
        (tn - tbn) * (1 + s5) + s5 = 2 * t1 - 2 * t2 by A4,XCMPLX_1:87; then
        tn*(1+s5)-(tbn+tbn*s5)+s5=2*(tn*tau to_power 1)
        - 2 * t2 by Th2; then
        tn*(1+s5)-tbn*(1+s5)+s5= 2*(tn*tau to_power 1)
        -2*(tbn*tau_bar to_power 1) by Th2;then
        tn*(1+s5)-tbn*(1+s5) + s5 = 2 * (tn * tau) -
        2 * (tbn * tau_bar to_power 1); then
        (tn - tbn) * (1+s5) + s5 = 2 * (tn * tau) -
        2 * (tbn * tau_bar); then
        tn*(1+s5 - 2*tau) + tbn* (2*tau_bar - (1+s5))+s5 = 0; then
        (-2 * tbn + 1)*sqrt 5 = 0 by FIB_NUM:def 1,def 2; then
        -2 * tbn + 1 = 0 by A4;
        hence contradiction by Th8,A1;
      end;
      hence thesis by A3,XXREAL_0:1;
    end;
    hence thesis by A2,INT_1:def 7;
  end;
