
theorem
  for n being Nat holds
    [/ (tau to_power (2*n+1)) / sqrt 5 \] = Fib (2*n+1)
  proof
    let n be Nat;
A1: Fib (2 * n + 1) =
    ((tau to_power (2*n+1)) - (tau_bar to_power (2*n+1)))/(sqrt 5) by FIB_NUM:7
    .= (tau to_power (2*n+1))/sqrt 5 - (tau_bar to_power (2*n+1))/sqrt 5
    by XCMPLX_1:120;
A2: sqrt 5 > 0 by SQUARE_1:17,27;
    tau_bar to_power (2*n+1) < 0
    proof
      set t = - tau_bar;
A3:   tau_bar to_power (2*n+1) = (-t) to_power (2*n+1)
      .= - (t to_power (2*n+1)) by Th4;
      t to_power (2*n+1) > 0 by POWER:34;
      hence thesis by A3;
    end; then
A4:(tau to_power (2*n+1)) / sqrt 5 <= Fib (2*n+1) by A1,A2,XREAL_1:46;
   (tau_bar to_power (2*n+1)) / sqrt 5 + 1 / 2 > 0 by Th17; then
(tau_bar to_power (2*n+1)) / sqrt 5 + 1/2 - 1/2 > 0 - 1/2 by XREAL_1:9;
   then (tau_bar to_power (2*n+1)) / sqrt 5 > -1 by XXREAL_0:2; then
   - ((tau_bar to_power (2*n+1)) / sqrt 5) < - -1 by XREAL_1:24; then
   - ((tau_bar to_power (2*n+1)) / sqrt 5) + (tau to_power (2*n+1)) / sqrt 5 <
   1 + (tau to_power (2*n+1)) / sqrt 5 by XREAL_1:8; then
   (tau to_power (2*n+1)) / sqrt 5 - ((tau_bar to_power (2*n+1)) / sqrt 5) <
   1 + (tau to_power (2*n+1)) / sqrt 5; then
   (tau to_power (2*n+1) - tau_bar to_power (2*n+1)) / sqrt 5 <
   1 + (tau to_power (2*n+1)) / sqrt 5 by XCMPLX_1:120; then
   Fib (2*n+1) < (tau to_power (2*n+1)) / sqrt 5 + 1 by FIB_NUM:7;
   hence thesis by A4,INT_1:def 7;
  end;
