
theorem
  for n being Nat holds [\ (tau to_power n )/sqrt 5 + 1/2 /] = Fib n
  proof
    let n be Nat;
    set tn = tau_bar to_power n;
A1: Fib n = ((tau to_power n) - tn)/(sqrt 5) + 1/2 - 1/2 by FIB_NUM:7
    .= (tau to_power n)/sqrt 5 - tn / sqrt 5 + 1/2 - 1/2 by XCMPLX_1:120
    .= ((tau to_power n)/sqrt 5 + 1/2) - (tn/sqrt 5 + 1/2);
    tn/sqrt 5 + 1/2 > 0 by Th17; then
A2: tn/sqrt 5 + 1/2 + Fib n > 0 + Fib n by XREAL_1:6;
    tn/sqrt 5 + 1/2 < 1 by Th17; then
    tn/sqrt 5 + 1/2 - 1/2 < 1 - 1/2 by XREAL_1:9; then
    - (tn/sqrt 5) > - (1/2) by XREAL_1:24; then
    - (tn/sqrt 5) + (tau to_power n)/sqrt 5 >
    - (1/2) + (tau to_power n)/sqrt 5 by XREAL_1:6; then
    (tau to_power n)/sqrt 5 - (tn/sqrt 5) >
    - (1/2) + (tau to_power n)/sqrt 5; then
    ((tau to_power n) - tn)/sqrt 5 >
    - (1/2) + (tau to_power n)/sqrt 5 by XCMPLX_1:120; then
    Fib n  > (tau to_power n )/sqrt 5 + 1/2 - 1 by FIB_NUM:7;
    hence thesis by A2,A1,INT_1:def 6;
  end;
