
theorem Th17:
  for n being Nat holds (tau_bar to_power n)/sqrt 5 + 1/2 > 0 &
  (tau_bar to_power n)/sqrt 5 + 1/2 < 1
  proof
  let n be Nat;
    set b = (1-sqrt 5)/2;
A1: |.b.| = -b by ABSVALUE:def 1,FIB_NUM:def 2;
A2: (|.b.| to_power n) * (1/sqrt 5) < 1/2 by Lm11,FIB_NUM:def 2;
    (b to_power n)*(1/sqrt 5) + 1/2 > 0 & (b to_power n)*(1/sqrt 5) + 1/2 < 1
    proof
      per cases;
      suppose  n is even;
        then
|.b.| to_power n = b to_power n by Th3,A1,FIB_NUM:def 2; then
        (b to_power n)*(1/sqrt 5) > 0 & (b to_power n)*(1/sqrt 5) < 1/2
        by Lm10,Lm11,FIB_NUM:def 2; then
        (b to_power n) * (1/sqrt 5) > 0  &
        (b to_power n) * (1/sqrt 5) + 1/2 < 1/2 + 1/2 by XREAL_1:8;
        hence thesis;
      end;
      suppose n is odd; then
        |.b.| to_power n = - (b to_power n) by Th4,A1,FIB_NUM:def 2; then
        (-(b to_power n))*(1/sqrt 5) > 0 & - (b to_power n)*(1/sqrt 5) < 1/2
        by Lm10,A2,FIB_NUM:def 2; then
        -(b to_power n)/sqrt 5 > 0 & - (b to_power n)/sqrt 5 < 1/2
        by XCMPLX_1:99; then
        (b to_power n)/sqrt 5 < 0 & - -(b to_power n)/sqrt 5 > - (1/2)
        by XREAL_1:24; then
        (b to_power n)/sqrt 5 + 1/2 < 0 + 1/2 &
        (b to_power n)/sqrt 5 + 1/2 > -1/2 + 1/2 by XREAL_1:8; then
        (b to_power n)/sqrt 5 + 1/2 < 1 & (b to_power n)/sqrt 5 + 1/2 > 0
        by XXREAL_0:2;
        hence thesis by XCMPLX_1:99;
      end;
      end;
    hence thesis by FIB_NUM:def 2,XCMPLX_1:99;
  end;
