
theorem THTU3:
  log(tau,10) < 5
  proof
    [\ (tau to_power 5 )/sqrt 5 + 1/2 /] = 5 by FIB_NUM4:18,LFIB5; then
    5 -1/2 <= (tau to_power 5 )/sqrt 5 + 1/2 -1/2
      by XREAL_1:9,INT_1:def 6; then
    9/2*(sqrt 5) <= ((tau to_power 5 )/(sqrt 5))*(sqrt 5) by XREAL_1:64;
    then 9/2*(sqrt 5) <= (tau to_power 5 )/((sqrt 5)/(sqrt 5)) by XCMPLX_1:81;
    then WEQ:
    9/2*(sqrt 5) <= (tau to_power 5 )/(1) by XCMPLX_1:60;
    9/2*(sqrt 5) -10 =(9*(sqrt 5) -20)/2 /((9*(sqrt 5) +20)/(9*(sqrt 5) +20))
    by XCMPLX_1:51
    .=(9*(sqrt 5) -20) /((9*(sqrt 5) +20)/(9*(sqrt 5) +20))/2 by XCMPLX_1:48
    .=((9*(sqrt 5) -20) *(9*(sqrt 5) +20))
    /(9*(sqrt 5) +20)/2 by XCMPLX_1:77
    .=(81*(sqrt 5)^2 -400)/(9*(sqrt 5) +20)/2
    .=(81*5 -400)/(9*(sqrt 5) +20)/2 by SQUARE_1:def 2
    .=5/(9*(sqrt 5) +20)/2; then
    10 < 9/2*(sqrt 5) -10 +10 by XREAL_1:29; then
    10 < (tau to_power 5) by WEQ,XXREAL_0:2; then
    log(tau,10) < log(tau,(tau to_power 5)) by POWER:57,THTU; then
    log(tau,10) < 5*log(tau,tau) by POWER:55,THTU; then
    log(tau,10) < 5*1 by POWER:52,THTU;
    hence thesis;
  end;
