Is this Riemann zeta zero counting function equality simpler to prove than the Franca-Leclair equality since there is no big Oh term?

Question

Introduction:
In André Leclair's document The Riemann Hypothesis for physicists there is on page 30 the following:

Quote:
"How to prove the Riemann hypothesis

Recall our main result:

$\boxed{\text{The n - th zero is of the form $\;\;\;\;\;$ $\rho=\frac{1}{2}+iy_{n}$}\\ \\ \;\;\;\;\;\;\; \frac{y_{n}}{2\pi}\log\left(\frac{y_{n}}{2\pi e}\right)+\lim\limits_{\delta\rightarrow 0^{+}}\frac{1}{\pi}\arg\zeta\left(\frac{1}{2}+\delta+iy_{n}\right)=n-\frac{11}{8} \;\;\;\;\; (n = 1,2,...)}$

$\boxed{\textit{If there is a unique solution to this equation for every n}}$, since they are enumerated by n, we can count how many zeros are on the line up to a height y=T.

$N_{o}(T)$ = number of zeros on the line with ordinate $y<T$. The above formula implies:

$$N_{0}(T)= \frac{T}{2\pi}\log\left(\frac{T}{2\pi e}\right)+\frac{7}{8}+\frac{1}{\pi}\arg\zeta\left(\frac{1}{2}+iT\right)+O\left(T^{-1}\right)$$

(Notice the big Oh term above which I ask about at the end.)

Now: N(T) = number of zeros on the entire critical strip has been known for over 100 years by performing a certain contour integral (argument principle) around the strip (Riemann, Backlund).

$$\Large\boxed{\text{our N$_0$(T) = the known N(t)}}$$

$$\Large \text{Thus: all zeros are on the line}$$ "

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Imitating the above (hoping to learn something in the process) by first removing the number $\delta$ and trying to preserve the form by writing:

$\textit{(1)}$ $\boxed{\text{The n - th zero is of the form $\;\;\;\;\;$ $\rho=\frac{1}{2}+iy_{n}$}\\ \\ \;\;\;\;\;\;\; \left\lfloor \frac{y_{n}}{2 \pi } \log \left(\frac{y_{n}}{2 \pi e}\right)+\frac{11}{8}\right\rfloor +\frac{1}{2} (\text{sgn}(\Im(\zeta \left(\frac{1}{2}+iy_{n}\right)))-1) = n - \frac{3}{2} \;\;\;\;\; (n = 1, 3, 5, 8, 10, 11, 14, 16, 17, 18, 20, 21, 23,...)}$

https://oeis.org/A282897

Mathematica 8 code to back it up:

Monitor[a = 
  Table[N[Floor[
       Im[ZetaZero[n]]/(2*Pi)*Log[Im[ZetaZero[n]]/(2*Pi*Exp[1])] + 
        11/8] + (Sign[Im[Zeta[ZetaZero[n]]]] - 1)/2, 20] - n + 
    3/2, {n, 1, 300}], n]
Flatten[Position[Abs[Round[a]], 0]]

$\textit{(2)}$ $\boxed{\text{The n - th zero is of the form $\;\;\;\;\;$ $\rho=\frac{1}{2}+iy_{n}$}\\ \\ \;\;\;\;\;\;\; \left\lfloor \frac{y_{n}}{2 \pi } \log \left(\frac{y_{n}}{2 \pi e}\right)+\frac{11}{8}\right\rfloor +\frac{1}{2} (\text{sgn}(\Im(\zeta \left(\frac{1}{2}+iy_{n}\right)))-1) = n - \frac{1}{2} \;\;\;\;\; (n = 2, 4, 6, 7, 9, 12, 13, 15, 19, 22, 24, 26, 27, 30, 31,...)}$

https://oeis.org/A282896

Mathematica 8 code to back it up:

Monitor[a = 
  Table[N[Floor[
       Im[ZetaZero[n]]/(2*Pi)*Log[Im[ZetaZero[n]]/(2*Pi*Exp[1])] + 
        11/8] + (Sign[Im[Zeta[ZetaZero[n]]]] - 1)/2, 20] - n + 
    1/2, {n, 1, 300}], n]
Flatten[Position[Abs[Round[a]], 0]]

$\boxed{\textit{If there is a unique solution to these two equations (1) and (2) for every n}}$,since they are enumerated by n, we can count how many zeros are on the line up to a height y=T.

$N_{o}(T)$ = number of zeros on the line with ordinate $y<T$. The above formula implies:

$$N_{0}(T) = \left\lfloor \frac{T}{2 \pi } \log \left(\frac{T}{2 \pi e}\right)+\frac{7}{8}\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i T+\frac{1}{2}\right)\right)\right)\right)$$

Mathematica 8:

Table[N[(Floor[
     T/(2*Pi)*Log[T/(2*Pi*Exp[1])] + 
      7/8] + (Sign[Im[Zeta[1/2 + I*T]]] - 1)/2), 10], {T, 1, 100}]

Now: N(T) = number of zeros on the entire critical strip has been known for over 100 years by performing a certain contour integral (argument principle) around the strip (Riemann, Backlund).

$$\Large\boxed{\text{our N$_0$(T) = the known N(t)}}$$

$$\Large \text{Thus: What? Nothing?}$$

Question: Is it any easier to show that:

$$N_{0}(T) = \left\lfloor \frac{T}{2 \pi } \log \left(\frac{T}{2 \pi e}\right)+\frac{7}{8}\right\rfloor +\frac{1}{2}\left(-1+\text{sgn}\left(\Im\left(\zeta \left(iT+\frac{1}{2}\right)\right)\right)\right)$$

is equal to:

$$N(T)=\frac{\vartheta (t)+\arg \left(\zeta \left(it+\frac{1}{2}\right)\right)}{\pi }$$

since there is no Big Oh term?

$\vartheta (t)$ is the Riemann-Siegel Theta function

Mathematica 8:

a3 = N[Table[((RiemannSiegelTheta[t] + Im[Log[Zeta[1/2 + I*t]]])/
     Pi), {t, 1, 100}]]
a5 = N[Table[(Floor[
      t/(2*Pi)*Log[t/(2*Pi*Exp[1])] + 
       7/8] + (Sign[Im[Zeta[1/2 + I*t]]] - 1)/2), {t, 1, 100}]]
Chop[a3 - a5]