I am partly repeating myself here again. Is this a correct relation between the Riemann zeta function and the Prime zeta function?

$$ \zeta (s) = \sum\limits\_{n=1}^{\infty}\frac{1}{n^{s}}$$

$\displaystyle \log(\zeta (s)) = \sum\limits\_{n=1}^{\infty}\frac{a\_{n}}{n^{s}}$ where $\displaystyle a\_{1} = 0\;$, $\displaystyle a\_{n} = \frac{\Lambda(n)}{\log(n)}$ and $\Lambda(n)$ is the Mangoldt function defined by $\displaystyle \Lambda(n) = \log(p)$ if $n = p^{k}$ for $p$ a prime.

$\displaystyle ω\_{1}(\log(\zeta (s))) = \sum\limits\_{n=1}^{\infty}\frac{\log(a\_{n})}{n^{s}},$ where $ω\_{1}(\;)$ is an operation and $\displaystyle \log(a\_n) = 0$ if $a\_n = 0$
$$ω\_{1}(\log(\zeta (s)))\zeta (s) = \sum\limits\_{n=1}^{\infty}\frac{b\_{n}}{n^{s}}$$

$\displaystyle ω\_{2}(ω\_{1}(\log(\zeta (s)))\zeta (s)) = \sum\limits\_{n=1}^{\infty}\frac{e^{b\_{n}}}{n^{s}}\;$ where $ω\_{2}(\;)$ is another operation and $\displaystyle e^{b\_{n}}=\exp(b\_{n})$

$$\sum\limits\_{p\;\text{prime}} \frac1{p^s} = \log(ω\_2(ω\_1(\log(\zeta (s)))\zeta (s)))$$