Wiener Filter Additive Noise Uncorrelated
Let me change the notation slightly to make things clear. Let $d[n]$ be the desired signal, $x[n]$ is the noisy input signal, and $v[n]$ is the noise. Furthermore, we assume that

$$x[n]=d[n]+v[n]\tag{1}$$

and that the desired signal $d[n]$ and the noise $v[n]$ are uncorrelated and at least one of them has zero mean, which implies that their cross-correlation equals zero:

$$R_{dv}[n]=E\{d^*[k]v[k+n]\}=0\tag{2}$$

From $(2)$ it follows that the cross-correlation between the desired signal $d[n]$ and the input signal $x[n]$ is given by

$$\begin{align}R_{dx}[n]&amp;=E\{d^*[k]x[k+n]\}\\&amp;=E\{d^*[k]\left(d[k+n]+v[k+n]\right)\}\\&amp;=E\{d^*[k]d[k+n]\}+E\{d^*[k]v[k+n]\}\\&amp;=E\{d^*[k]d[k+n]\}=R_{d}[n]\tag{3}\end{align}$$

From $(3)$ it follows that the cross-power spectral density of $d[n]$ and $x[n]$ equals the power spectral density of $d[n]$:

$$S_{dx}(\omega)=S_{d}(\omega)\tag{4}$$

With the given assumptions, the frequency response of the non-causal Wiener filter can be written as

$$H(\omega)=\frac{S_{dx}(\omega)}{S_x(\omega)}=\frac{S_d(\omega)}{S_d(\omega)+S_v(\omega)}\tag{5}$$