[@Gisin; @PR]. Now, the quantum prediction for the correlation function $E^{+}_{12}$ ($E^{-}_{12}$) associated with subensemble $+$ ($-$) is given by $$\begin{aligned}
E^{\pm}_{12}(\vec{e}_1,\vec{e}_2) &=& \langle \Psi^{\pm}_{12}|\sigma(\vec{e}_1) \otimes \sigma(\vec{e}_2) |\Psi^{\pm}_{12}\rangle  \nonumber \\
&=& \gamma \cos\vartheta_1 \cos\vartheta_2 \pm z_3 (c_1 c_2 /p_{\pm}) \sin\vartheta_1
\sin\vartheta_2  \nonumber  \\
&& \times\sin\vartheta_3 \cos (\varphi_1 +\gamma\varphi_2 +z_1 z_3 \varphi_3),\end{aligned}$$ where $\gamma$ is a sign factor with value $+1$ ($-1$) for $\text{sgn}z_1 = \text{sgn}z_2$ ($\text{sgn}z_1 \neq \text{sgn}z_2$). Incidentally, it will be noted that for $\vartheta_3 = n\pi$ we have $E^{\pm}_{12} = \gamma \cos\vartheta_1 \cos\vartheta_2$, and hence it is not possible a violation of Bell’s inequality for any choice of directions $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, $\vec{e}\,^{\prime}_{2}$. Of course this arises from the fact that the superposition state (2) reduces to the unentangled term $|z_1 \rangle |z_2 \rangle |z_3 \rangle$ or $|-z_1 \rangle |-z_2 \rangle |-z_3 \rangle$ whenever a spin measurement along the $z$ axis is performed on particle 3. Further, as expected, for either $c_1 =0$ or $c_2 =0$ (i.e., for product states) no violation can happen. Therefore, for general directions $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, $\vec{e}\,^{\prime}_{2}$, and $\vec{e}_3$, the relevant predictions by quantum mechanics will violate inequality (10) if $$\begin{aligned}
|\gamma &[& \cos\vartheta_1 \cos\vartheta_2 + \cos\vartheta_1 
\cos\vartheta^{\prime}_{2} + \cos\vartheta^{\prime}_{1} \cos\vartheta_2  \nonumber \\
