= Solution
A putative simultaneous assignment of classical values $\widetilde s_x^{(i)},\widetilde s_y^{(i)}\in\{\pm a\}$ reproducing the first three certain quantum outcomes must obey, almost surely,
$$
\widetilde s_x^{(1)}\widetilde s_y^{(2)}\widetilde s_y^{(3)}=\widetilde s_y^{(1)}\widetilde s_x^{(2)}\widetilde s_y^{(3)}=\widetilde s_y^{(1)}\widetilde s_y^{(2)}\widetilde s_x^{(3)}=a^3.
$$
Multiply the three equalities. Each $y$ value occurs twice and its square is $a^2$, leaving $a^6\widetilde s_x^{(1)}\widetilde s_x^{(2)}\widetilde s_x^{(3)}=a^9$. Thus
$$
\boxed{\widetilde s_x^{(1)}\widetilde s_x^{(2)}\widetilde s_x^{(3)}=+a^3,}
$$
contrary to the quantum prediction $-a^3$.
This is the <GHZ theorem> against predetermined local measurement values. Experimentally prepare many copies of the three-spin state and choose independently among the four measurement settings on separated subsystems. The three mixed correlations force the classical prediction for the fourth, whereas quantum mechanics predicts the opposite sign. Since $s_x,s_y$ on the same subsystem do not commute, the four settings are measured on separate copies; the classical contradiction relies precisely on the common assignment of values across settings.
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