= Solution
The equal mixture of $|\Phi^+\rangle$ and $|\Phi^-\rangle$ cancels their off-diagonal coherence, leaving the <separable quantum state>
$$
\rho(0)=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle11|.
$$
A trivial <LHV model> chooses a shared random bit $j$, uniformly zero or one, and prepares both local <qubits> in $|j\rangle$. Each party uses the local <Born rule> for its own chosen spin direction, with independent local random seeds if a deterministic description is desired. This reproduces the entire joint measurement distribution, not merely the $z$-axis correlation.
Since the state has an explicit <local hidden-variable theory> for these local measurements, \b[it cannot violate the CHSH inequality], by the derivation in part (b). Classical correlation of the shared preparation bit is sufficient; the cancellation of coherence removes the <entanglement>.
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