= Solution
Let $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$ be the <Pauli X eigenstates>. Encode
$$
\boxed{|0_L\rangle=|+ -\rangle,\qquad |1_L\rangle=|- +\rangle.}
$$
Both have zero total $X$ eigenvalue, so $R_x(\phi)\otimes R_x(\phi)$ is the identity on their span. This is the <decoherence-free subspace> obtained from the previous code by applying a <Hadamard gate> to each physical <qubit>.
Since $Z|+\rangle=|-\rangle$ and $Z|-\rangle=|+\rangle$, a correlated $Z\otimes Z$ error exchanges the codewords. It acts as logical $X_L$, not as harmless identity. The <Pauli Y gate> obeys $Y|+\rangle=-i|-\rangle$ and $Y|-\rangle=i|+\rangle$, so
$$
\boxed{(Z\otimes Z)|\psi_L\rangle=(Y\otimes Y)|\psi_L\rangle=X_L|\psi_L\rangle.}
$$
The phases cancel within each codeword. Thus either unknown correlated gate gives a logical bit error on a general encoded state, though a logical $X_L$ <eigenstate> is unchanged as a physical ray. The code prevents collective $x$ rotations, not every possible correlated error.
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