= Solution
Write the arbitrary initial <pure state> as
$$
|\chi\rangle_{QR}=|0\rangle_Q|r_0\rangle_R+|1\rangle_Q|r_1\rangle_R,
\qquad\langle r_0|r_0\rangle+\langle r_1|r_1\rangle=1.
$$
The reference vectors need not be normalized or <orthogonal>. By <linearity>, the same <quantum teleportation> calculation, with the <identity operator> on $R$, gives the post-gate <quantum state>
$$
\frac12\sum_{m,n=0}^1|mn\rangle_{QA}
\left[(X_B^nZ_B^m)\otimes I_R\right]
\left(|0\rangle_B|r_0\rangle_R+|1\rangle_B|r_1\rangle_R\right).
$$
Thus each outcome still has <probability> $1/4$. Bob's <Pauli gate> correction leaves the normalized joint <quantum state>
$$
\boxed{|\chi\rangle_{BR}=|0\rangle_B|r_0\rangle_R+|1\rangle_B|r_1\rangle_R.}
$$
All operations act trivially on $R$, so the result is independent of its holder or location. This establishes <teleportation as an identity channel on a reference>, preserving <quantum entanglement> with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint <density matrix>.
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