= Solution
Put $|0\rangle=|\uparrow_z\rangle$ and $|1\rangle=|\downarrow_z\rangle$. The <total-spin sector> with spin zero consists of the <spin singlet state> $|\Psi^-\rangle=(|01\rangle-|10\rangle)/\sqrt2$, not the whole subspace with zero $z$ component. These must be distinguished: the <Bell state> $|\Psi^+\rangle$ has zero $z$ component but total spin one.
There are two meanings of verification to separate here. A one-pair <entanglement-assisted statistical singlet verification> is possible: choose a shared random axis from $x,y,z$, measure that axis's two-spin parity by the meter circuit below, and accept anticorrelation. An $x$ or $y$ test uses local basis changes before and after the computational-axis circuit. The singlet always passes and is unchanged. Averaging the three acceptance projectors gives
$$
\Omega=\frac13\sum_{j=x,y,z}\frac{I-\sigma_j\otimes\sigma_j}{2}
=P_s+\frac13(I-P_s).
$$
Every triplet state passes with probability $1/3$, so a failure excludes the singlet, while repeated tests on independently prepared copies can give statistical confidence. A single pass does not certify the singlet with certainty.
\b[Exact single-shot verification cannot be implemented with just one shared Bell pair.] Here exact verification means a yes outcome with probability one on the <spin singlet state>, zero on its orthogonal complement, and preservation of the singlet on yes. If that stronger meaning is intended, the question needs an extra resource. The <Bell-pair cost of exact nondemolition singlet verification> gives a short proof. For a fully resolved tuple $\mu,\nu$ of local measurement records, contraction of the shared $|\Phi^+\rangle$ gives a system <Kraus operator>
$$
K_{\mu\nu}=\frac1{\sqrt2}\left(A_{\mu0}\otimes B_{\nu0}+A_{\mu1}\otimes B_{\nu1}\right).
$$
Local ancillas and locally adaptive operations are included in these operators. Refining any unobserved local environment gives the same form, so the <operator Schmidt rank> is at most two. On each nonzero yes branch, zero false positives forces $K_{\mu\nu}$ to vanish on the entire triplet subspace, and singlet preservation forces
$$
K_{\mu\nu}=c_{\mu\nu}P_s,\qquad
P_s=\frac14(I\otimes I-X\otimes X-Y\otimes Y-Z\otimes Z).
$$
The four <Pauli matrices>, including the identity, are an orthogonal operator basis. Thus $P_s$ has <operator Schmidt rank> four, contradicting the rank-two bound. At least one yes branch is nonzero because the singlet must pass with certainty. Shared classical randomness cannot evade this branchwise argument.
With \b[two shared Bell pairs], an explicit corrected protocol is available. On the first meter pair, apply local system-to-meter <CNOT gates> and measure both meters in the computational basis. If their records are $u,v$, the <Kraus operator> is $P_z/\sqrt2$, where $z=(-1)^{u\oplus v}$ and $P_z=(I+zZ_AZ_B)/2$. On the second pair, measure $X_AX_B$ by the same circuit conjugated with local <Hadamard gates> on the system. The two parities commute, and their joint projectors are the four <Bell-state nondemolition measurement> projectors. The singlet is exactly the outcome $(z,x)=(-1,-1)$ and remains unchanged. Local interactions and readouts fit within the stated interval; the combined verdict becomes available only after the records are compared by <classical communication>.
This corrected exact protocol resolves the triplet into three <Bell states>. It preserves the singlet and every <Bell state>, but generally destroys triplet superpositions. A binary <Lüders rule> measurement preserving all triplet coherence is a different operation: it violates the <singlet-triplet measurement causality obstruction>. The one-pair protocol used below measures a single parity and supplies the statistical test above; it does not give exact single-shot singlet verification.
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