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Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

A Bell inequality is a constraint on observable correlations satisfied by the specified class of local hidden-variable models. In finite-setting scenarios such constraints bound linear combinations of probabilities or expectation values. The CHSH inequality is one example, following from Bell factorization, measurement independence and bounded outcomes. A violation rules out the conjunction of the assumptions defining that model class; it does not alone identify a unique false premise.

Consider two spacelike-separated measurement regions $R_A$ and $R_B$, with locally chosen settings $a,b\in \{0,1\}$ and recorded outcomes $A,B\in \{-1,1\}$. A relevant past region supplies a sufficient specification $\lambda$ of the common preparation and other hidden physical information needed to screen the outcomes. This is not necessarily just a quantum state written at the source, nor a conditioning on arbitrary records which already contain the future choices. The spacetime shielding and sufficiency requirements are substantive causal assumptions.

Use Bell local causality: once the relevant past is sufficiently specified, learning facts in the remote measurement region changes no local outcome probability. In particular, requirewhere conditional probabilities are defined. The probability chain rule givesThis factorization combines parameter independence with outcome independence. Neither probability is required to be zero or one: the argument includes stochastic local hidden-variable theories.

Also assume measurement independence, $\rho (\lambda \mid a,b)=\rho (\lambda )$, so all four setting pairs average over one normalized nonnegative past-variable distribution. This is a statistical condition on the setting procedure and relevant hidden information, not a proof of a metaphysical doctrine of free will. Finally the recorded outcomes must represent the same ensemble of trials. One can include nondetections in a specified bounded-outcome rule; if instead one discards events depending on setting or outcome, a further sampling assumption is required. Setting-dependent postselection is not covered silently by the argument.

Let ${\alpha }_a(\lambda )={\sum }_{A}AP(A\mid a,\lambda )$ and ${\beta }_b(\lambda )={\sum }_{B}BP(B\mid b,\lambda )$. They lie in $[-1,1]$, and the factorization givesFor each $\lambda$,Averaging with the same $\rho$ proves the CHSH inequalityNo separate assumption of simultaneous predetermined values for incompatible quantum observables was needed. The common probability model and its screening relations did the work.

A Reichenbach common cause principle can motivate the same calculation: a complete classical common cause in the past screens the two outcomes, while each setting affects only its own wing. Together with setting-independent sampling this gives the displayed factorization. It is not enough to provide a different screening variable for each pair of settings: those constructions need not form one compatible common-cause model for the four correlations.

A stochastic Einstein locality formulation instead constrains objective chances by facts in the appropriate causal past. To use it as a sufficient premise here, specify both local dependence on settings and a joint screening condition for the two outcomes. Merely requiring local marginal chances not to depend on a remote setting gives parameter independence, which is weaker than factorization. Thus the precise stochastic locality version matters; one should not infer a Bell inequality from every condition loosely described as no superluminal influence.

If a successful preparation flag is fixed before freely selected Bell inequality settings, conditioning a local hidden-variable theory on that flag only replaces its hidden-variable distribution by the conditional distribution. The later response functions remain local and the distribution is independent of the later settings, so the CHSH inequality still holds. Setting-dependent rejection during the Bell measurements does not satisfy this argument.