= Solution
{disambiguate=Hardy contradiction}
In a <deterministic local hidden-variable model>, write $C^+(R,\lambda)$ or $D^+(I,\lambda)$ without an argument for the remote splitter setting. That omission is the <parameter independence> assumption: the local response for a fixed complete hidden state does not change when the other wing changes its apparatus. <Measurement independence> also requires the hidden-state ensemble to be independent of the freely selected settings, allowing the same $\lambda$ assignments to be compared across configurations.
The zero probabilities derived above then constrain those assignments, apart from sets of hidden states of zero measure:
* Case $RR$ requires $C^+(R,\lambda)C^-(R,\lambda)=0$.
* Case $IR$ requires $D^+(I,\lambda)=1\Rightarrow C^-(R,\lambda)=1$.
* Case $RI$ requires $D^-(I,\lambda)=1\Rightarrow C^+(R,\lambda)=1$.
For a trial with $D^+(I,\lambda)D^-(I,\lambda)=1$, the last two implications <force> $C^+(R,\lambda)=C^-(R,\lambda)=1$, contradicting the first condition. Thus these deterministic local assignments predict zero <probability> for the double-$D$ event, whereas quantum theory predicts $1/16$. This is <Hardy's paradox>, an inequality-free ideal contradiction. With imperfect apparatus the zero-event conditions become small nonzero rates and require a statistical inequality rather than declaring a contradiction from one imperfect detection.<Quantum mechanics, local realistic theories, and Lorentz-invariant realistic theories by Lucien Hardy>[1992 abstract: a two-particle proof of Bell's theorem without inequalities]{ref}
The logical contradiction combines counterfactual \b[hidden-variable assignments]; it does not assign simultaneously observed path properties to a <quantum state>. Nor does it allow a message to be sent by changing a distant splitter: the observable local marginal obeys <quantum no-signalling>. For example, fixing the plus-side splitter to $I$ gives $P(D^+)=1/8$ for either minus-side setting in the results above.
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