 Solution (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:

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.^[ref]

The logical contradiction combines counterfactual 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.