CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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NumberSAT is not a Boolean membership predicate

Statement

With declared variables $(x, y)$ , let $φ = x \lor y$ and $ψ = (x \lor y) \land (\neg x \lor \neg y)$ . Then NumberSAT returns $3$ and $2$ , respectively, although the SAT membership predicate returns yes on both.

Facts & Assumptions

Given: the two displayed formulas and their declared-variable lists.

[L1]

NumberSAT is a numerical function rather than a Boolean predicate. by NumberSAT.

Counterexample

technique · direct

1.1L1givenalgebra

Exactly three assignments satisfy $x \lor y$ . Exactly the two assignments with unequal bits satisfy $(x \lor y) \land (\neg x \lor \neg y)$ . Hence [L1] gives outputs $3$ and $2$ , neither in the Boolean codomain ${0, 1}$ .

2.1step 1.1

Both counts are positive, so the associated SAT predicate maps both inputs to yes. It therefore discards information that NumberSAT retains.

3.1L1step 1.1step 2.1∎

These explicit values show that NumberSAT is a numerical function, not a Boolean membership predicate.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Arora and Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Lance Fortnow, Counting Complexity (standard reference, not scraped)