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2-satisfiability restricts SAT to CNF in which each Boolean clause has at most two Boolean literals. The example in Boolean satisfiability problem is therefore a 2-SAT instance. Unlike general SAT and 3-SAT, 2-SAT has a polynomial-time algorithm; its tractability concerns finding a satisfying assignment or deciding that none exists, not merely checking a candidate. 3-SAT allows up to three Boolean literals per Boolean clause and is NP-complete, even though it includes many individually easy instances such as this one.^[ref]

The key logical equivalence iswhere $a,b$ are Boolean literals. It expresses a two-Boolean literal Boolean clause as two logical implications. Representing these logical implications as edges in a directed graph gives a route to the polynomial-time algorithm. In the example, $C_2$ says both $x\to z$ and $\neg z\to \neg x$: enabling the feature requires its component, and an absent component excludes the feature.

For a CNF input with $n$ Boolean variables, $m$ Boolean clauses, and $L$ Boolean literal occurrences, first check that the supplied Boolean valuation gives each input Boolean variable exactly one truth value. Evaluate each Boolean literal by lookup or logical negation, each Boolean clause by logical disjunction, and the Boolean formula by logical conjunction. Indexed Boolean variables and constant-cost lookups give an $O(n+m+L)$ operation count; with a finite binary string encoding the check still takes polynomial time. Thus a satisfying assignment is a certificate in computational complexity for SAT, proving that SAT belongs to NP. Finding it by exhaustive checking can require $2^n$ candidate Boolean valuations; this enumeration cost is not a lower bound on every possible SAT algorithm.^[ref]