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The Boolean satisfiability problem asks whether a Boolean formula has an assignment making it true. A complete assignment is a Boolean valuation: it gives every Boolean variable a truth value, consistently at all occurrences. A satisfying assignment, also called a model of a Boolean formula, makes the entire Boolean formula true. A solver's answer SAT asserts that at least one satisfying assignment exists; UNSAT asserts that the input is an unsatisfiable Boolean formula. These are statements about existence, rather than the truth value under one particular Boolean valuation.

For conjunctive normal form, or CNF, writeEach Boolean literal $\ell$ is a Boolean variable or its logical negation. Each Boolean clause uses inclusive logical disjunction: one true Boolean literal suffices. The outer logical conjunction requires every Boolean clause to succeed under the same Boolean valuation. Choosing a different Boolean valuation for each Boolean clause would answer a different question.

Consider the original three-Boolean variable exampleAt the Boolean valuation $(x,y,z)=(0,1,0)$, the Boolean literals in $C_1$ have values $(0,1)$, those in $C_2$ have values $(1,0)$, and those in $C_3$ have values $(0,1)$. Every Boolean clause therefore has truth value one, so $F$ has truth value one: this is a satisfying assignment.

The complete truth table has $2^3=8$ Boolean valuations. Exactly two are satisfying assignments: $(0,1,0)$ and $(1,0,1)$. In particular, $(1,1,1)$ satisfies the first two Boolean clauses but falsifies the third, so satisfying most Boolean clauses does not suffice for SAT. That different objective belongs to maximum satisfiability.

A short proof of completeness avoids listing all eight rows. If $x=0$, the Boolean clause $C_1$ forces $y=1$, after which $C_3$ forces $z=0$; $C_2$ is already true. If $x=1$, $C_2$ forces $z=1$, after which $C_3$ forces $y=0$; $C_1$ is already true. These two cases exhaust the truth values of $x$ and produce precisely the two satisfying assignments above.

Verifying a satisfying assignment and finding a satisfying assignment are different tasks. Given a candidate, evaluate the Boolean formula once; finding a candidate requires choosing compatible truth values across all its Boolean clauses. The Cook-Levin theorem says that general SAT is NP-complete, even though a proposed satisfying assignment is efficiently checkable. Its proof encodes a bounded nondeterministic computation as a Boolean formula whose satisfying assignments describe accepting computations. Stephen A. Cook's 1971 paper established this connection between computational complexity theory and propositional logic.^[ref]

A concrete use is checking finite configuration rules. Let $x$ mean that an optional feature is enabled, $y$ that a fallback is selected, and $z$ that the feature's required component is installed. Require a feature or fallback, require the component whenever the feature is enabled, and forbid using the fallback with that component. Logical implication rewrites the second rule $x\to z$ as $\neg x\vee z$; De Morgan's laws rewrite the third rule $\neg (y\wedge z)$ as $\neg y\vee \neg z$. Their logical conjunction is exactly $F$, so its two satisfying assignments give the two allowed configurations. The usefulness of SAT lies in expressing local rules uniformly while demanding one globally consistent Boolean valuation. The conclusion concerns exactly these encoded rules.

Encode each truth value by $t_k\in \{0,1\}$. A positive Boolean literal has value $t_k$ and its negation has value $1-t_k$. The sum of literal values in a clause is at least one exactly when that clause is true. Thus the displayed inequalities, with objective zero, express Boolean satisfiability as a binary integer programming feasibility problem. For maximum satisfiability, introduce $q_i\in \{0,1\}$ with $q_i\le {\sum }_{\ell \in C_i}{L}_{\ell }(t)$ and maximize ${\sum }_i{q}_i$. False clauses force zero indicators; true clauses permit indicators one, and maximization selects all of them. Hence the optimal objective equals the maximum number of true clauses.

Let $t_k\in \{0,1\}$ encode the truth value of $X_k$. For a Boolean literal $\ell$, write $L_{\ell }(t)=t_k$ if $\ell =X_k$ and $L_{\ell }(t)=1-t_k$ if $\ell ={\overline{X}}_k$. A clause is true precisely when the sum of its literal values is at least one. Thus the integer programming formulation of satisfiability is the feasibility problemwith objective zero. Every feasible binary vector gives a satisfying assignment of the Boolean satisfiability problem, and conversely. An empty clause gives $0\ge 1$ and hence infeasibility.

For maximum satisfiability, introduce binary variables $q_i$ marking clauses selected as satisfied and solveA false clause forces $q_i=0$, while a true clause permits $q_i=1$. At an optimum every permitted $q_i$ is one, since increasing it improves the objective without affecting another constraint. Therefore the objective is exactly the maximum number of simultaneously true clauses. Repeated literals cause no difficulty, and a clause containing both signs of a variable is always true.

For the literal-frequency greedy approximation for MAX-SAT, keep only clauses not already satisfied and remove literals made false. Count occurrence in clauses, rather than multiplicity within a clause. At a step let $a$ be the number of current clauses containing the chosen literal $z$, and $b$ the number containing its opposite $\overline{z}$. Since $z$ has maximum clause frequency among all remaining literals, $b\le a$.

Setting $z$ true permanently satisfies exactly the $a$ clauses containing it. Some other clauses may become empty and are then permanently lost. Every newly empty clause previously contained $\overline{z}$, so their number $d$ satisfiesA clause containing both signs is satisfied and cannot be one of the lost clauses. Repeat this accounting at every step. Satisfied clauses are removed, so no success is counted twice; each lost nonempty original clause becomes empty at exactly one step. If $G$ is the total number satisfied and $D$ the number of initially nonempty clauses lost, summing the inequalities gives $D\le G$. Every initially nonempty clause eventually falls into one of these two classes. Writing their number as $N_{+}$, we getEven an optimal assignment can satisfy at most $N_{+}$ clauses. ConsequentlyThus Greedy is a $1/2$-approximation in the convention stated in the question. The proof also covers initial empty clauses by excluding them from $N_{+}$; they are unsatisfiable for every algorithm.