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3-SAT is the Boolean satisfiability problem restricted to Boolean formulas in conjunctive normal form with at most three Boolean literals per Boolean clause. It is NP-complete. A nonempty short Boolean clause can be padded to three positions by repeating a Boolean literal, without changing satisfiability.

A Boolean formula is a finite expression built from Boolean variables and Boolean operations. A choice of truth values determines its value. The Boolean satisfiability problem asks whether some choice makes it true.

Maximum satisfiability asks for a truth assignment maximizing the number of true clauses of a conjunctive normal form formula. The Boolean satisfiability problem asks whether all its clauses can be made true. Binary clause indicators give an integer programming formulation, and the literal-frequency greedy approximation for MAX-SAT guarantees at least half the optimum. Counts of clauses, rather than multiplicities of repeated literals within one clause, are relevant to this objective.

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.

The Cook-Levin theorem states that the Boolean satisfiability problem is NP-complete under polynomial-time many-one reductions. Membership in NP follows by guessing an assignment and evaluating the formula in time polynomial in its description length.

For hardness, let $L\in \mathrm{NP}$ have a deterministic polynomial-time verifier $M(x,y)$, with a polynomial witness-length bound. Use a fixed polynomial certificate length: a short certificate is encoded by its length followed by padded data, and the verifier checks this encoding. Pad the computation to exactly $T=T(|x|)$ steps, with accepting and rejecting states absorbing. Enlarge $T$ polynomially if necessary to cover input/witness initialization. A standard polynomial slowdown permits a single-tape Turing machine, so it suffices to handle that model.

Encode a tape cell by a fixed number of bits recording its alphabet symbol and either no head or the head's finite control state. Starting with one head, a cell's next label depends only on its own label and the two neighboring labels: a head can change the symbol where it sits and can enter only a neighbor. Each such finite local function has a constant-size Boolean circuit. The initial row fixes $x$, blanks and the starting head, leaving only the witness bits as Boolean circuit inputs. There are $O(T)$ relevant cells with blank margins beyond every possible head position, and $T$ updates. Repeating these local Boolean circuits produces a Boolean circuit $C_x(y)$ of size $O(T^2)$, with an output detecting an accepting head in the final row. The construction is computable in polynomial time. Induction on rows shows that every assignment to $y$ produces exactly the verifier's valid computation; invalid local encodings can be assigned arbitrary Boolean circuit behavior because they never arise from the valid initial row.

Convert this Boolean circuit to conjunctive normal form using a Tseitin transformation. Introduce one variable for each wire, and encode each gate output $z$ by:

Unit clauses fix constant sources and assert the final output. Each input assignment has exactly one extension to its gate values, so the resulting formula $F_x$ is satisfiable precisely when some witness makes $M(x,y)$ accept. It has $O(T^2)$ clauses of bounded length and is produced in polynomial time. ThusThis proof also gives hardness for clauses of at most three literals, without needing a separate satisfiability assumption.

This is Ladner's theorem. Assume $\mathbf{P}\ne \mathbf{NP}$ and let $S$ be the Boolean satisfiability problem. Enumerate all clocked polynomial-time deterministic machines as $D_0,D_1,\dots$ and all clocked polynomial-time candidate reductions as $R_0,R_1,\dots$. A standard delayed-diagonalization schedule gives a nondecreasing, unbounded, polynomial-time computable function $s:\mathbb{N}\to \mathbb{N}$ that increases only by one. DefineThe schedule alternates two requirements. At even stage $2j$, it holds $s$ fixed while searching successively larger finite sets of strings for an $x$ on which $D_j(x)$ disagrees with membership in $L$; after finding one it increments $s$. At odd stage $2j+1$, it searches for an $x$ such thatand then increments $s$. Length and simulation budgets are increased slowly enough that each finite search is eventually exhaustive but computing $s(n)$ still takes polynomial time. This is achieved, for example, by permitting only $n$ simulation steps and searches on strings of logarithmic length before deciding $s(n+1)$.

Every stage must finish. If an even stage $2j$ remained forever, then $s$ would eventually be a fixed even number, so $L$ and $S$ would differ on only finitely many strings. If $D_j$ decided $L$, those finitely many exceptions could be hardwired to decide $S$ in P, contradicting $\mathbf{P}\ne \mathbf{NP}$. If an odd stage $2j+1$ remained forever, then $s$ would eventually be odd and $L$ would be finite. A correct polynomial-time many-one reduction $R_j$ from $S$ to $L$ would again put $S$ in P. Thus every $D_j$ fails to decide $L$, and every $R_j$ fails to reduce $S$ to $L$.

Finally, $L\in \mathbf{NP}$: compute $s(|x|)$ and, when it is even, use the usual polynomial-time certificate for satisfiability. Hence