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Satisfiability of conjunctive normal form formulas whose clauses have at most two unnegated variables. It is NP-complete: binary AND/OR and unary NOT gate equivalences in a Tseitin transformation already satisfy that restriction. Clauses may contain arbitrarily many negative literals; the total clause width is not restricted to two.

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.

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.

A Boolean formula is in conjunctive normal form when it is a logical conjunction of Boolean clauses. It is true precisely when every Boolean clause is true under the same Boolean valuation; an empty logical conjunction is true. For example, $(x\vee \neg y)\wedge (z)$ has two Boolean clauses with widths two and one. Width counts Boolean literal occurrences within a Boolean clause, whereas the Boolean clause count counts the conjuncts. 2-SAT bounds width by two, and 3-SAT bounds width by three. A Boolean formula with an empty Boolean clause is an unsatisfiable Boolean formula.

Every NP verifier can be compiled into a polynomial-size Boolean circuit whose inputs encode its certificate, and then into an equisatisfiable conjunctive normal form by a Tseitin transformation. This gives a polynomial-time many-one reduction from every NP language to SAT. Checking a guessed assignment proves membership, so SAT is NP-complete.

A Boolean formula is in disjunctive normal form when it is a disjunction of conjunctions of Boolean literals. Width bounds the number of literals in each conjunction, not the number of conjunctions. A depth-$s$ decision tree gives such a formula of width at most $s$ by taking the disjunction of its accepting leaf conditions; rejecting leaves give the dual conjunctive normal form.

A restriction requiring exactly one member of a set of Boolean variables to be true. A direct conjunctive normal form encoding combines one at-least-one Boolean clause, $p_1\vee \cdots \vee p_q$, with $\left(\genfrac{}{}{0ex}{}{q}{2}\right)$ at-most-one Boolean clauses, $\neg p_i\vee \neg p_j$ for $i<j$. The first supplies existence and the second supplies uniqueness. For $q=0$, the empty at-least-one Boolean clause correctly makes the condition unsatisfiable.

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.

Use a sequential SATPlan encoding with times $t=0,\dots ,T$. The fluent universe isThere are nine support Boolean variables and three clearance Boolean variables per time layer. Static block types, parameter inequalities and permanent table availability are handled during action grounding. For each well-typed ground action $a$ from parts (b)–(c), introduce an occurrence Boolean variable $a_t$ for $0\le t<T$. Allow at most one action per layer, including no action, so that a horizon of $T$ encodes at most $T$ physical actions.

The complete initial-state formula isEvery well-typed fluent not initially true is explicitly false. The final goal formula is simply $G_T=\mathrm{On}(A,B{)}_T\wedge \mathrm{On}(B,C{)}_T$. Leaving unrelated final facts unconstrained is appropriate; leaving unrelated initial facts unconstrained would introduce imaginary starting states.

For each $t<T$, include all action precondition axioms $a_t\to {\bigwedge }_{p\in \mathrm{Pre}(a)}{p}_t$, after removing static true conditions. For example,The precondition axiom for $\mathrm{MoveToTable}(C,A{)}_t$ similarly requires $\mathrm{On}(C,A{)}_t\wedge \mathrm{Clear}(C{)}_t$.

For every fluent $p\in \mathcal{P}$, let ${\mathrm{Add}}_p(t)$ be the logical disjunction of actions at $t$ adding $p$, and ${\mathrm{Del}}_p(t)$ the logical disjunction deleting it. Include the successor-state axiomIt combines action effects with the frame axiom, preventing unsupported changes. A fully grounded example isApply the same construction to every support and clearance fluent, using the exact add/delete lists. If no action occurs, both effect logical disjunctions are false, so all fluents persist.

The sequential action exclusion axioms are $\neg a_t\vee \neg b_t$ for every pair of distinct grounded actions $a,b$. For example,forbids selecting those two occurrences together. Do not add a requirement that some action occur in every layer: that would remove idle padding and change “at most $T$” into an exact-length requirement. One may additionally include true planning state invariants, such as a block having at most one support and a block support carrying at most one block; the unlimited table must be exempt from the latter restriction.

Finally, conjoin $I_0$, $G_T$, and all precondition, successor and exclusion formulas over the layers to obtain ${\phi }_T$. It is a propositional formula; if a solver requires conjunctive normal form, use a satisfiability-preserving Tseitin transformation with auxiliary Boolean variables. The equivalence between this encoding and real action sequences follows from sequential SATPlan correctness: a satisfying assignment induces exactly the stated STRIPS trajectory, and every sufficiently short valid plan supplies such an assignment.

Use an exactly-one constraint for each finite-domain variable and a Boolean clause forbidding each disallowed constraint tuple. Let the variables be $X_1,\dots ,X_n$, with explicitly listed finite constraint domains $D_i$, and let constraint $C$ have distinct-variable scope $S_C$ and allowed relation $R_C\subseteq {\prod }_{i\in S_C}{D}_i$. Introduce a Boolean variable $p_{i,v}$ for every $v\in D_i$, interpreted as $X_i=v$. The direct SAT encoding of finite-domain constraints isIn the middle logical conjunction, each unordered distinct pair is included once. The first Boolean clause requires at least one value, the pairwise Boolean clauses permit at most one, and a final Boolean clause is false exactly when all coordinates of one forbidden tuple are selected. The formula is in conjunctive normal form. An empty constraint domain produces an empty Boolean clause, correctly making the instance unsatisfiable.

For soundness of the reduction, a solution $x$ of the constraint satisfaction problem gives the Boolean valuation $p_{i,v}=1$ exactly when $v=x_i$. Every exactly-one constraint holds; a forbidden tuple differs from $x$ on at least one scoped coordinate, making its negative Boolean clause true. Thus $x$ supplies a satisfying assignment for ${\phi }_P$.

Conversely, a satisfying assignment selects a unique value $x_i$ for every $X_i$. If a constraint failed, the selected scoped tuple would belong to $F_C$, and its entire negative Boolean clause would be false. That contradicts satisfiability. Hence $x$ is a solution. Decoding takes $O({\sum }_i|D_i|)$ by scanning the value indicators. This proves $P$ has a solution if and only if ${\phi }_P$ is satisfiable.

For explicit size accounting, put $q_i=|D_i|$, $r_C=|S_C|$ and $f_C=|F_C|$. There are ${\sum }_i{q}_i$ Boolean variables,Boolean clauses, andliteral occurrences. These formulas count nonempty scopes; a false zero-arity constraint contributes an empty Boolean clause, while a true one contributes none. Formula storage is $O(K+L)$ literal-sized records; a bit-level encoding also stores indicator indices. With maximum constraint domain size $q$, $m$ constraints and maximum arity $r$, this is bounded by $O(n{q}^2+mr{q}^r)$ for nontrivial domains and scopes. In the common binary case it is $O((n+m)q^2)$.

The input representation matters: listing all forbidden tuples costs up to ${\prod }_{i\in S_C}{q}_i$ tests per constraint. This is polynomial for bounded arity and explicitly represented domains, but can be exponential in the description length of a succinct high-arity constraint. The translation is valid in that case without constituting a polynomial-size reduction. If a compact constraint is supplied as a polynomial-size Boolean circuit, one may instead encode that circuit using auxiliary Boolean variables and local conjunctive normal form equivalences, with size proportional to the circuit. No compact representation is silently assumed.

For the instance in (c), after unary filtering there are $17$ indicators. The five at-least-one Boolean clauses, $22$ pairwise at-most-one Boolean clauses and $24$ forbidden-equality Boolean clauses give $51$ Boolean clauses. Setting only $p_{a,1},p_{b,2},p_{c,3},p_{d,2},p_{e,1}$ true encodes the displayed solution.

Use a Boolean circuit of unbounded-fan-in AND and OR gates, with negations pushed to the input Boolean literals. Its circuit depth is the largest number of AND/OR gates on an input-to-output path, and its circuit size $M$ counts these gates. Constants and Boolean literals may be supplied as inputs. Equivalent standard size conventions only enlarge the absolute constants. One can merge adjacent gates of the same type; the depth-reduction argument below keeps the two possible normal forms available at each gate.

A random restriction with parameter $p$ independently leaves each variable live with probability $p$ and otherwise fixes it to zero or one with probability $(1-p)/2$ each. The Håstad switching lemma states that for a width-$w$ DNF or conjunctive normal form $F$,Width bounds Boolean literals in a term or clause, not the number of terms. A decision tree of depth at most $s$ has both a DNF and a conjunctive normal form of width at most $s$ and has real multilinear degree at most $s$. Indeed every leaf condition is a product of at most $s$ factors $x_i$ or $1-x_i$; sum the accepting leaves. The switching lemma controls the canonical decision tree, hence implies this representation bound.

We detail the Boolean circuit simplification parameters. First use a $1/10$-random restriction. For a bottom AND gate, let $X$ count its live Boolean literals on the event that no fixed Boolean literal kills it. For $w$ distinct Boolean literals,Thus the probability of retaining more than $s$ live Boolean literals without becoming constant is at most $5^{-s}$. The same calculation applies to bottom OR gates, with the opposite killing value. Repeated Boolean literals are removed and contradictory Boolean literals make a constant gate. A union bound makes all nonconstant bottom gates width at most $s$, except with probability at most $M_0{5}^{-s}$.

Now apply $d-1$ successive restrictions, each retaining probability $1/(10s)$ of the remaining variables. At a two-layer gate, the current bottom forms are width-$s$ DNFs or conjunctive normal forms. The Håstad switching lemma bounds its probability of failing to reduce to decision-tree depth at most $s$ by $2^{-s}$. Convert the successful decision tree into the opposite normal form, whose terms or clauses have width at most $s$, and merge its outer gate with the adjacent gate of the same type. This removes one layer. Repeat upwards until the output has a depth-$s$ decision tree.

The bookkeeping matters: although a decision tree may give $2^s$ clauses, each subsequent switching event is charged to the corresponding original Boolean circuit gate, not to each newly created clause. The lemma permits any number of terms. Each original gate is charged once, so the sum of the failure probabilities is at most $M{2}^{-s}$, without an extra exponential size factor. Conditioning on successful preceding stages is harmless because each next restriction is fresh and the width bound holds for every preceding success. For a general Boolean circuit the conjunctive normal form or DNF representation at a shared gate can be chosen separately for each outgoing use; no extra gates need to be counted as switching targets.

The composed restriction has live-variable probabilityFixed values are still unbiased and coordinates independent. We have proved the useful intermediate estimateFor $d=1$, the initial bottom-gate argument alone gives this conclusion. This is a depth-reduction estimate, not the desired Fourier-tail theorem used as a black box.

Next derive exactly how restrictions affect the Fourier-Walsh transform. Let $J$ be the random live set and $a$ the fixed assignment outside $J$. With ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$, the restricted coefficient at $U\subset J$ isOrthogonality of the fixed-coordinate characters eliminates cross terms after averaging over $a$. Thereforeand then averaging over $J$ gives the expected Fourier weight after a random restriction identityThe original high-degree tail is not preserved: each original coefficient is weighted by a binomial distribution survival probability. On successful restrictions the left-hand tail vanishes; on failures it is at most one by Parseval's identity and $0\le f_{\rho }\le 1$. Hence its expectation is at most $M{2}^{-s}$.

Put $u=t^{1/d}$. If $u\ge 40$, choose $s=\lfloor u/40\rfloor \ge 1$ and the preceding value of $p$. For every $|S|>t$, the binomial distribution mean satisfiesIts variance is at most $\mu$. The Chebyshev inequality givesThus the exact expectation identity bounds at least half the original tail above $t$, andWhen $u<40$ the same bound follows from the trivial total-weight bound ${\sum }_S|\hat{f}(S){|}^2\le 1$. We obtain explicit absolute constants,This supplies the requested overview of the Linial-Mansour-Nisan theorem with the actual switching scale, gate-count bookkeeping and restriction-to-Fourier calculation. The argument uses the unbounded-fan-in AND/OR Boolean circuit basis; allowing arbitrary gates would make the theorem false, since one parity gate could have all its nonconstant Fourier weight at level $n$. The original theorem is Linial, Mansour and Nisan's circuit Fourier bound.

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.

The given logical implication is false exactly when $A=B$ and $C\ne D$. For every literal on $A,B,C,D$, there is such a false Boolean valuation making that literal true. For a prescribed value of $A$ or $B$, give the other the same value and choose $C\ne D$. For a prescribed value of $C$ or $D$, give the other the opposite value and choose $A=B$.

Consequently no literal implies the whole propositional formula: there is no one-literal implicant. If an equivalent disjunctive normal form contained a one-literal conjunction term $\ell$, then every Boolean valuation satisfying $\ell$ would satisfy that disjunctive normal form, contradicting its equivalence to the given propositional formula. This proves the claim. Even if auxiliary Boolean variables are allowed in the equivalent expression, a one-literal term on one of them fails too: fix any false Boolean valuation of the original four and set the auxiliary literal true. The word “conjunct” here refers to a conjunction term in the disjunctive normal form, not a factor in a conjunctive normal form.

Use $0$ and $1$ for false and true truth values. The truth table is:

The two satisfying rows with $A=B=1$ combine by eliminating $C$, while the remaining satisfying row has $A=B=0,C=1$. An equivalent disjunctive normal form is thereforeExpanding the logical biconditional into its two implications, and using $\neg B\to C\equiv B\vee C$, gives the conjunctive normal formEach form has exactly the same three satisfying assignments as the original propositional formula.