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A Boolean formula is a finite expression built from Boolean variables and Boolean operations, equivalently a propositional formula with two-valued semantics. A Boolean valuation determines its truth value. In $(x\vee y)\wedge \neg x$, logical negation acts only on $x$, and the outer logical conjunction combines $x\vee y$ with $\neg x$. Occurrences of the same Boolean variable share one truth value. The Boolean satisfiability problem asks whether some Boolean valuation makes the Boolean formula true.

A Boolean literal is a Boolean variable $u$ or its logical negation $\neg u$. The two Boolean literals have opposite truth values: a Boolean valuation $v$ gives $v(\neg u)=1-v(u)$ when false and true are encoded as zero and one. Thus $u=0$ makes the positive Boolean literal $u$ false and the negative Boolean literal $\neg u$ true. A Boolean clause combines Boolean literals, and a Boolean-pair colouring gadget represents a Boolean variable and its logical negation by two vertices with opposite Boolean colours.

A Boolean operation is a Boolean function taking truth values encoded as bits: $0$ means false and $1$ means true. These operations supply the concrete two-valued interpretation of a Boolean algebra. A Boolean valuation fixes the inputs; a truth table lists every possible input. With two input bits there are four rows.

The basic logical connectives are logical conjunction (AND, $P\wedge Q$, both true), logical disjunction (OR, $P\vee Q$, at least one true), exclusive or (XOR, $P\oplus Q$, exactly one true), and logical negation (NOT, $\neg P$, the opposite value). OR is inclusive: it accepts the case where both inputs are true.^[ref]

For an original program-condition example, let $P$ mean “the backup is recent” and $Q$ mean “the backup is verified”. A rule requiring a recent, verified backup uses logical conjunction; one accepting either qualification uses logical disjunction. A rule flagging a mismatch between the two qualifications uses exclusive or. A recent, verified backup gives $(P,Q)=(1,1)$: AND and OR accept it, while XOR does not flag it. The last row of the truth table makes this difference explicit.

The same small Boolean operations build larger tests and Boolean circuits. In a programming language, the value of a condition and the procedure for evaluating it both matter. For Boolean operands, Python's and, or, and not implement these logical connectives. Evaluation proceeds from the left; and skips its right operand if the left is false, and or skips it if the left is true. More generally, and and or return an operand, whereas not returns a Boolean.^[ref]

For example, the Python condition denominator != 0 and numerator / denominator > 2 avoids division by zero because the logical conjunction's second test is evaluated only when the first succeeds. Reversing the tests preserves the abstract truth table for defined operands but can raise a division-by-zero error handled through exception handling. A rewrite involving a side effect must also preserve the needed evaluation behavior.

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 variable takes one of two truth values, false or true, encoded as zero or one. It is an input to a Boolean formula; a Boolean literal uses the Boolean variable either positively or with logical negation. A Boolean valuation assigns one truth value to each distinct Boolean variable, so three distinct Boolean variables give eight possible complete assignments regardless of how often each appears.

A bounded arithmetic sentence is an arithmetic bounded formula with no free variables. Every quantifier is restricted by an arithmetic first-order term. In the standard natural numbers such a first-order sentence has a definite truth value determined by finite bounded searches.

A Boolean clause is an expression ${\ell }_1\vee \cdots \vee {\ell }_k$ formed from Boolean literals, true when at least one Boolean literal is true. The empty Boolean clause is false. Repeating a Boolean literal does not change the truth value. A CNF Boolean formula is a logical conjunction of these Boolean clauses.

For $(\neg x\vee z)$, the only failing pair is $(x,z)=(1,0)$: its two Boolean literals then both have truth value zero. The one-Boolean literal Boolean clause $(x)$ requires $x=1$. A Boolean clause containing both $x$ and $\neg x$ is a tautology, since one of those Boolean literals is true under every Boolean valuation. These observations separate Boolean clause satisfaction from the simultaneous satisfaction required by SAT.

A congruence relation on a first-order structure is an equivalence relation compatible with each interpreted function and relation. Equivalent input tuples give equivalent function outputs, and an atomic relation has the same truth value on equivalent input tuples. This compatibility makes the functions and relations well-defined on the quotient set. In a Henkin term model with equality, membership of a formal equality in the maximal consistent theory gives precisely the equivalence relation used for this quotient.

De Morgan's laws exchange logical conjunction and logical disjunction when a logical negation moves across them:Here $\equiv$ denotes logical equivalence: the two propositional formulas agree under every Boolean valuation.^[ref]

A complete verification uses the four rows of a truth table. In input order $(0,0),(0,1),(1,0),(1,1)$, both sides of the first logical equivalence have output sequence $(1,1,1,0)$; both sides of the second have $(1,0,0,0)$. A proof in words is equally short: failure of “both” means at least one failure, whereas failure of “at least one” means both fail. The differing middle rows explain why simply pushing logical negation inward without changing AND to OR is wrong.

For the events in Boolean operations on events, the corresponding set identities areOne can prove these identities by checking an arbitrary set element $u\in U$: its two membership truth values obey De Morgan's laws. The same reasoning works for any two subsets of a common universal set.

For a nested program test, let $R$ mean “backup recent”, $V$ mean “backup verified”, and $O$ mean “manual override authorized”. Rejecting when neither an acceptable backup nor an override is present givesThe transformed condition reads “backup stale or unverified, and no override”. Apply the second of De Morgan's laws at the outside, then the first at the inner logical conjunction. For $(R,V,O)=(1,0,0)$ it rejects; changing only $O$ to one makes it accept. With Boolean operands in Python, not ((recent and verified) or override) and ((not recent) or (not verified)) and (not override) have the same result and inspect the operands in the same order, with the same skipped operands.

Exclusive or returns one exactly when its two input bits differ. Its truth table gives $0\oplus 0=0$, $0\oplus 1=1$, $1\oplus 0=1$, and $1\oplus 1=0$. The identitiesexpress exclusive or using logical conjunction, logical disjunction, and logical negation. The first expression accepts at least one input and then excludes both; the second explicitly lists the two accepted input pairs.

For numerical bits, exclusive or is addition in modular arithmetic modulo two:The binary operation is commutative and associative, because addition modulo two has those properties. It satisfies $x\oplus 0=x$, $x\oplus 1=\neg x$, and $x\oplus x=0$. These identities explain why a bit can be toggled and restored with the same operation.

On binary strings of equal length, bitwise exclusive or applies the one-bit rule separately at each position. For example,A bit mask of $0110$ toggles exactly the middle two bits. Applying the mask twice restores the original binary string: each selected bit changes twice, and each unselected bit stays fixed. This positional operation is distinct from taking the exclusive or of the overall truth values of two integers.

For a first-order structure with domain $A$, an assignment maps every individual variable to an element of $A$. It extends recursively to first-order terms using the interpreted functions. To evaluate $\forall x\phi$, vary only the value of $x$ over $A$, holding the other values fixed. The truth value of a first-order formula depends only on its free variables; a first-order sentence is therefore independent of the assignment.

A first-order structure whose domain is the Herbrand universe, with constants denoting themselves and function symbols acting by syntactic term formation. Its predicates are specified by assigning truth values to ground atoms. With first-order equality, ordinary equality of syntactic terms imposes additional restrictions; more general term models may require quotienting by a congruence.

Equivalently, the predicate meanings of a Herbrand interpretation correspond to a subset of its Herbrand base.

A choice of a domain and meanings for the symbols of a formal language, used to evaluate its sentences. The same sentence can have different truth values under different interpretations, unless it is valid under every admitted interpretation.

A binary logical connective true exactly when its two arguments have the same truth value. It is logically equivalent to $(P\to Q)\wedge (Q\to P)$, and hence to $(\neg P\vee Q)\wedge (\neg Q\vee P)$. Its truth table has value one on input pairs $(0,0)$ and $(1,1)$, and zero on $(0,1)$ and $(1,0)$.

Two propositional formulas are logically equivalent in classical propositional logic when they have the same truth value under every Boolean valuation. Identical output columns in a truth table prove logical equivalence; one differing row refutes it. De Morgan's laws give useful examples.

A logical variable is a symbol whose value is supplied by the semantics of a logical language. A propositional variable receives a truth value under a Boolean valuation. An individual variable of first-order logic receives an element of a structure under an assignment. These are different kinds of variable and should not be identified.

A logical matrix specifies truth values, designated values, and operations interpreting the primitive connectives. To prove nonderivability from an axiom set, it suffices that every axiom instance is designated and every inference rule preserves designation, while the proposed conclusion is not designated under some valuation. The primitive connective symbols need not receive their intended classical meanings: preservation by the displayed rules is what makes the syntactic obstruction valid. Rules with discharged assumptions, such as a deduction theorem for propositional logic, must also preserve consequence for arbitrary assumption sets.

A partial Boolean assignment gives truth values to some of a Boolean formula's Boolean variables. An extension is a complete Boolean valuation agreeing on the assigned Boolean variables. In CNF, a Boolean clause containing a true assigned Boolean literal is already satisfied; a Boolean clause whose every Boolean literal is assigned false rules out all extensions. A Boolean clause with no true assigned Boolean literal and at least one unassigned Boolean variable is unresolved. An unresolved Boolean clause need not have a compatible extension with the other Boolean clauses.

For $F$ in Boolean satisfiability problem, setting $x=y=0$ already falsifies $C_1$, so both choices of $z$ can be discarded. Setting only $x=0$ leaves $C_1$ unresolved and forces $y=1$ if a satisfying assignment is to survive; then $C_3$ forces $z=0$. Branching on an unassigned Boolean variable into its two truth values and discarding branches with a false Boolean clause is sound: every satisfying assignment follows one branch and cannot extend a false Boolean clause. If all branches are eliminated, the original Boolean formula is an unsatisfiable Boolean formula.

The clique decision problem belongs to NP. First reject $k>|V(G)|$. Otherwise a complexity certificate lists $k$ distinct vertices. Check distinctness and test that each of the $\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ pairs is an edge. Under a conventional explicit graph encoding, the complexity certificate length and verification time complexity are polynomial in the input length. The early rejection is relevant when $k$ is given in the binary numeral system.

For NP-hardness, first transform a SAT instance into an equisatisfiable conjunctive normal form $F=C_1\wedge \cdots \wedge C_m$ of polynomial size using a Tseitin transformation. This step avoids the possibly exponential increase caused by distributing arbitrary Boolean formulas into conjunctive normal form. Introduce a fresh Boolean variable for each subformula, constrain it to have the connective's truth value using a constant number of Boolean clauses, and require the top variable to be true. A Boolean valuation of the original Boolean variables satisfies the original Boolean formula exactly when it extends to a satisfying assignment of these constraints.

Create one vertex for each occurrence of a Boolean literal in each Boolean clause. Join two such vertices by an edge exactly when they belong to different Boolean clauses and their Boolean literals are not a variable and its logical negation. Set $k=m$. There are at most as many vertices as Boolean literal occurrences and at most quadratically many edges, so this construction takes polynomial time.

If $F$ has a satisfying assignment, choose one true Boolean literal occurrence from each Boolean clause. No two selected Boolean literals are contradictory, so their $m$ vertices form a clique. Conversely, an $m$-clique contains at most one vertex from each Boolean clause, because there are no edges within a Boolean clause. It therefore selects exactly one occurrence from each of the $m$ Boolean clauses. The selected Boolean literals are pairwise noncontradictory, so assign their Boolean variables the indicated truth values and assign any other Boolean variables arbitrarily. This Boolean valuation satisfies every Boolean clause.

An empty Boolean clause gives no vertex in its group, correctly preventing an $m$-clique. An empty logical conjunction can instead be mapped to a fixed yes-instance, avoiding any convention about $k=0$. Malformed SAT descriptions can be mapped to a fixed no-instance. We have proved

It suffices to eliminate one existential quantifier from a quantifier-free formula $\theta (y,\bar{x})$. Indeed, logical connectives preserve equivalence, and a universal quantifier can be rewritten using $\forall y\alpha \equiv \neg \exists y\neg \alpha$. An induction on the construction of a first-order formula then eliminates all its quantifiers from the inside out.

Convert $\theta$ to disjunctive normal form. Since existential quantification distributes over a finite logical disjunction, consider a single logical conjunction of atomic first-order formulas and their negations. There are no function or constant symbols in this first-order signature, so its atoms are equalities between variables and instances of $E$ between variables.

For a tuple $\bar{x}=(x_1,\dots ,x_r)$, enumerate its possible complete atomic diagrams $\delta (\bar{x})$. Such a quantifier-free formula specifies which variables are equal and, between distinct equality classes, exactly which pairs are adjacent. Keep only the diagrams in which first-order equality is an equivalence relation and $E$ is symmetric and irreflexive. There are finitely many diagrams for each $r$, and every tuple in a first-order model of $T$ satisfies exactly one of them. Every retained diagram is realizable: insert one vertex for each equality class successively, using the extension axioms to prescribe its adjacency to earlier vertices.

Fix one diagram $\delta$. A potential existential witness for $y$ has one of two forms.

All first-order formulas involving only $\bar{x}$ are already decided by $\delta$. Therefore, for each diagram, the existence of a suitable $y$ has a definite answer, determined by a finite syntactic test. Let $\mathcal{D}$ be the set of diagrams for which at least one disjunct and at least one of the two witness cases succeeds. ThenThe right side is a quantifier-free formula. An empty logical disjunction means false, and for an empty tuple the empty atomic diagram means true. These logical truth constants also handle closed first-order sentences in this constant-free first-order signature.

For a concrete illustration of the fresh-witness case,If $x_1=x_2$, the edge requirements conflict. Otherwise the extension property supplies a vertex outside the finite set of parameters with the desired pattern; the extra avoidance of $x_3$ causes no obstruction. The general diagram construction handles all such equality coincidences systematically and proves quantifier elimination.