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A polynomial-time many-one reduction represents each variable by an adjacent literal pair and each clause by a triangle in a graph. Connect each occurrence vertex to its corresponding literal vertex. A vertex cover of size $n+2m$ must choose one literal vertex per variable and two vertices per clause, and its omitted clause vertex forces a true literal. Conversely a satisfying Boolean valuation supplies such a cover. This establishes NP-hardness of the vertex-cover decision problem from 3-SAT.

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 valuation $v$ assigns each primitive proposition a value in $\{\mathrm{F},\mathrm{T}\}$, equivalently $\{0,1\}$, and extends recursively to every propositional formula using the classical logical connectives. The same primitive proposition always receives the same value within one Boolean valuation. For compound propositional formulas $A$ and $B$,Every other case receives the opposite value. Thus $v(p)=\mathrm{T},v(q)=\mathrm{F}$ gives $v((p\to q)\wedge p)=\mathrm{F}$. A truth table varies the assignment; recursive evaluation computes one row. The truth-functional restriction means that changing what the primitive propositions describe, while retaining their values, cannot change the value of the compound propositional formula. See forall x: Calgary, chapter 9 and section 10.1.

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 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 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.

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.

If a consistent propositional theory locally omits every member of a countable family of propositional types, it has a Boolean valuation omitting them all. One may also impose any finite condition consistent with the theory. Successively extend the condition by a negated member of the next type; local omission preserves consistency. The propositional compactness theorem then supplies the valuation. Countability of the ambient language and decidability of consistency are not needed for this argument.

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.

An argument has logical validity in classical propositional logic when no Boolean valuation makes all premises $\Gamma$ true and the conclusion $A$ false: $\Gamma \models A$. With no premises, $\models A$ says that $A$ is a tautology. See forall x: Calgary, sections 12.4–12.6.

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.

A propositional type is a set $\Sigma$ of propositional formulas; a Boolean valuation realizes it if every member is true, and omits it if at least one member is false. A consistent propositional theory $T$ locally omits $\Sigma$ if every formula $\theta$ which implies all members of $\Sigma$ modulo $T$ is refutable modulo $T$. Equivalently, whenever $T\cup \{\theta \}$ is consistent, there is a $\sigma \in \Sigma$ for which $T\cup \{\theta ,\neg \sigma \}$ is consistent. For a consistent propositional type, this says that it is a nonprincipal propositional type.

The extended omitting types theorem for propositional logic says that if a consistent $T$ locally omits each of a countable family ${\Sigma }_0,{\Sigma }_1,\dots$, there is one Boolean valuation satisfying $T$ and omitting them all. It also works inside any prescribed finite condition consistent with $T$.

To prove it, start with such a condition ${\theta }_0$, taking $\top$ when none is prescribed. At stage $i$, choose ${\sigma }_i\in {\Sigma }_i$ such thatSuch a choice exists by local omission. More explicitly, failure would imply $T⊢{\theta }_i\to \sigma$ for every $\sigma \in {\Sigma }_i$, hence $T⊢\neg {\theta }_i$, contradicting the stage invariant. Every finite subset of $T\cup \{{\theta }_0\}\cup \{\neg {\sigma }_i:i\in \mathbb{N}\}$ lies inside a consistent stage. By the propositional compactness theorem and the completeness theorem for propositional logic, it has a Boolean valuation. That Boolean valuation satisfies $T$ and makes the selected member ${\sigma }_i$ of every type false. All types are therefore omitted simultaneously. No effective test for consistency is assumed, and the argument does not require the ambient propositional language to be countable. Only the family of omission requirements is countable.

Use the specified construction as a polynomial-time many-one reduction from 3-SAT. There are $2n+3m$ vertices and $n+6m$ edges: one edge per variable pair, three edges per clause triangle in a graph, and three occurrence edges per clause. Any vertex cover must take at least one vertex from every variable pair and at least two from every clause triangle. Therefore every cover has at least $n+2m$ vertices.

Given a satisfying Boolean valuation, include the vertex corresponding to the true Boolean literal in each variable pair. In each clause choose a true occurrence and omit its clause vertex, including the other two. Every variable edge and triangle edge is covered. An occurrence edge with an included clause endpoint is covered automatically; the only omitted occurrence vertex is joined to the included true-literal vertex. This is a vertex cover with exactly $n+2m$ vertices.

Conversely, a cover of that size must use exactly one vertex in each variable pair and exactly two in each triangle. Declare a variable true precisely when its positive-literal vertex is included. Each clause has one omitted vertex. Its occurrence edge forces the corresponding literal vertex into the cover, so that literal is true. Every clause is therefore satisfied. ThusThe construction and target size are polynomial in the input length. Since 3-SAT is NP-complete, the decision problem is NP-hard. The usual “at most $k$” version has the same equivalence here because $n+2m$ is an unavoidable lower bound. More generally, a cover with fewer than $k\le |V|$ vertices can be padded to exactly $k$. Verifying a proposed cover is polynomial, so the usual vertex-cover decision problem is also in NP and hence NP-complete.

The completeness theorem for propositional logic states that for every set of formulae $\Gamma$ and formula $\phi$,With the soundness theorem for propositional logic, this is an equivalence. It is also equivalent to saying that every consistent set of formulae has a model.

The usual countable proof lists all formulae and extends a consistent theory one formula at a time. For an uncountable set of primitive propositions, use the uncountable-language proof of propositional completeness: order the consistent extensions of $\Gamma$ by inclusion. The union of every chain is consistent, since a proof of a contradiction would use only finitely many assumptions and hence would already lie in one member of the chain. Zorn lemma gives a maximal consistent set in propositional logic $M$. Define a Boolean valuation by $v(p)=1$ exactly when $p\in M$. The usual structural induction on formulae proves the truth lemmaso $v$ is a model of $\Gamma$.

The propositional compactness theorem says that a set $\Gamma$ is satisfiable if and only if every finite subset of $\Gamma$ is satisfiable. Only the forward implication is immediate. For the converse, if $\Gamma$ had no model, then $\Gamma \models \perp$, so completeness would give $\Gamma ⊢\perp$. Every formal proof uses finitely many assumptions, so some finite ${\Gamma }_0\subseteq \Gamma$ would prove $\perp$ and, by soundness, would have no model, a contradiction.

The decidability theorem for propositional logic says that there is an algorithm deciding whether a propositional formula is a theorem. A formula contains only finitely many primitive propositions, even when the full language is uncountable. Its finite truth table decides whether it is valid, and completeness says that validity is equivalent to theoremhood.

Now let $(X,<)$ be the given partially ordered set. For every distinct $x,y\in X$, introduce propositions $p_{xy}$ and $q_{xy}$. Form a propositional theory $T$ containing the following clauses:

The first two families say that $p$ and $q$ encode strict total orders. The third makes both orders extend $<$, while the fourth makes them disagree on every incomparable pair.

Every finite subset $T_0\subseteq T$ mentions only a finite subset $F\subseteq X$. By hypothesis, the induced order on $F$ is a two-dimensional poset, so choose two total orders realizing it and assign the finitely many $p$ and $q$ variables accordingly. This satisfies $T_0$. Thus every finite subset of $T$ is satisfiable, and propositional compactness gives a model of all of $T$.

Define $x{<}_{1}y$ when $p_{xy}$ is true and $x{<}_{2}y$ when $q_{xy}$ is true. The clauses make ${<}_1$ and ${<}_2$ total orders on $X$. They both contain the original order, and their intersection contains no incomparable pair. Hencewhich proves the finite-local characterization of two-dimensional partially ordered sets.

A propositional formula is built from atomic propositions using logical connectives. Unlike a first-order formula, it has no quantifiers over individuals. A Boolean valuation interprets its atoms as truth values and evaluates its connectives recursively.

Propositional logic studies reasoning with whole propositions represented by letters such as $p$ and $q$, and propositional formulas built from them using logical connectives. The classical two-valued, truth-functional presentation below assigns each primitive proposition either true or false: each compound value depends only on its component values. This is classical propositional logic; other systems, including Intuitionistic propositional logic, have different semantics and rules.

Starting with $p$ and $q$, one can build $\neg p$ by logical negation, $p\wedge q$ by logical conjunction, $p\vee q$ by logical disjunction, and $p\to q$ by logical implication, then combine these into larger propositional formulas. Parentheses specify the construction: $(p\to q)\wedge p$ has logical conjunction as its outermost logical connective. Primitive propositions have no internal structure in this language; a first-order formula can additionally describe objects and quantify over them. The formation rules are given in forall x: Calgary, chapter 6.

The following truth table shows the four basic logical connectives. Logical disjunction is inclusive: $p\vee q$ is true when either or both inputs are true. Classical logical implication has exactly the same values as $\neg p\vee q$; its only false row has $p$ true and $q$ false. These truth conditions are specified in forall x: Calgary, chapter 9.

Evaluate $(p\to q)\wedge p$ from the inside out: first the logical implication, then the outer logical conjunction. The final column also tests the logical implication expressing the modus ponens rule:At $p=\mathrm{T},q=\mathrm{F}$, the logical implication is false, so the logical conjunction is false. The abstract syntax tree below separates the construction of the propositional formula from the values propagated through it. Both occurrences of $p$ receive the same value from the Boolean valuation. With the opposite assignment $p=\mathrm{F},q=\mathrm{T}$, the logical implication is true but the logical conjunction is still false because its second input $p$ is false. Recursive evaluation and exhaustive truth tables are explained in forall x: Calgary, chapter 11.

The last column is always true: $((p\to q)\wedge p)\to q$ is a tautology. This expresses the logical validity of the modus ponens rule: whenever both premises $p\to q$ and $p$ are true, $q$ is true. It does not say the premises are true under every Boolean valuation; their logical conjunction is true only on the first row. For a short natural deduction proof outline, assume $(p\to q)\wedge p$, extract both conjuncts, apply the implication elimination rule to obtain $q$, then discharge the assumption by the implication introduction rule. See forall x: Calgary, sections 17.3–17.5.

Three tests answer different questions: Boolean satisfiability requires at least one true row, a tautology requires every row true, and a contradiction has every row false. Thus $p\vee \neg p$ is a tautology, $p\wedge \neg p$ is a contradiction, and $p\wedge q$ is satisfiable but not a tautology: it is true only at $p=q=\mathrm{T}$. See forall x: Calgary, sections 12.1–12.4.

Propositional logic is useful because an inference pattern can be checked once and reused for many choices of propositions. Its beauty is that a finite truth table completely settles classical truth-functional questions about a fixed propositional formula. But logical validity does not verify real atomic claims. If $p$ means “the switch is closed” and $q$ means “the lamp is lit”, a valid use of the modus ponens rule still requires establishing its premises about the apparatus; the truth table supplies neither an observation nor a causal explanation. See forall x: Calgary, chapter 2 and chapter 10.

A propositional theory is a set of propositional formulas, sometimes identified with its deductive closure. It is a consistent set of formulae when contradiction is not derivable from it. The completeness theorem for propositional logic identifies consistency with existence of a satisfying Boolean valuation.