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A finite game represented succinctly by propositional logic. Each player controls a disjoint collection of Boolean variables and chooses their truth values. Together those choices determine a Boolean valuation. A player's goal is a propositional formula, with payoff one if satisfied and zero otherwise. The control sets normally form a set partition of all variables.

A pure strategy strategy profile is a Nash equilibrium if every unsatisfied player remains unsatisfied after every change to that player's controlled variables. Checking a candidate can therefore require solving a Boolean satisfiability problem, even though evaluating its current payoffs is easy.

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.

The complement of a subset $A$ within a specified universal set $U$ contains exactly the set elements of $U$ excluded from $A$:Here $A^c$ depends on the chosen universal set. Within $U$, taking a complement performs logical negation (NOT) on membership. Writing $U\setminus A$ keeps the ambient set visible when there could be ambiguity.

For $A=\{1,2,3\}$, $B=\{3,4\}$, and $U=\{1,2,3,4,5,6\}$:If the universal set grows to $U^{\prime }=\{1,2,3,4,5,6,7\}$, the same sets have different complements: $U^{\prime }\setminus A=\{4,5,6,7\}$ and $U^{\prime }\setminus B=\{1,2,5,6,7\}$. In a database illustration, “records that fail this condition” therefore requires deciding which records are eligible to begin with; those eligible records form $U$.

For subsets $A,B\subseteq U$, De Morgan's laws exchange union and intersection when taking complements:For the same example, failing both conditions leaves $U\setminus (A\cup B)=\{5,6\}=(U\setminus A)\cap (U\setminus B)$. Failing at least one condition leaves $U\setminus (A\cap B)=\{1,2,4,5,6\}=(U\setminus A)\cup (U\setminus B)$. This is useful for rewriting exclusion rules into separate membership tests. The symmetry is also pleasing: exchanging union with intersection turns one rule into its partner.

A brief membership proof works in classical propositional logic. For any $x\in U$, let $P$ mean $x\in A$ and $Q$ mean $x\in B$. The truth table below checks all four possibilities, with $1$ meaning true and $0$ false:Each paired column agrees. Translating logical disjunction (OR), logical conjunction (AND), and logical negation (NOT) back into membership proves the two set equalities. For the general method of reducing set equalities to equivalences in propositional logic, see Lehman, Leighton and Meyer, Mathematics for Computer Science (2015), §4.1.5 and Problem 4.5.

A historical perspective appears in Augustus De Morgan's ^[ref]: he describes treating each name together with its contradictory name. In the membership viewpoint here, a set and its complement likewise pair a condition with its logical negation.

There is an algorithm deciding whether any finite propositional formula is a theorem. A formula contains only finitely many atoms, so a finite truth table decides whether it is valid; the completeness theorem for propositional logic identifies validity with provability.

The finite valuation space of propositional logic gives the decision procedure; this theorem is a property of that logic rather than an alternative name for it.

In propositional logic with the usual two truth values, negating a proposition twice preserves its truth value. Thus the complement of a set taken twice, in one fixed universal set, is the original set.

A first-order clause containing no variables. Its first-order terms are ground terms, so resolution between such first-order clauses reduces to propositional logic on the ground atoms.

A deduction from premises $\Gamma$ in propositional logic is a finite sequence of formulas, each an allowed logical axiom instance, a member of $\Gamma$, or obtained from earlier formulas by an inference rule, such as modus ponens. For example a Hilbert system can use the schemas $p\to (q\to p)$, $[p\to (q\to r)]\to [(p\to q)\to (p\to r)]$, and $(\neg q\to \neg p)\to (p\to q)$, with modus ponens as its rule. Its last formula is the conclusion. A finite deduction uses only finitely many premises. The completeness theorem for propositional logic says $\Gamma \models v$ implies $\Gamma ⊢v$; together with soundness, syntactic consistency is equivalent to existence of a satisfying valuation.

For compactness, if $\Gamma$ had no model, completeness would give a deduction of a contradiction. That deduction uses a finite subset of $\Gamma$, which therefore also has no model. Contrapositively, if every finite subset has a model, so does $\Gamma$.

Fix the given valuation on $Q$ and let $L_v$ be its collection of true literals. Suppose $\{s\}\cup L_v$ were inconsistent. A finite subset of its literals, with conjunction $\ell \in L(Q)$, would satisfy $s⊢\neg \ell$ by the deduction theorem and classical logic. But $\neg \ell \in U$ whereas the given valuation makes $\ell$ true, contradicting that it satisfies all of $U$. Hence $\{s\}\cup L_v$ is consistent.

If $U\cup \{\neg t\}$ had a model, restrict that valuation to $Q$ and apply the preceding consistency result. Completeness supplies an extension to $P\cup Q$ making $s$ true and preserving those $Q$ values. Combine its $P$ values with the original model's $Q\cup R$ values. Disjointness of $P,Q,R$ makes this a well-defined valuation satisfying $s$ and $\neg t$, contradicting soundness of $s⊢t$. Thus $U\cup \{\neg t\}$ is inconsistent.

Compactness now supplies $u_1,\dots ,u_k\in U$ with $u_1,\dots ,u_k⊢t$. Put $u={\bigwedge }_j{u}_j$. ThenThis proves propositional interpolation. The empty conjunction is the truth constant; if the language omits truth and falsehood symbols, use an equivalent tautology when $Q$ is nonempty, or adjoin those constants for the empty-common-language case.

Define a valuation $v$ on primitive propositions byThe assumptions say exactly that $S$ is a maximal consistent set in propositional logic. A structural induction on formulae provesFor example, consistency and the decision property give $\neg t\in S$ exactly when $t\notin S$, while deductive closure gives $s\wedge t\in S$ exactly when both $s,t\in S$; the other connectives follow similarly. Every member of $S$ is therefore true under $v$, so

Take a first-order sentence $Y$ refutable in $S$ and suppose, toward contradiction, that a first-order consistent $S$ proved $\neg P_S(┌Y┐)$. Propositional logic would then giveLöb's inference, applied with $X=Y$, would give $S⊢Y$, contradicting the existing formal proof of $\neg Y$. Hence first-order consistency excludes the proposed formal proof of non-provability.

Thus the familiar unprovability of the first-order consistency statement associated with a refutable first-order sentence is a special consequence of the general ruleThe general rule applies to every first-order sentence, not just a chosen false or refutable one.

Every complement of a set is taken in the same universal set $\mathcal{U}$. Apply De Morgan's laws for propositional logic to the membership propositions. For every $x\in \mathcal{U}$,Equality of membership for every element proves the set identities

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 substitution maps each atomic propositional variable to a propositional formula and extends recursively through every logical connective. Replacements are simultaneous: a variable appearing inside a replacement formula is not replaced again in the same substitution step. Since propositional logic has no variable-binding quantifiers, this operation has no variable-capture problem.

An inference rule deriving a clause from two clauses containing complementary literals. In propositional logic the displayed rule eliminates the atom $A$. For first-order clauses, complementary atoms can first be made identical by a suitable first-order substitution.