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For each $0\le k\le 2^n$, a propositional formula over $n$ specified variables can have exactly $k$ satisfying assignments and length $O(n^2)$. For $0<k<2^n$, interpret a Boolean valuation as the binary number $x_1\cdots x_n$ and test whether it is less than $k=b_1\cdots b_n$.

For every position $i$ with $b_i=1$, use a conjunction asserting that all earlier bits equal $b_j$ and $x_i=0$, leaving later bits unrestricted. Take the disjunction of these terms. Each satisfying assignment has a unique first differing position, so the terms are disjoint, and their counts sum to ${\sum }_{i:b_i=1}{2}^{n-i}=k$. Their combined length is $O(1+2+\cdots +n)=O(n^2)$. Use logical falsity and logical truth at the two endpoints. The count is over all $n$ variables, including those absent from an individual term or from the whole formula.

Replacing a designated subformula by a logically equivalent formula preserves logical equivalence of the surrounding propositional formulas. For any Boolean valuation the two expressions at the replaced node have the same truth value. Each logical connective preserves this agreement when its other inputs are unchanged. Induction from that node to the root proves agreement of the whole formulas under every valuation.

A satisfying assignment is a complete Boolean valuation $v$ for which a Boolean formula $F$ has truth value one, written $v\models F$. The notation describes a relation between the Boolean valuation and the Boolean formula. For the example in Boolean satisfiability problem, $(0,1,0)$ is a model of a Boolean formula $F$, whereas $(0,0,0)$ is not. SAT asks for existence of at least one satisfying assignment, rather than for every Boolean valuation to work; the latter condition makes $F$ a tautology.

For distinct propositional variables, eliminate each right-associated logical implication to obtain $\neg A_n\vee \cdots \vee \neg A_2\vee A_1$. Exactly one Boolean valuation falsifies all these disjuncts: $A_1=0$ and $A_2=\cdots =A_n=1$. Of the $2^n$ assignments, the other $2^n-1$ are satisfying assignments. Left-associated implication chains have a different meaning and need not have the same count.

A propositional formula $\phi$ semantically entails $\psi$ when every Boolean valuation satisfying $\phi$ satisfies $\psi$. Equivalently, $\phi \wedge \neg \psi$ is an unsatisfiable Boolean formula. Deciding semantic entailment is co-NP-complete: a counterexample Boolean valuation certifies failure in polynomial time, while $\theta \mapsto ⟨\theta ,p\wedge \neg p⟩$ reduces unsatisfiability to semantic entailment.

A tautology is a propositional formula true under every Boolean valuation in classical propositional logic, such as the law of excluded middle $p\vee \neg p$. A primitive proposition $p$ is not a tautology, even if it represents a true statement. See forall x: Calgary, section 12.1.

A truth table lists a propositional formula's value for every Boolean valuation of its distinct primitive propositions. With $k$ primitive propositions, there are $2^k$ rows: each independently receives T or F. Evaluate compound columns from their subexpressions using the classical rules for logical connectives. This is exhaustive checking, so a formula true on every row is valid in classical propositional logic.

For a computed example over the natural numbers, let $A(n)$ mean that $n$ is an even number and $B(n)$ mean $n>2$. The choices $n=4,2,3,1$ realize respectively $(A,B)=(\mathrm{T},\mathrm{T}),(\mathrm{T},\mathrm{F}),(\mathrm{F},\mathrm{T}),(\mathrm{F},\mathrm{F})$: all four possible inputs.

The logical implication $A\to B$ and its contrapositive agree on every row; the converse implication $B\to A$ differs on two rows. A truth table therefore checks equivalence or supplies a counterexample. Classical truth conditions are given in forall x: Calgary, section 31.2. These two-valued truth tables do not characterize validity in Intuitionistic propositional logic.

In classical propositional logic, a truth value is false or true, encoded by a bit as zero or one. A Boolean valuation assigns these values to primitive propositions and extends to compound propositional formulas through logical connectives.

If all but one literal of a Boolean clause are false under a partial Boolean valuation, the remaining literal must be true in any extension satisfying the clause. Repeatedly make such forced assignments, remove satisfied Boolean clauses and remove false literals from the others. An empty Boolean clause proves that the current partial Boolean valuation has no satisfying extension. This is sound because every step is forced by an original or already implied Boolean clause; it is not by itself a complete SAT decision procedure.

A logical formula or a collection of formulas is unsatisfiable when no permitted interpretation makes all of it true. The permitted interpretations come from the chosen semantics: Boolean valuations for propositional logic, or structures and variable assignments for first-order logic. Unsatisfiability is therefore a semantic property; a refutation is a syntactic proof of that property.

For example, $p\wedge \neg p$ is unsatisfiable because neither truth value of $p$ makes both parts true. The logical formula $p\vee q$ is satisfiable, although it is false in the valuation assigning false to both letters. Thus being false in one interpretation does not establish unsatisfiability. In classical propositional logic, $\phi$ is valid exactly when $\neg \phi$ is unsatisfiable. Likewise, $\Gamma \models \phi$ exactly when $\Gamma \cup \{\neg \phi \}$ is unsatisfiable. This turns entailment questions into searches for countermodels or contradiction certificates.

An unsatisfiable Boolean formula has no satisfying assignment. Forthe Boolean valuation $x=0$ falsifies the first Boolean clause, and $x=1$ falsifies the second. Both possible truth values fail, proving UNSAT. Each Boolean clause alone has a satisfying assignment, but their logical conjunction does not: compatibility is the central issue. By contrast, finding just one failed Boolean valuation for a general Boolean formula only rejects that candidate. Finding one successful Boolean valuation proves SAT; proving UNSAT requires excluding every candidate, possibly with a compact argument rather than a full truth table.

For a CNF input with $n$ Boolean variables, $m$ Boolean clauses, and $L$ Boolean literal occurrences, first check that the supplied Boolean valuation gives each input Boolean variable exactly one truth value. Evaluate each Boolean literal by lookup or logical negation, each Boolean clause by logical disjunction, and the Boolean formula by logical conjunction. Indexed Boolean variables and constant-cost lookups give an $O(n+m+L)$ operation count; with a finite binary string encoding the check still takes polynomial time. Thus a satisfying assignment is a certificate in computational complexity for SAT, proving that SAT belongs to NP. Finding it by exhaustive checking can require $2^n$ candidate Boolean valuations; this enumeration cost is not a lower bound on every possible SAT algorithm.^[ref]

A finite rule base pairs propositional formulas with weights and sums the weights of the formulas satisfied by a Boolean valuation. Over a finite set $S$, associate one Boolean variable to membership of each element. Any utility function $u:2^S\to \mathbb{R}$ is represented by giving each subset's minterm weight $u(T)$. Exactly one rule fires at each valuation, so at most $2^{|S|}$ rules suffice. This bound counts rules, not the symbols required to write all formulas.

For computational questions, the weight encoding matters. Binary rational numbers permit exact addition and comparison in polynomial time; unrestricted real constants do not specify an ordinary finite input encoding.