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The succinct input is the players' variable-control sets, their goal propositional formulas, and a complete candidate Boolean valuation; an exponentially expanded payoff matrix is not part of the input. This representation is essential to candidate equilibrium checking in a Boolean game.

For membership in co-NP, consider the complement: the candidate is not a Nash equilibrium. A complexity certificate consists of a player $i$ and an alternative Boolean valuation of that player's controlled Boolean variables. Evaluate the current goal and the goal after this unilateral deviation, keeping every other player's variables fixed. A profitable unilateral deviation exists precisely when the current goal is false and the new goal is true. The certificate has at most the number of input variables plus the player index, and both propositional formula evaluations take polynomial time. Thus the complement belongs to NP, and the original decision problem belongs to co-NP.

For hardness, start with a propositional formula $F(X)$ and introduce a fresh Boolean variable $z$. Construct a one-player Boolean game in which that player controls $X\cup \{z\}$ and has goalChoose the candidate pure strategy assigning every variable false. Its payoff is zero because $z$ is false, regardless of $F$. If $F$ has a satisfying assignment, changing $X$ to that assignment and $z$ to true is a profitable unilateral deviation. If $F$ is an unsatisfiable Boolean formula, no unilateral deviation can satisfy the goal. ThereforeThe construction increases the formula length by only a constant and writes a candidate of linear length, so it is a polynomial-time many-one reduction. Unsatisfiability is co-NP-complete; combining hardness and membership proves that candidate Nash equilibrium checking for Boolean games is co-NP-complete. If a convention requires at least two players, add a player controlling a fresh unused variable and having the always-true goal; this changes neither direction of the proof.

For the intended finite representation, put $m=|S|<\infty$. For each $T\subseteq S$, form the mintermA Boolean valuation satisfies ${\chi }_T$ exactly when its selected subset is $T$. Therefore the weighted formula utility representationrepresents $u$ exactly: at the Boolean valuation for $T$, only that minterm contributes, with weight $u(T)$. This proves completeness and the bound of $2^{|S|}$ rules simultaneously. Zero-weight rules may be omitted. The bound counts rules; writing each minterm also requires $O(m)$ symbols in Big O notation.

For the computational question, assume rule weights are rational numbers encoded in binary. This makes exact equality of sums a finite computational task. With this conventional encoding, equivalence of weighted formula utility representations is co-NP-complete.

For membership, the complement has a short witness: a Boolean valuation on the variables occurring in either rule base. Evaluate each propositional formula, sum the weights of satisfied rules, and accept the witness if the sums differ. Propositional formula evaluation takes polynomial time. Exact sums of binary rational numbers also have polynomially bounded bit lengths: the product of their denominators has bit length at most the sum of the input denominator lengths. Thus non-equivalence lies in NP, and equivalence lies in co-NP.

For hardness, take a Boolean satisfiability problem instance $\theta$ and constructThe first utility function is one at satisfying Boolean valuations and zero elsewhere, while the second is always zero. ConsequentlyThe unsatisfiable Boolean formula problem is co-NP-complete, as the complement of the NP-complete Boolean satisfiability problem. The construction is a polynomial-time reduction, proving hardness even with the single weight $1$.

Two qualifications matter in the literal wording. First, unrestricted real weights have no specified finite encoding, so a standard bit-complexity classification needs the encoding assumption above. Second, the problem says only that $S$ is nonempty. If infinite $S$ is allowed while rule bases remain finite, its completeness claim is false. For $S=\mathbb{N}$, take $u(T)=1$ when $T$ is infinite and $u(T)=0$ when $T$ is finite. A finite rule base mentions only a finite subset $F$ of the variables. The Boolean valuations for $\varnothing$ and $\mathbb{N}\setminus F$ agree on all those variables but have different required values. No such rule base represents this $u$. Thus the construction answers the intended finite-set question and identifies the missing hypothesis in the unrestricted statement.

The definition in this part uses a single Boolean valuation, so the relevant Nash equilibria are pure strategy equilibria of a Boolean game. Use its standard convention that the players' control sets form a set partition of the variables.

Membership follows from an existential choice of a strategy profile and a universal check of all unilateral deviations. Assign payoff one to satisfying a player's goal and zero otherwise. A strategy profile $\sigma$ is a Nash equilibrium precisely when, for every player $i$ and every alternative Boolean valuation ${\tau }_i$ of that player's variables,is true: either the goal is already satisfied or no deviation can satisfy it. Thus the requested property isThe strings encoding $\sigma$, $i$ and ${\tau }_i$ have polynomial length, and evaluating the bracketed propositional formulas takes polynomial time. Invalid encodings of a player index can simply be ignored by the test. This is the existential second level of the polynomial hierarchy, ${\Sigma }_2^{\mathrm{P}}$.

For hardness, reduce the canonical complete problem of deciding a two-block quantified Boolean formulaMake a two-player Boolean game: player 1 controls $X$ and has goal ${\gamma }_1=\top$; player 2 controls $Y$ and has goal ${\gamma }_2=\neg \psi$. Set the query formula to $\varphi =\psi$. The construction has size linear in the input formula.

If an assignment $x$ satisfies $\forall Y\psi (x,Y)$, combine it with any assignment $y$. Player 1 already satisfies its goal, and player 2 cannot satisfy $\neg \psi$ by any unilateral deviation. Thus $(x,y)$ is a Nash equilibrium satisfying the query.

Conversely, let $(x,y)$ be a Nash equilibrium satisfying $\psi$. Player 2 currently has payoff zero. If some alternative $y^{\prime }$ made $\psi (x,y^{\prime })$ false, that deviation would satisfy player 2's goal and increase its payoff to one. The Nash equilibrium condition therefore implies $\forall Y\psi (x,Y)$. This proves the required equivalence and the polynomial-time reduction.

Together, membership and hardness prove ${\Sigma }_2^{\mathrm{P}}$-completeness of existential equilibrium checking in a Boolean game. The reduction works with only two players and one constant goal.

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 structural induction on the propositional formula. The induction claim is that its truth value under the all-one Boolean valuation is one. A variable has this truth value by definition. For the induction step, suppose the immediate subformulas $\alpha ,\beta$ each have truth value one. ThenThese are all permitted construction cases. Thus every such propositional formula evaluates to one. Equivalently, the three logical connectives define one-preserving Boolean functions, and composition preserves this property. The variable-only base grammar matters: admitting a constant false formula would invalidate the claim.

Let $F=(A\to C)\vee (B\to C)$ and $H=(A\vee B)\to C$, so the two displayed propositional formulas are $\phi =F\to H$ and $\psi =H\to F$. Their complete truth tables areTherefore $\psi$ is a tautology; $\phi$ is not. For example $(A,B,C)=(0,1,0)$ makes the antecedent $F$ true via $A\to C$, while $H$ is false because $B$ is true and $C$ is false. This explains why having just one of the two logical implications is insufficient for the forward direction.

Only $\psi$ requires a natural deduction proof. Assume $h:(A\vee B)\to C$. Within that assumption, temporarily assume $a:A$. By disjunction introduction, obtain $A\vee B$; applying $h$ by the implication elimination rule gives $C$. Discharge $a$ by the implication introduction rule to derive $A\to C$. Disjunction introduction now gives $(A\to C)\vee (B\to C)$. Finally discharge $h$ by the implication introduction rule:All temporary assumptions are discharged. No classical rule is needed; the same proof works in Intuitionistic propositional logic. The false propositional formula $\phi$ has no proof in this sound natural deduction system.

Reading the abstract syntax tree recursively gives the eight distinct subformulasA subformula includes the complete propositional formula itself as well as the operands of its connectives. There are nine occurrences in the abstract syntax tree, because the atomic formula $A$ occurs twice, but only eight different propositional formulas in the set.

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.

Apply structural induction to the formation of a propositional formula. Let $L(\theta )$ and $R(\theta )$ count its opening and closing parentheses, and use the usual grammar generated by propositional variables, any allowed truth constants, negation, and parenthesized binary logical connectives.

An atomic variable or truth constant contains neither kind of parenthesis, so $L=R=0$. If $L(\alpha )=R(\alpha )$, prefixing negation leaves both counts unchanged. If the notation writes a negated formula with an extra pair of parentheses, it adds one to each and still preserves equality.

For a binary construction $(\alpha \circ \beta )$, where $\circ$ is an allowed binary logical connective,The two induction hypotheses make these equal. Every grammar construction preserves the property, so every propositional formula contains equally many left and right parentheses. This is balanced parentheses in propositional syntax.

Interpret $x_1,\dots ,x_n$ as the bits of a binary number, with $x_1$ most significant. Construct a propositional formula saying that this number is strictly less than $k$. This gives a quadratic-size formula with prescribed model count.

For $0<k<2^n$, write $k=(b_1\cdots b_n{)}_2$. For each $i$ with $b_i=1$, define the conjunctionThe empty prefix for $i=1$ is omitted. Let ${\phi }_k={\bigvee }_{i:b_i=1}{T}_i$. A Boolean valuation satisfies $T_i$ exactly when its first difference from the binary representation of $k$ is at position $i$, changing a one of $k$ to a zero. All later bits are unrestricted. Hence these cases are disjoint and cover exactly the numbers $0,\dots ,k-1$.

Alternatively, counting the disjoint cases directly givesEach $T_i$ has length $O(i)$, including its logical connectives and parentheses; the total length is at most a constant multiple of ${\sum }_{i=1}^{n}i=O(n^2)$. The endpoints are ${\phi }_0=\perp$ and ${\phi }_{2^n}=\top$. If constants are unavailable and $n\ge 1$, use $x_1\wedge \neg x_1$ and $x_1\vee \neg x_1$ instead. Assignments are counted over all $n$ specified variables, including any absent from the formula. For $n=0$, the constants describe the two possible counts, zero and one. Thus the construction covers $0\le k\le 2^n$, including the endpoint zero despite the source's word “positive”.

For example, $n=4,k=11=(1011{)}_2$ gives the disjoint cases $0\ast \ast \ast$, $100\ast$ and $1010$, with $8+2+1=11$ satisfying assignments.

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 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.

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.

A propositional type is a collection of propositional formulas to be jointly realized by a Boolean valuation. A valuation realizes it when every member is true and omits it when some member is false. For propositional omission arguments, the collection need not itself be consistent.

The displayed identity follows by testing the two possible truth values of $p$. Existentially forgetting $p$ gives $A[p:=\top ]\vee A[p:=\perp ]$. Repeating this operation produces a propositional formula on a specified remaining vocabulary. Logical truth constants are needed if that vocabulary is empty.

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.

A subformula is a constituent propositional formula in a formula's recursive construction, including the whole formula itself. For an atomic variable the only subformula is itself; for $\neg A$, add $\neg A$ to the subformulas of $A$; for a binary connective, add the combined formula to the union of the two operands' subformulas. The set of distinct subformulas can have smaller cardinality than the number of occurrences in the abstract syntax tree when the same expression appears more than once.

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.