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Take precisely the propositional tautologies as axioms and supply no inference rules. A one-line proof then establishes exactly the valid propositional formulas, so this calculus is sound and complete for theoremhood. Its axiom set is decidable by finite truth tables. This construction does not by itself give strong completeness for consequence from arbitrary assumptions.

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.

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.

Put $T=S+\{X\}$ and choose its Gödel sentence $Z$, so $T⊢Z\leftrightarrow \neg B_{T}Z$. The assumed reflection scheme gives the particular theoremThe given extension-provability implication is $S⊢B_{T}Z\to B_S(X\to Z)$. The fixed-point equivalence gives $T⊢\neg Z\to B_{T}Z$, so these implications compose toThe supplied propositional tautology $((\neg Z\to (X\to Z))\to (X\to Z))$ gives $T⊢X\to Z$, and $T⊢X$ then yields $T⊢Z$. The first of the derivability conditions for $T$ gives $T⊢B_{T}Z$, while its Gödel equivalence gives $T⊢\neg B_{T}Z$. Hence $T$ is inconsistent. By the deduction theorem for propositional logic,This proof uses diagonalization and proof transfer for the extension, and invokes neither Löb's theorem nor the second incompleteness theorem.

Take $Y(p)=p\to q$ and put $\top =(\perp \to \perp )$, $H=Y(\top )$ and $F_0=Y(\square H)=\square H\to q$. The first supplied substitution identity readsUse the supplied replacement of equivalents in the context $D(p)=p\to q$. It givesBut $H=(\top \to q)$ is equivalent to $q$. Replacement under a box gives $\square H\leftrightarrow \square q$, so $F_0$ is equivalent toReplacing equivalents once more transfers the fixed-point equation to $F$, proving $\mathrm{GL}⊢F\leftrightarrow (\square F\to q)$.

A direct check of this GL fixed point for a provability implication is useful: the Löb axiom gives $\square F\to \square q$, and necessitating the tautology $q\to F$ gives $\square q\to \square F$ by $\mathrm{K}$. Thus $\square F\leftrightarrow \square q$, and replacing the antecedent in $\square F\to q$ recovers $F$.

Yes, for soundness and completeness of theoremhood. An axiom-only complete propositional calculus is obtained as follows. Declare every propositional tautology, and only such a tautology, to be an axiom; supply no inference rules. A formal proof then consists of a single axiom. Every proved logical formula is valid, giving soundness, and every valid logical formula is already an axiom, giving completeness.

This is not merely an unspecified collection: whether a finite propositional formula is a tautology is decidable by its finite truth table, so the axiom set can be recognized effectively. It is a deliberately uninformative Hilbert proof system, since all the work has been put into axiom recognition. If one instead asked for strong completeness of derivations from arbitrary additional assumptions, a system whose only formal proof lines are axioms or assumptions would not suffice; the affirmative construction here concerns $⊢\varphi$ versus $\models \varphi$.

Capture-safe equality replacement gives the corrected scheme:where ${\varphi }^{\prime }$ replaces selected free occurrences of $x_i$ by $x_j$, without capture. One may equivalently use full capture-avoiding substitution, after renaming conflicting binders. Replacing bound occurrences as though they were free is not permitted.

Here is a syntactic proof. Write $e$ for $x_i=x_j$. First derive symmetry under $e$: A6 and A4 give $x_i=x_i$, and A7 applied to the first position in that atomic equality gives $e\to (x_i=x_i\to x_j=x_i)$. Modus ponens and the deduction theorem give $⊢e\to x_j=x_i$. Atomic replacement in the forward direction is A7 itself; in the reverse direction apply A7 with $i,j$ exchanged and use this symmetry. Thus under $e$, an atomic formula and its designated replacement imply each other.

Induct on the formation of $\varphi$, proving both directions simultaneously. For negation, the reverse implication for the immediate subformula gives the forward implication of its negations by contraposition, and vice versa. For $\theta \to \eta$, the two induction equivalences combine propositionally to $(\theta \to \eta )\leftrightarrow ({\theta }^{\prime }\to {\eta }^{\prime })$. This uses only the propositional tautologies allowed in this part.

For $\varphi =\forall z\theta$, first take $z$ distinct from $x_i,x_j$. From assumptions $e$ and $\forall z\theta$, A4 gives $\theta$, the induction hypothesis gives ${\theta }^{\prime }$, and generalization gives $\forall z{\theta }^{\prime }$: $z$ is free in neither open assumption. Discharge the two assumptions to obtain $e\to (\forall z\theta \to \forall z{\theta }^{\prime })$. The reverse direction is identical. If the binder is $x_i$, no free occurrence of that variable lies beneath it and the replacement there is empty. If it is $x_j$, capture avoidance likewise forbids a changed free $x_i$ below that binder unless the binder is first renamed to a fresh variable. Bound-variable renaming is derived from A4 followed by admissible generalization and A5, with the same freshness restriction. This completes the structural induction and proves the scheme for arbitrary formulas.

Choose a fresh variable $x_3$. In A7 take the equality antecedent $x_3=x_1$, the atomic formula $\varphi =(x_3=x_2)$, and replace its occurrence of $x_3$ by $x_1$. This gives directly the theoremGeneralize on $x_3$, with no open assumptions. A4 then instantiates that quantifier with the term $f(x_0)$. The substitution is free for $x_3$, since the matrix contains no quantifiers. One modus-ponens step yieldsFinally generalize, in order, on $x_2$, $x_1$, and $x_0$. The result isEvery step is A7, generalization, A4, or modus ponens. Neither completeness, adequacy, nor an unproved appeal to all propositional tautologies is needed. In particular the proof does not silently upgrade the variable-only equality axiom to an arbitrary-term axiom: quantification and instantiation perform that upgrade explicitly.

Assign to each propositional formula its finite symbol length. The set of lengths of formulas logically equivalent to $\alpha$ is a nonempty subset of the natural numbers, because it contains the length of $\alpha$. The well-ordering principle for the natural numbers gives a least such length, and hence a shortest equivalent formula.

A shortest equivalent formula need not be unique. For example $(p_0\to p_0)$ and $(p_0\leftrightarrow p_0)$ are distinct tautologies of the same minimal length. A shorter formula would contain no binary connective and hence be a variable with some number of negations, which is not a tautology. This counts each propositional variable as one symbol, as in the stated language, and proves that a shortest equivalent formula need not be unique.

Logical equivalence means that the two propositional formulas receive the same truth value under every Boolean valuation of the propositional variables. Equivalently, $\alpha \leftrightarrow \beta$ is a tautology:

GL has all classical propositional tautologies, the normal distribution axiom $\square (A\to B)\to (\square A\to \square B)$ and the Löb axiom $\square (\square A\to A)\to \square A$, closed under uniform substitution, modus ponens and necessitation of theorems. We use its derived positive-introspection theorem $\square A\to \square \square A$, normal boxed reasoning, and the boxed replacement principle supplied in the question.

For completeness, positive introspection follows from these axioms. Put $C=A\wedge \square A$. Distribution gives $\square C\to \square A$, so propositional logic proves $A\to (\square C\to C)$. Necessitate and distribute to get $\square A\to \square (\square C\to C)$; Löb's axiom gives $\square A\to \square C$, and boxing $C\to \square A$ finishes $\square A\to \square \square A$.

For a GL fixed point by truth substitution, set $\top =(\perp \to \perp )$, $U=\square Y(\top )$ and $V=\square Y(U)$. From the tautology $U\to (\top \leftrightarrow U)$, boxed reasoning gives $\square U\to \square (\top \leftrightarrow U)$. The supplied replacement principle and distribution then giveSince $U$ is itself boxed, positive introspection gives $U\to \square U$, and hence $U\to V$.

For the reverse direction, the displayed implication entails the theorem $V\to (\square U\to U)$. Necessitate it and distribute to obtain $\square V\to \square (\square U\to U)$. Since $V$ is boxed, $V\to \square V$ holds. The Löb axiom therefore yields $V\to \square U$, and the same displayed implication then yields $V\to U$. ConsequentlyAll occurrences of the unknown letter have been removed from $U$ by substitution of truth; any other parameter letters remain available. The proof does not assume the invalid reflection principle $\square U\to U$ as an unconditional axiom.

The language of Gödel-Löb logic has propositional variables, $\perp$, implication, and the unary operator $\square$; the other logical connectives are abbreviations. Its axioms are all classical tautologies, the distribution scheme $\square (A\to B)\to (\square A\to \square B)$, and the Löb scheme $\square (\square A\to A)\to \square A$. Its rules are modus ponens rule, necessitation rule of theorems, and uniform substitution.

For a single boxed equation $p\leftrightarrow \square C(p)$, set $F=\square C(\top )$. The identity supplied in the source gives directly $GL⊢F\leftrightarrow \square C(F)$. If $p$ does not occur, take $F=\square C$ instead.

Proceed by induction for a system of $m$ equations, allowing other propositional variables as parameters. Treat $p_1,\dots ,p_{m-1}$ as parameters in the last equation and putThe one-variable result proves $D\leftrightarrow \square C_m(p_1,\dots ,p_{m-1},D)$. Substitute this $D$ for $p_m$ in the first $m-1$ equations. They are still boxed equations, so induction supplies solutions $F_1,\dots ,F_{m-1}$. Define $F_m=D(F_1,\dots ,F_{m-1})$.

By construction, the first $m-1$ equations hold after this substitution; substituting into the theorem for $D$ proves the last. HenceThe elimination introduces no new parameter letters and removes every unknown letter. This gives a constructive simultaneous boxed fixed-point elimination in GL.

One can takeFor the first formula, the Löb axiom with conclusion $\perp$ gives $\square (\square \perp \to \perp )\to \square \perp$, that is, $\square F_1\to \square \perp$. Conversely, the tautology $\perp \to F_1$, necessitation rule, and distribution give $\square \perp \to \square F_1$. Thus $GL⊢\square F_1\leftrightarrow \square \perp$. Negating both sides gives $GL⊢\neg \square F_1\leftrightarrow F_1$.

For the second formula, $\neg F_2=\top$ is a theorem, so $GL⊢\square \neg F_2$. Hence $\neg \square F_2\wedge \neg \square \neg F_2$ is false in the logic and is equivalent to $\perp =F_2$. A fixed point is a formula satisfying the equivalence; it need not itself express a nontrivial undecidable assertion.

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.

Apply the propositional substitution simultaneously to the atomic variables:By distributivity and the complementary literals $Y,\neg Y$,The other disjunct simplifies using logical implication and absorption:Thus $\phi S\equiv X\vee \neg X\equiv \top$, a tautology.

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 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.

Lengths of formulas logically equivalent to a fixed propositional formula form a nonempty subset of the natural numbers, so a minimum exists. With variables counted as atomic symbols and no truth constants, $(p\to p)$ and $(p\leftrightarrow p)$ are distinct shortest tautologies. A formula with no binary connective is a variable preceded by negations and cannot be a tautology. Thus minimization need not give a canonical representative.

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.