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One presentation of classical propositional logic uses the schemes $A\to (B\to A)$, $(A\to (B\to C))\to ((A\to (\neg B\to C))\to (A\to C))$, and $(A\to B)\to (A\to (\neg B\to C))$, together with modus ponens rule and the case rule from $A\to B$ and $\neg A\to B$ to $B$. Its deduction theorem for propositional logic follows by induction on logical derivations: conditional modus ponens handles a modus ponens rule line, and the second scheme handles a case-rule line.

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.

The contrapositive of a logical implication $A\to B$ reverses the direction and applies logical negation to both sides: $\neg B\to \neg A$. In classical propositional logic they are equivalent, so either can be proved in place of the other. The contrapositive of “divisibility by four implies being an even number” says that a number that is not an even number cannot be divisible by four.

In Intuitionistic propositional logic, $A\to B$ still implies its contrapositive: assumptions $A$ and $\neg B$ would yield $B$ and hence a contradiction. The reverse direction is not generally valid. From $\neg B\to \neg A$ and $A$, one obtains only $\neg \neg B$; obtaining $B$ additionally requires double-negation elimination.

A logical connective forms a compound proposition from component propositions; logical truth and logical falsity are the zero-input cases. Logical negation has one input, while logical conjunction, logical disjunction, and logical implication each have two. In classical propositional logic, these logical connectives are truth-functional: their output value is determined by their input values, as listed in a truth table. This does not characterize every kind of logical connective in every logic. See forall x: Calgary, section 10.1.

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.

Logical falsity, written $\perp$, is a zero-input logical connective interpreted as false under every Boolean valuation. In standard Intuitionistic propositional logic and classical propositional logic, it has no introduction rule that produces it without premises. The logical falsity elimination rule permits deriving any logical formula from it.

A formal proof can derive $\perp$ from contradictory assumptions: $P$ together with $\neg P$ gives it by the negation elimination rule. A first-order theory is first-order consistent when it has no assumption-free proof of $\perp$ from its own axioms. Thus having no proof is a consistency property of a specified formal proof system, rather than an unconditional definition of logical falsity.

Logical implication forms the proposition $A\to B$, read “if $A$, then $B$”. In classical propositional logic, it is false exactly when $A$ is true and $B$ is false. Equivalently, it has the same truth table as $\neg A\vee B$, using logical negation and logical disjunction. A false hypothesis makes the logical implication true regardless of its conclusion; thus it asserts neither the hypothesis nor a causal connection.

For example, for every natural number $n$, divisibility by four implies that $n$ is an even number: $n=4k=2(2k)$. At $n=8$, both statements are true. At $n=6$, the hypothesis is false and the conclusion true; at $n=3$, both are false. The only forbidden case is a multiple of four that is not an even number.

In Intuitionistic propositional logic, a proof of $A\to B$ consists of a construction transforming any proof of $A$ into a proof of $B$. The implication introduction rule proves $B$ under a temporary assumption $A$; the modus ponens rule applies the result when $A$ is available. The classical identification with $\neg A\vee B$ is not generally intuitionistically valid.

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.

Mathematical logic studies formal languages, formal proofs, model theory, computability theory, and the foundations of mathematics. It makes the rules of reasoning explicit enough that reasoning itself becomes an object of mathematics. Proof theory studies deductions; model theory studies interpretations in which statements hold.

Three questions organize the subject: is an expression a well-formed first-order formula, is it true in a specified first-order structure, and does it have a formal proof from specified axioms? These questions are distinct. For example, $\exists n(n+n=1)$ is a first-order sentence in an appropriate first-order language; it is false in the natural numbers but true in the rational numbers, with $n=1/2$.

The beauty of mathematical logic is that the same logical connectives and rules of universal quantification and existential quantification organize arguments across mathematics. Classical propositional logic uses two truth values; Intuitionistic propositional logic interprets logical implication through constructions on proofs, without unrestricted law of excluded middle. This proof-based framework is developed in Frank Pfenning, Natural Deduction, sections 1–3.

The Gödel second incompleteness theorem says that a consistent, effectively axiomatized first-order theory with enough arithmetic and its standard derivability and diagonalization properties cannot prove its formal consistency statement ${\mathrm{Con}}_S=\neg B\perp$.

If a consistent $S$ proved ${\mathrm{Con}}_S$, it would prove $B\perp \to \perp$ by classical propositional logic. Apply Löb's theorem to $A=\perp$. This would give $S⊢\perp$, contradicting consistency. ThereforeFor an inconsistent first-order theory the conclusion need not hold, since it proves every first-order sentence.

Use just the assumed local reflection instance with $Y=\neg X$:By the deduction theorem for propositional logic, $S⊢X\to (B_S(\neg X)\to \neg X)$. Classical propositional logic makes this imply the unconditional reflection statement $S⊢B_S(\neg X)\to \neg X$: if $X$ is false the consequent is already true, and if $X$ is true the displayed implication supplies it. Löb's theorem applied to $\neg X$ now givesThus full local reflection in a one-sentence extension is inconsistent; in this argument even the single instance concerning $\neg X$ suffices.

One Hilbert axiomatization of classical propositional logic uses the schemas $A\to (B\to A)$, $[A\to (B\to C)]\to [(A\to B)\to (A\to C)]$ and $(\neg B\to \neg A)\to (A\to B)$, with modus ponens as its inference rule. Other connectives can be defined from implication and negation. Syntactic entailment $\Gamma ⊢A$ means that a finite formal proof derives $A$ from the schemas and members of $\Gamma$. Consistency means that no formula and its negation are both derivable, equivalently no contradiction is derivable.

Use the standard deduction theorem, closure of derivability under modus ponens, and the maximal-consistency properties: a maximal consistent set contains exactly one of $A,\neg A$, is closed under derivability, and satisfies $A\to B\in \Phi$ exactly when $A\notin \Phi$ or $B\in \Phi$. These follow from the deduction theorem and the classical schemas; in particular adjoining a missing formula to a maximal set must yield a contradiction. Define $v(p)=T$ iff $p\in \Phi$. Induction on formula construction now provesFor negation use the exactly-one property, and for implication use its displayed membership rule, which is precisely the Boolean truth table. Thus every member of $\Phi$ is true under $v$, providing the required valuation.

If $S⊢G$, theoremhood gives $S⊢\square G$, while the fixed-point equivalence also gives $S⊢\neg \square G$. Therefore first-order consistency implies $S⊬G$.

If $S\cup \{\neg G\}$ were inconsistent, the deduction theorem and classical propositional logic would give $S⊢\neg \neg G$, hence $S⊢G$. This is impossible by the preceding argument. ThereforeThis is a syntactic first-order consistency result. It does not claim that the extra axiom is true in the standard natural numbers or that the extended first-order theory is sound.

The Hilbert-Bernays-Löb derivability conditions are: if $S⊢A$ then $S⊢\square A$; $S⊢\square (A\to B)\to (\square A\to \square B)$; and $S⊢\square A\to \square \square A$. The first is an external rule about provability; the latter two are internal theorems. Standard numerical correctness and the stipulated ${\Sigma }_1$ form accompany the predicate in this question. These conditions do not include reflection $\square A\to A$.

Choose a fixed refutable sentence $\perp$ and identify ${\mathrm{Con}}_S$ with $\neg \square \perp$. Different fixed refutable choices have provably equivalent consistency sentences, because $S$ proves their equivalence to a contradiction. From $\perp \to G$, necessitation and distribution give $\square \perp \to \square G$. Since $G\to \neg \square G$, this yields $G\to {\mathrm{Con}}_S$.

For the converse, necessitate $G\to \neg \square G$ and distribute to get $\square G\to \square \neg \square G$. Positive introspection also gives $\square G\to \square \square G$. The two boxed contradictories imply $\square \perp$ by formalized propositional reasoning. Thus $\square G\to \square \perp$, and contraposition gives ${\mathrm{Con}}_S\to \neg \square G\to G$. ConsequentlyFinally the deduction theorem says that $S+G$ is inconsistent exactly when $S⊢\neg G$. The internal equivalence identifies $S+G$ and $S+{\mathrm{Con}}_S$ in their deductive consequences. Hence $S⊬\neg G$ exactly when $S+{\mathrm{Con}}_S$ is consistent. Consistency of $S$ alone need not ensure this stronger consistency condition.

Suppose $S+A$ proves every specified reflection instance. Use only the instance with $X=\neg A$. The deduction theorem givesClassical propositional reasoning turns this into $S⊢\square \neg A\to \neg A$: if $\neg A$ already holds the implication is true, while otherwise $A$ holds and the displayed theorem supplies it. Löb's theorem now yields $S⊢\neg A$, making $S+A$ inconsistent. Therefore no sentence consistent with $S$ can supply all the local reflection instances in a one-sentence extension. The proof uses just one well-chosen instance, and never applies necessitation to a derivation with the extra assumption still open.

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.

In proof by contradiction, assume the negation of the desired conclusion and derive logical falsity; discharge that assumption and infer the desired conclusion. Its sequent rule isThis is a rule of classical propositional logic. It follows from the paper's negation introduction rule, which first yields $\neg \neg \varphi$, and double-negation elimination, which yields $\varphi$. This is stronger than merely deriving $\neg \varphi$ by showing that $\varphi$ leads to logical falsity.

Write $D(\varphi ,\psi )=\neg (\neg \varphi \wedge \neg \psi )$. We derive the two disjunction introduction rules and the disjunction elimination rule without using any disjunction rule inside the derivations.

For left introduction, start with $\varphi$ and temporarily assume $\neg \varphi \wedge \neg \psi$. The conjunction elimination rule gives $\neg \varphi$, and the negation elimination rule gives $\perp$. Discharge the temporary conjunction by the negation introduction rule, concluding $D(\varphi ,\psi )$. For right introduction, start with $\psi$ and use the right conjunction elimination rule to obtain $\neg \psi$ instead. Thus either input derives the required encoded disjunction.

For elimination, suppose $D(\varphi ,\psi )$ is available, together with a derivation of $\kappa$ from an additional assumption $\varphi$ and a derivation of $\kappa$ from an additional assumption $\psi$. Keep their common external context fixed. Temporarily assume $\neg \kappa$ under label $a$. Temporarily assume $\varphi$ under label $b$ and insert the supplied first branch derivation to obtain $\kappa$. Negation elimination rule gives $\perp$, so negation introduction rule discharges $b$ and gives $\neg \varphi$. Repeat with a temporary assumption $\psi$ under label $c$ and the supplied second branch derivation, discharging $c$ to obtain $\neg \psi$.

Conjunction introduction rule now gives $\neg \varphi \wedge \neg \psi$. Its negation $D(\varphi ,\psi )$ gives $\perp$ by negation elimination rule. The supplied reductio ad absurdum rule discharges $a$ and yields $\kappa$. This discharges the two branch assumptions as required and proves the encoded elimination rule. Its last step uses classical propositional logic; the same encoded elimination principle is not generally available in Intuitionistic propositional logic.

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 refutation of a collection of logical assumptions is a derivation of a contradiction from them. To prove a logical formula $\phi$ by refutation in classical propositional logic, add its negation $\neg \phi$ to the assumptions and derive falsity. The correctness of this inference follows from the rules of the chosen proof system.

For example, the clauses $p\vee q$, $\neg p$ and $\neg q$ have a resolution refutation: the first two resolve to $q$, and $q$ with $\neg q$ yields the empty clause. No truth assignment satisfies the three original clauses. This distinguishes the semantic claim of unsatisfiability from a syntactic certificate establishing it. In automated proving, producing a refutation is useful because another program can check the individual inference steps without repeating the search.