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Logic: propositional and predicate

Philosophy (BA) · Logic · free study notes for exams, placements and interviews

In short

Propositional (sentential) logic treats whole statements as atoms joined by connectives — not, and, or, if-then, if-and-only-if — and asks whether a form is a tautology, a contradiction or contingent, and whether an argument is valid. Predicate (first-order, quantificational) logic looks inside the statement at individuals, predicates, relations and the quantifiers "all" and "some", so it can validate arguments propositional logic cannot. It is core to BA Philosophy logic papers under UGC CBCS, to UGC-NET Paper 1 reasoning and Philosophy Paper 2, and to CUET-PG; the same material appears as Discrete Mathematics ("propositional and first-order logic") in GATE CS, and in CLAT/judiciary legal-reasoning sections. Employers use it in software testing and SQL conditions, in rule engines at Indian banks and insurers, in SAT/model-checking tools at chip-design centres (Intel, NVIDIA, Synopsys, Cadence in Bengaluru, Noida, Pune), and in any interview question about boolean conditions.

Explained

1. Propositions, connectives and well-formed formulas

2. Truth tables, status of formulas and validity

3. Equivalences, implication forms and valid argument forms

4. Normal forms, resolution and adequate sets

5. Predicate logic: symbols, quantifiers and translation

6. Predicate rules, quantifier equivalences and the big results

Key points

Formulas and examples

Rows in a truth table = 2ⁿ, where n = number of distinct propositional variables
Distinct truth functions of n variables = 22ⁿ (n = 2 → 16; n = 3 → 256)
p → q ≡ ¬p ∨ q ; ¬(p → q) ≡ p ∧ ¬q ; p → q ≡ ¬q → ¬p (transposition)
p ↔ q ≡ (p → q) ∧ (q → p) ≡ (p ∧ q) ∨ (¬p ∧ ¬q)
¬(p ∧ q) ≡ ¬p ∨ ¬q ; ¬(p ∨ q) ≡ ¬p ∧ ¬q (De Morgan)
p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) ; p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) (distribution)
Argument P₁…Pₙ ⊢ C is valid ⟺ (P₁ ∧ … ∧ Pₙ) → C is a tautology ⟺ P₁ ∧ … ∧ Pₙ ∧ ¬C is unsatisfiable
¬∀x Fx ≡ ∃x ¬Fx ; ¬∃x Fx ≡ ∀x ¬Fx (quantifier negation)
∀x(Fx ∧ Gx) ≡ ∀xFx ∧ ∀xGx ; ∃x(Fx ∨ Gx) ≡ ∃xFx ∨ ∃xGx (the ∧/∃ and ∨/∀ versions are one-way only)
Exactly one F: ∃x(Fx ∧ ∀y(Fy → y = x))

Example 1 — truth table and models of (p ∨ q) ∧ (¬q ∨ r). n = 3, so 8 rows.

pqrp∨q¬q∨rformula
TTTTTT
TTFTFF
TFTTTT
TFFTTT
FTTTTT
FTFTFF
FFTFTF
FFFFTF

Four models out of eight, so the formula is contingent and satisfiable. Full DNF = (p∧q∧r) ∨ (p∧¬q∧r) ∨ (p∧¬q∧¬r) ∨ (¬p∧q∧r) — four minterms, matching the four T rows.

Example 2 — valid versus fallacious conditional reasoning. "If the GST return is filed late, a late fee applies. A late fee applies. So the return was filed late." Let p = filed late, q = late fee. Form: p→q, q ⊢ p — affirming the consequent. Counterexample row: p = F, q = T → premises p→q = T and q = T, conclusion p = F. Premises true, conclusion false, so invalid. Now modus tollens: p→q, ¬q ⊢ ¬p. Test (p→q) ∧ ¬q → ¬p by the short method: assume conclusion ¬p false, so p = T; premise ¬q true, so q = F; then p→q = T→F = F, so a premise fails. No row makes premises true with the conclusion false — valid.

Example 3 — CNF conversion and a resolution refutation. Prove p→q, q→r, ¬r ⊢ ¬p. Clauses: (1) ¬p ∨ q, (2) ¬q ∨ r, (3) ¬r, and the negated conclusion (4) p.

The negation of the conclusion is inconsistent with the premises, so the argument is valid. The same result by natural deduction: 1, 2 give p→r by hypothetical syllogism; with ¬r, modus tollens gives ¬p.

Example 4 — symbolisation and quantifier negation. Domain: people. Sx = x is a student, Lx = x is a lecture, A(x,y) = x attends y, E(z) = z is an exam, P(x,z) = x passes z. "Every student who attends all lectures passes some exam": ∀x[(Sx ∧ ∀y(Ly → A(x,y))) → ∃z(Ez ∧ P(x,z))]. Its negation, driven inwards by QN and ¬(A→B) ≡ A∧¬B: ∃x[Sx ∧ ∀y(Ly → A(x,y)) ∧ ∀z(Ez → ¬P(x,z))] — "some student attends every lecture and passes no exam." Also note "No student passed" = ¬∃x(Sx ∧ Qx) ≡ ∀x(Sx → ¬Qx) — the two standard forms of the E proposition.

Example 5 — proving a predicate argument invalid by counter-model. Claim: ∀x(Fx → Gx), ∃x Gx ⊢ ∃x Fx. Take domain D = {1, 2}, extension of F = { } (empty), extension of G = {1}.

Premises true, conclusion false in this interpretation, so the argument is invalid. (Swapping the second premise with the conclusion — ∀x(Fx → Gx), ∃x Fx ⊢ ∃x Gx — gives a valid argument by EI, UI, modus ponens and EG.)

Propositional logicPredicate (first-order) logic
Unit of analysisWhole statements as atomsIndividuals, predicates, relations
Symbolsp, q, r, connectivesConstants, variables, predicates, quantifiers, =
SemanticsTruth assignment (2ⁿ rows)Interpretation: domain + extensions
Decision procedureTruth table, trees, resolution — always terminatesTrees/resolution; may not terminate
Decidable?Yes (but NP-complete for SAT)No (semi-decidable); monadic fragment is decidable
Proving invalidityOne row of the truth tableA counter-model with a small domain
Expressive limitCannot handle "all"/"some" inferenceCannot quantify over predicates or express "most"

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Related topics

Syllogisms · Critical reasoning: arguments and inferences · Set theory and Boolean algebra · Informal fallacies · Modal and many-valued logics · Philosophy of language and meaning.

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