Logical Cases MCQs: Tautology, Contradiction, Contingency, Satisfiability and Validity

Students often know the definitions of tautology and contradiction but still mix up valid, satisfiable and contingent when a formula appears in an MCQ. The cure is not another list of definitions, but one repeatable test: enumerate every valuation, inspect the final column, and classify the formula from that column.

The same final-column test also separates validity from satisfiability. The Types of Logical Cases practice questions page keeps the topic's exercises together, including items without standalone lesson pages.

Related reading: Propositions and laws and Propositional logic MCQs.

The three-row classification test

The number of rows doubles with each variable. One variable gives 2^1 = 2 rows, two variables give 2^2 = 4 rows, and three variables give 2^3 = 8 rows. Do not classify from one convenient row. Complete the final column first.

Questions 1 to 3: identify the definition before calculating

Question 1

A proposition that is always true is called:

• A. Tautology

• B. Contradiction

• C. Contingency

• D. None of the above

Answer: A, Tautology. "Always true" means that every row in the final column is T. For p ∨ ¬p, the two valuations of p produce T ∨ F = T and F ∨ T = T, so the output sequence is T, T. A contradiction is always false, while a contingency is true on some rows and false on others.

Question 2

A proposition that is always false is called:

• A. Tautology

• B. Contradiction

• C. Contingency

• D. None of the above

Answer: B, Contradiction. Take p ∧ ¬p. When p = T, its negation is F, giving T ∧ F = F. When p = F, its negation is T, giving F ∧ T = F. The output sequence is F, F, with zero satisfying valuations.

Question 3

A contingent statement is:

• A. Always true

• B. Always false

• C. Sometimes true and sometimes false

• D. None of the above

Answer: C, Sometimes true and sometimes false. For p ↔ q, use the valuations (T,T), (T,F), (F,T), (F,F). A biconditional is true when both sides match, so its final column is T, F, F, T. It is true in 2 of 4 valuations. Therefore it is satisfiable but not valid, which makes it contingent.

The key is to classify the complete final column, not any single row that happens to be true or false.

Questions 4 to 6: classify formulas with a two-row truth table

Question 4

If a statement is a contradiction, then its truth value is:

• A. Always true

• B. Always false

• C. Sometimes true

• D. Sometimes false

Answer: B, Always false. "Sometimes false" is too weak because a contingency is also false sometimes. A contradiction must be false under every valuation, so its final column contains only F.

Question 5

What is the truth value of $p \land \neg p$?

• A. Tautology

• B. Contradiction

• C. Contingency

• D. None of the above

Answer: B, Contradiction. Work both valuations instead of relying only on the law name. If p = T, then ¬p = F, so T ∧ F = F. If p = F, then ¬p = T, so F ∧ T = F. Since the result is false in both rows, p ∧ ¬p is a contradiction.

Question 6

Which of the following represents a tautology?

• A. \( p \lor \neg p \)

• B. \( p \land \neg p \)

• C. \( p \rightarrow \neg p \)

• D. \( \neg p \land q \)

Answer: A, p ∨ ¬p. For p = T, it becomes T ∨ F = T. For p = F, it becomes F ∨ T = T. Option B is a contradiction. Option C is false at p = T, because T → F = F. Option D is false at p = T, q = T, because F ∧ T = F. A single false valuation is enough to reject a tautology candidate.

The excluded-middle check in Question 6 is the two-row baseline. For broader argument translation, inference and counterexample drills that also use excluded middle, continue with Questions and Practice Problems on Propositions MCQs. The remaining questions here stay focused on formula classification, satisfiability and semantic entailment.

Questions 7 and 8: separate validity from semantic entailment

Question 7

A valid sentence or tautology is a sentence that is:

• A. True under all interpretations, no matter what the world is actually like or what the semantics is

• B. False under all interpretations

• C. Both A and B

• D. None of the above

Answer: A. In propositional logic, the operational test is precise: a valid formula is true under every valuation of its variables. If a formula is true under only one valuation, or even under several but not all valuations, it is satisfiable but not valid.

Question 8

The mathematical notation to describe logical entailment of a sentence “α entails another sentence β” is : -

• A. α ⊨ β

• B. α ⊆ β

• C. β ⊨ α

• D. β ⊆ α

Answer: A, α ⊨ β. Entailment means that every valuation satisfying α also satisfies β. Let α = p ∧ q and β = p. For (p,q) = (T,T), (α,β) = (T,T). For (T,F), it is (F,T). For (F,T) and (F,F), it is (F,F). No row has α = T and β = F, so α ⊨ β. The symbol ⊨ expresses semantic entailment; ⊆ is the subset symbol.

Validity classifies one formula. Entailment compares a condition with a conclusion. Rows where α is false do not violate α ⊨ β; only a row with true α and false β would do that.

Questions 9 and 10: biconditional contingency and a valid argument form

Question 9

The proposition (P⇒Q)⋀(Q⇒P) is a

• A. tautology

• B. contradiction

• C. contingency

• D. absurdity

Answer: C, contingency. The expression (P⇒Q)⋀(Q⇒P) is equivalent to P ↔ Q. At (T,T), both implications are true. At (T,F), P⇒Q is false, so the conjunction is false. At (F,T), Q⇒P is false, so the conjunction is again false. At (F,F), both implications are true because each has a false antecedent. The final column is therefore T, F, F, T. Both truth values occur, so the formula is contingent. The step-by-step construction is covered in Implication and Biconditional Operators in Logic.

Question 10

Choose the correct statement about “ P ∧ (P-> Q) ->Q “

• A. Not valid

• B. Tautology

• C. Not satisfiable

• D. Contradiction

Answer: B, Tautology. The intended grouping is (P ∧ (P→Q))→Q, the propositional form of modus ponens. For (P,Q) = (T,T), the antecedent is T, so the result is T. For (T,F), P→Q = F, making the antecedent F and the whole implication T. For (F,T) and (F,F), the antecedent is F, so each implication is T. The final column is T, T, T, T.

The only row with a true antecedent also has a true conclusion. Every other row makes the outer implication true through a false antecedent, so no counterexample exists.

Questions 11 and 12: simplify a tautology and map the boundary cases

Question 11

The proposition [(p Λ q) -> (p v q) v ∼p v q]

• A. tautology

• B. satisfiable but not tautology

• C. a contradiction

• D. none of the above

Answer: A, tautology. Normalise the source symbols as [(p ∧ q) → (p ∨ q)] ∨ ¬p ∨ q, then simplify:

[(p ∧ q) → (p ∨ q)] ∨ ¬p ∨ q
≡ [¬(p ∧ q) ∨ (p ∨ q)] ∨ ¬p ∨ q
≡ [(¬p ∨ ¬q) ∨ (p ∨ q)] ∨ ¬p ∨ q
≡ (¬p ∨ p) ∨ (¬q ∨ q) ∨ ¬p ∨ q
≡ T ∨ T ∨ ¬p ∨ q
≡ T

The first step replaces A → B with ¬A ∨ B; the next uses De Morgan's law. The complementary pairs then make the entire expression true under every valuation.

Question 12

Consider the statements: S1: Every unsatisfiable formula is a contingency. S2: Every valid formula is a contingency. S3: Every invalid formula is a contradiction. Which option identifies the correct statements?

• A. S1 and S3

• B. S2 and S3

• C. S1 only

• D. None of these

Answer: D, None of these. All three statements are false:

• S1 fails because an unsatisfiable formula has 0 satisfying valuations, while a contingency has at least one true valuation and at least one false valuation.

• S2 fails because a valid formula is true in all valuations, while a contingency is false in some valuations.

• S3 fails because invalid formulas include both contradictions and contingencies, not contradictions alone.

The model counts make the boundaries visible. p ∨ ¬p is true in 2/2 valuations, so it is valid and satisfiable. p ↔ q is true in 2/4, so it is invalid but satisfiable. p ∧ ¬p is true in 0/2, so it is invalid and unsatisfiable.

The short version and next practice step

Keep this three-line memory rule beside every truth table:

• All T means valid and satisfiable.

• All F means invalid and unsatisfiable.

• Mixed T and F means invalid but satisfiable.

Write the final-column sequence before looking at the options. With two variables, T, F, F, T proves contingency, T, T, T, T proves tautology, and F, F, F, F proves contradiction. This habit prevents definition words from blurring together under time pressure.

With three variables, apply the same rule across all eight rows.

For a structured course route, continue with GATE Guidance by Sanchit Sir. The GATE CS Exam Preparation Courses & Test Series category helps you choose the next topic.