Students know → and ↔ truth tables but lose marks when English hides them. They confuse converse with contrapositive, mistranslate “only if” and “unless”, or forget p → q ≡ ¬p ∨ q. These 11 MCQs come from KnowledgeGate's bank of about 55 published questions on this subtopic. Each shows the deciding step.
Lock these five facts before attempting the questions
Implication: p → q is false only for p=T, q=F. It is true otherwise, including whenever p is false, called vacuously true.
Core identity: p → q ≡ ¬p ∨ q.
Negation: ¬(p → q) ≡ p ∧ ¬q, a conjunction.
Three relatives: converse q → p, inverse ¬p → ¬q, contrapositive ¬q → ¬p. Only the contrapositive equals p → q. Converse and inverse equal each other.
Biconditional: p ↔ q is true when p and q match. Its forms are (p → q) ∧ (q → p), (¬p ∨ q) ∧ (p ∨ ¬q), and (p ∧ q) ∨ (¬p ∧ ¬q).
For full theory, read Propositional and Predicate Logic.
Translating English into implication: converse, “only if”, “unless”
Q1. Converse of “only if”
Stem: What is the converse of the following assertion? I stay only if you go.
Options: A) I stay if you go B) If I stay then you go C) If you do not go then I do not stay D) If I do not stay then you go
Correct answer: A. With S for “I stay” and G for “you go”, the assertion is S → G. Its converse G → S reads “I stay if you go”. Option B is the original. See this converse question solved.
Q2. Translating “cannot be given when”
Stem: p: Fail grade can be given. q: Student scores more than 50% marks. Statement: “Fail grade cannot be given when student scores more than 50% marks.”
Options: A) q → ¬p B) q → p C) p → q D) ¬p → q
Correct answer: A. The condition is q and the result is ¬p, so the formula is q → ¬p. Reversing the arrow is the trap. See this sentence-translation question solved.
Q3. Translating “unless”
Stem: “If X, then Y unless Z” in propositional logic.
Options: A) (X ∧ ¬Z) → Y B) (X ∧ Y) → ¬Z C) X → (Y ∧ ¬Z) D) (X → Y) ∧ ¬Z
Correct answer: A. “Y unless Z” means ¬Z → Y. X is also required, giving (X ∧ ¬Z) → Y. Attaching ¬Z to the result is the trap. See this unless question solved.
Equivalence and the contrapositive trap
Q4. Inverse, converse, and equivalent form
Stem: The statement (¬p) → (¬q) is logically equivalent to which of the statements below? I. p → q II. q → p III. (¬q) ∨ p IV. (¬p) ∨ q.
Options: A) I only B) I and IV only C) II only D) II and III only
Correct answer: D. The contrapositive of ¬p → ¬q is q → p, statement II. Expanding it gives ¬q ∨ p, statement III. The trap is equating this inverse with p → q. See this logical-equivalence question solved.
Q5. Contrapositive in ordinary language
Stem: P: Good mobile phones are not cheap. Q: Cheap mobile phones are not good. L: P implies Q. M: Q implies P. N: P is equivalent to Q. Which is CORRECT?
Options: A) Only L is TRUE B) Only M is TRUE C) Only N is TRUE D) L, M and N are TRUE
Correct answer: D. With g for good and c for cheap, P is g → ¬c and Q is its contrapositive c → ¬g. They are equivalent, so L, M, and N hold. The trap is missing the contrapositive in words. See this contrapositive question solved. Also revise Set Theory and Relations.
Biconditional equivalence questions
Q6. Spot the XOR
Stem: Which one of the following is NOT equivalent to p ↔ q?
Options: A) (¬p ∨ q) ∧ (p ∨ ¬q) B) (¬p ∨ q) ∧ (q → p) C) (¬p ∧ q) ∨ (p ∧ ¬q) D) (¬p ∧ ¬q) ∨ (p ∧ q)
Correct answer: C. A and B express (p → q) ∧ (q → p), while D covers matching cases. C is XOR, true when p and q differ, so it negates p ↔ q. Missing that reversal is the trap. See this biconditional-equivalence question solved.
Q7. Exactly two variables are true
Stem: Which one of the following propositional logic formulas is TRUE when exactly two of p, q and r are TRUE?
Options: A) ((p ↔ q) ∧ r) ∨ (p ∧ q ∧ ¬r) B) (¬(p ↔ q) ∧ r) ∨ (p ∧ q ∧ ¬r) C) ((p → q) ∧ r) ∨ (p ∧ q ∧ ¬r) D) (¬(p ↔ q) ∧ r) ∧ (p ∧ q ∧ ¬r)
Correct answer: B. If r is true, p and q must differ: ¬(p ↔ q) ∧ r. If r is false, both must be true: p ∧ q ∧ ¬r. OR gives B. A wrongly makes p and q equal; D joins incompatible cases with AND. See this exactly-two-true question solved.
Evaluating a formula under a given premise
Q8. Can the target truth value be fixed?
Stem: If the proposition ¬p → q is true, then the truth value of the proposition ¬p ∨ (p → q), where ¬ is negation, ∨ is inclusive OR and → is implication, is
Options: A) True B) Multiple Values C) False D) Cannot be determined
Correct answer: D. The premise is ¬p → q ≡ p ∨ q. For p=T, q=T, the target is F ∨ (T → T) = T. For p=T, q=F, the premise F → F remains true, but the target is F ∨ (T → F) = F. Both satisfy the premise, so the target is undetermined. Testing one case is the trap. See this truth-value question solved.
Full-sentence translation and predicate-logic NAT questions
Q9. Translate both clauses before joining them
Stem: p, q, r denote “It is raining”, “It is cold”, “It is pleasant”. Statement: “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”
Options: A) (¬p ∧ r) ∧ (¬r → (p ∧ q)) B) (¬p ∧ r) ∧ ((p ∧ q) → ¬r) C) (¬p ∧ r) ∨ ((p ∧ q) → ¬r) D) (¬p ∧ r) ∨ (r → (p ∧ q))
Correct answer: A. The first clause is ¬p ∧ r. “A only if B” means A → B, so the second is ¬r → (p ∧ q). AND gives option A. Reversing “only if” or using OR is the trap. See this full-sentence translation solved.
Q10. Count the implied expressions
Stem: Consider the following expressions: (i) false (ii) Q (iii) true (iv) P ∨ Q (v) ¬Q ∨ P. The number of expressions given above that are logically implied by P ∧ (P → Q) is ____.
Correct answer: 4. The premise makes P true, then modus ponens makes Q true. Now (i) false is not implied; (ii) Q is; (iii) true always is; (iv) P ∨ Q = T; and (v) ¬Q ∨ P = F ∨ T = T. Items (ii) to (v) give 4. Forgetting that true is implied is the trap. See this implication-count NAT solved.
Q11. Simplify the negation before testing values
Stem: p: x ∈ {8, 9, 10, 11, 12}; q: x is a composite number; r: x is a perfect square; s: x is a prime number. Find the integer x ≥ 2 satisfying ¬((p → q) ∧ (¬r ∨ ¬s)) ____.
Correct answer: 11. Apply De Morgan's law and negate the implication:
¬((p → q) ∧ (¬r ∨ ¬s)) ≡ ¬(p → q) ∨ ¬(¬r ∨ ¬s) ≡ (p ∧ ¬q) ∨ (r ∧ s).
No x ≥ 2 is both prime and a perfect square, so r ∧ s is false. In the set, 8, 9, 10, and 12 are composite; only 11 satisfies p ∧ ¬q. Mishandling the two negations is the trap. See this predicate-logic NAT solved. Continue with Group Theory: Groups, Rings and Fields for more practice.
The traps that cost the most marks
Converse q → p and inverse ¬p → ¬q do not equal p → q. Only contrapositive ¬q → ¬p does.
“A only if B” means A → B. “A unless B” means ¬B → A. “A if B” means B → A.
Negating p → q gives p ∧ ¬q, never another implication.
p ↔ q is true on matching rows. Its negation, p XOR q, is true on differing rows.
For “cannot be determined”, test two premise-satisfying assignments.
The short version and where to practise next
Master p → q ≡ ¬p ∨ q and the direction of “only if” and “unless”. Most questions then need one short transformation.
Study with Engineering Mathematics for GATE, then use the GATE Test Series for timed practice. The GATE CS Exam preparation hub connects it to the wider syllabus. With about 55 solved questions on this subtopic, KnowledgeGate has enough practice to make these translations automatic.
