Propositions MCQs: 12 Solved Practice Problems with Explanations

Proposition questions look like symbol substitution, but errors usually come from reversing an implication, confusing a converse with a contrapositive, or testing too few assignments. Attempt these 12 solved MCQs and write one decisive equivalence or counterexample before checking each explanation. KnowledgeGate has over 25 practice questions on propositions alone. For truth-table basics, use the propositional and predicate logic refresher; this drill also fits GATE CS exam preparation.

The proposition rules to write before solving

Rule card: p → q is false only when p=T, q=F; its contrapositive is ¬q → ¬p; p ↔ q is true when both values match; and p → q is equivalent to ¬p ∨ q.

At p=T, q=F, p → q=F. Its contrapositive ¬q → ¬p is also T → F=F; the converse q → p is F → T=T, so it is not equivalent.

Translate, simplify, then prove every row or find a counterexample. Next, try the Propositional and Predicate Logic MCQs.

Questions 1-3: translate language and judge inferences

Question 1

Let game(ball, rugby) be true if the ball is used in rugby and false otherwise. Let shape(ball, round) be true if the ball is round and false otherwise. Consider the following logical sentences: s1: ∀ball ¬ game(ball, rugby) ⟹ shape(ball, round) s2: ∀ball ¬ shape(ball, round) ⟹ game(ball, rugby) s3: ∀ball game(ball, rugby) ⟹ ¬ shape(ball, round) s4: ∀ball shape(ball, round) ⟹ ¬ game(ball, rugby) Which of the following choices is/are logical representations of the assertion, “All balls are round except balls used in rugby”?

• A. 𝑠1 ∧ 𝑠3

• B. 𝑠1 ∧ 𝑠2

• C. 𝑠2 ∧ 𝑠3

• D. 𝑠3 ∧ 𝑠4

Correct answer: A.

s1 says non-rugby balls are round; s3 says rugby balls are not. Because s4 is s3's contrapositive, s3 ∧ s4 repeats one requirement and misses the other.

Question 2

Consider the following logical inferences. I1: If it rains then the cricket match will not be played. The cricket match was played. Inference: There was no rain. I2: If it rains then the cricket match will not be played. It did not rain. Inference: The cricket match was played. Which of the following is TRUE?

• A. Both I1 and I2 are correct inferences

• B. I1 is correct but I2 is not a correct inference

• C. I1 is not correct but I2 is a correct inference

• D. Both I1 and I2 are not correct inferences

Correct answer: B.

Let R mean rain and P played. I1 uses the contrapositive P → ¬R of R → ¬P. I2 denies the antecedent: ¬R cannot force P; poor light could still cancel play.

Question 3

Combinatorics deals with problems involving counting. For example, “How many distinct arrangements of N distinct objects in M spaces on a circle are possible?” is a typical problem in combinatorics. This kind of counting is sometimes used in the modeling of several physical phenomena. Often, in such models, the different combinatorial possibilities are assigned probability values. Assigning probabilities enables the computation of the average values of physical quantities. Consider the following statements: P: Combinatorics is always invoked in the modeling of physical phenomena. Q: Modeling some physical phenomena involves assigning probabilities to combinatorial possibilities in order to compute average values of physical quantities. Based on the passage above, what can be inferred about statements P and Q?

• A. P is False and Q is False

• B. P is False and Q is True

• C. P is True and Q is False

• D. P is True and Q is True

Correct answer: B.

“Sometimes” cannot support P's “always”, so P is false. Q restates the claim about some models, so it is true.

Questions 4-6: biconditionals and valid arguments

Question 4

In propositional logic P ↔ Q is equivalent to (Where ~ denotes NOT)

• A. ~(P ∨ Q) ∧ ~(Q ∨ P)

• B. (~P ∨ Q) ∧ (~Q ∨ P)

• C. (P ∨ Q) ∧ (Q ∨ P)

• D. ~(P ∨ Q) → ~(Q ∨ P)

Correct answer: B.

Expand P ↔ Q as (P → Q) ∧ (Q → P)= (~P ∨ Q) ∧ (~Q ∨ P). At P=T,Q=F, both are false.

Question 5

Let p, q, r and s be four primitive statements. Consider the following arguments: P: [((¬p ∨ q) ∧ (r → s) ∧ (p ∨ r))] → (¬s → q) Q: [((¬p ∧ q) ∧ (q → (p → r)))] → ¬r R: [((q ∧ r) → p) ∧ (¬q ∨ p)] → r S: [p ∧ (p → r) ∧ (q ∨ ¬r)] → q Which of the above arguments are valid?

• A. P and Q only

• B. P and R only

• C. P and S only

• D. P, Q, R and S

Correct answer: C.

For P, assume ¬s: r → s gives ¬r, p ∨ r gives p, and ¬p ∨ q gives q. For S, p → r gives r, so q ∨ ¬r forces q. Q fails at p=F,q=T,r=T; R fails at p=F,q=F,r=F: true premises, false conclusions.

Question 6

Let a, b, c, d be propositions. Assume that the equivalences a ↔ (b ∨ ¬b) and b ↔ c hold. Then the truth value of the formula (a ∧ b) → (a ∧ c) ∨ d is always

• A. True

• B. False

• C. Same as the truth value of b

• D. Same as the truth value of d

Correct answer: A.

Since b ∨ ¬b=T, a=T, and b ↔ c makes b=c. A true antecedent gives a ∧ c=T; a false antecedent makes the implication true. Thus d is irrelevant.

Questions 7-9: prove tautologies and build counterexamples

Question 7

Indicate which of the following well-formed formulae are valid:

• A. [(P ⇒ Q) ∧ (Q ⇒ R)] ⇒ (P ⇒ R)

• B. (P ⇒ Q) ⇒ (¬P ⇒ ¬Q)

• C. (P ∧ (¬P ∨ ¬Q)) ⇒ Q

• D. (P ⇒ R) ∨ (Q ⇒ R) ⇒ ((P ∨ Q) ⇒ R)

Correct answer: A.

A chains P → Q and Q → R into P → R. Counterexamples reject B at P=F,Q=T, C at P=T,Q=F, and D at P=T,Q=F,R=F when implication has lowest precedence.

Question 8

Let P, Q and R be three atomic propositional assertions. Let X denote (P ∨ Q) → R and Y denote (P → R) ∨ (Q → R). Which one of the following is a tautology?

• A. X ≡ Y

• B. X → Y

• C. Y → X

• D. ¬Y → X

Correct answer: B.

Simplify X to (¬P ∧ ¬Q) ∨ R and Y to ¬P ∨ ¬Q ∨ R; X true makes Y true. P=F,Q=T,R=F gives Y=T,X=F, rejecting A and C. P=T,Q=T,R=F gives Y=F,X=F, rejecting D.

Question 9

Which of the following is false? Read ∧ as AND, ∨ as OR, ¬ as NOT, → as one-way implication and ↔ as two-way implication

• A. ((x → y) ∧ x) → y

• B. ((¬x → y) ∧ (¬x → ¬y)) → x

• C. (x → (x ∨ y))

• D. ((x ∨ y) ↔ (¬x → ¬y))

Correct answer: D.

A is modus ponens, while C implies a weaker disjunction. In B, x=F requires both y and ¬y; at x=T, the conclusion holds. D fails at x=F,y=T: its sides are T and T → F=F.

Questions 10-12: positive integers, truth sets and excluded middle

Question 10

Let m and n are positive integers. Then (A) If n ≠ 1, then m < mn. (B) If k is composite, then k = mn where 1 < m, n > k. (C) If mn = 1, then m = 1 and n = 1. (D) If k is composite, then k = mn where 1 < m, n < k. Which of the following is correct:

• A. (A), (C), (D)

• B. (B), (C), (D)

• C. (A), (B)

• D. (A), (B), (C)

Correct answer: A.

For positive integers, n ≠ 1 gives n ≥ 2, so mn ≥ 2m > m; A holds. Only m=n=1 gives mn=1, proving C. Composite factors lie between 1 and the number, as 6=2×3 shows; D holds and B fails.

Question 11

Consider universe positive integer X={1≤n≤8} Proposition P= “n is an even integer” Q= “(3 ≤ n ≤ 7) ∧ (n ≠ 6)” then truth set of P↔Q is

• A. {1,4}

• B. {2,6}

• C. {3,4,5}

• D. {1}

Correct answer: A.

Here P={2,4,6,8} and Q={3,4,5,7}. Membership matches at n=1 (false) and n=4 (true), differing elsewhere. Thus the truth set is {1,4}.

Question 12

An example of a Tautology is:

• A. x ∨ y

• B. x ∨ ¬y

• C. x ∨ ¬x

• D. (x → y) ∧ (y → x)

Correct answer: C.

Excluded middle gives T ∨ F=T for x=T and F ∨ T=T for x=F. D fails at x=T,y=F because x → y is false.

Four traps this propositions set exposes

Separate converse from contrapositive. Question 2 derives P → ¬R from R → ¬P, not ¬R → P. One counterexample disproves a universal formula; successful rows do not prove validity. Use exact truth assignments to test the formulas.

Watch quantifier strength: “sometimes” cannot support “always,” while an “except” sentence may require two implications. Test both biconditional directions or compare complete truth sets. Use a full table only if needed. Continue with the Implication and Biconditional Operators MCQs for practice.

Short version and the next practice step

• Translate every sentence.

• Eliminate implications when useful.

• Try a decisive truth assignment.

• Distinguish proof from counterexample.

For a 20-minute follow-up, redo Questions 1, 5, 8 and 11 in four minutes each, then explain why one distractor fails in each.

For Discrete Mathematics study, continue with GATE Guidance by Sanchit Sir.