Quantifiers MCQs: 12 Solved Universal and Existential Questions

Most students remember that ∀ means "for all" and ∃ means "there exists". The marks are lost one step later, when an implication becomes a conjunction, a negation crosses a quantifier, or the order of two quantifiers changes who may depend on whom. All 12 worked questions are GATE previous-year questions. If you need a quick theory refresh first, read Propositional and Predicate Logic: Truth Tables to Proofs. The earlier Quantifiers and Predicate Logic MCQs: 10 Practice Problems Explained concentrates on diagnosing imperfect source notation and answer-key defects; these 12 GATE PYQs concentrate on translation, nested scope, uniqueness, validity, countermodels, and equivalence. Attempt each item before opening its explanation. Predict the formula, identify a witness or counterexample, and only then compare your reasoning with the worked answer.

1. Quantifiers in plain language: four moves to use before solving

On {Asha, Bharat}, ∀x P(x) checks both people, while ∃x P(x) needs one witness. Negation swaps the quantifier and negates the predicate:

• ¬∀x P(x) ≡ ∃x¬P(x)

• ¬∃x P(x) ≡ ∀x¬P(x)

Universal category restrictions normally use implication: ∀x(student(x) → studies(x)). Existential witnesses normally use conjunction: ∃x(student(x) ∧ studies(x)). ∀x∃y may choose a different y for every x, while ∃y∀x fixes one y for everyone.

Use four steps: identify the domain, mark every or some, choose implication or conjunction, then test a doubtful formula on one or two objects. The GATE CS Exam Preparation page is the wider preparation path.

2. Universal and existential translation MCQs

Q1. "Not all that glitters is gold" (2014)

GATE 2014 solved question.

Consider the statement “Not all that glitters is gold”. Let glitters(x) mean that x glitters and gold(x) mean that x is gold. Which logical formula represents the statement?

• A. $\forall x: \text{glitters} (x)\Rightarrow \neg \text{gold}(x)$

• B. $\forall x:\text{gold} (x)\Rightarrow \text{glitters}(x)$

• C. $\exists x: \text{gold}(x)\wedge \neg \text{glitters}(x)$

• D. $\exists x: \text{glitters}(x)\wedge \neg \text{gold}(x)$

Answer: D. ¬∀x(glitters(x) → gold(x)) ≡ ∃x¬(glitters(x) → gold(x)) ≡ ∃x(glitters(x) ∧ ¬gold(x)). "Not all" needs one counterexample, not a claim that every glittering object is not gold.

Q2. "None of my friends are perfect" (2013)

GATE 2013 solved question.

What is the logical translation of the following statement? “None of my friends are perfect.”

• A. $∃x (F (x)∧ ¬P(x))$

• B. $∃ x (¬ F (x)∧ P(x))$

• C. $∃x (¬F (x)∧¬P(x))$

• D. $¬∃ x (F (x)∧ P(x))$

Answer: D. The forbidden witness is someone who is both a friend and perfect, giving ¬∃x(F(x) ∧ P(x)). Equivalently, ∀x(F(x) → ¬P(x)).

Q3. "Some real numbers are rational" (2012)

GATE 2012 solved question.

What is the correct translation of the following statement into mathematical logic? “Some real numbers are rational”

• A. $\exists x (\text{real}(x) \lor \text{rational}(x))$

• B. $\forall x (\text{real}(x) \to \text{rational}(x))$

• C. $\exists x (\text{real}(x) \wedge \text{rational}(x))$

• D. $\exists x (\text{rational}(x) \to \text{real}(x))$

Answer: C. "Some" supplies one witness, and that witness must satisfy both predicates. Option B instead claims that every real number is rational.

Q4. Some boys are taller than all the girls (2004)

GATE 2004 solved question.

Identify the correct translation into logical notation of the following assertion. "Some boys in the class are taller than all the girls" Note : taller(x,y) is true if x is taller than y.

• A. (∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y)))

• B. (∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y)))

• C. (∃x) (boy(x) → (∀y) (girl(y) → taller(x,y)))

• D. (∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y)))

Answer: D. The witness must be a boy, so the outer connection is conjunction; the universal domain is restricted to girls by implication. Its scope is ∃x[boy(x) ∧ ∀y(girl(y) → taller(x,y))].

3. Nested quantifiers, dependency, and uniqueness MCQs

Q5. Every teacher is liked by some student (2005)

GATE 2005 solved question.

What is the first order predicate calculus statement equivalent to the following? Every teacher is liked by some student

• A. ∀(x) [teacher (x) → ∃ (y) [student (y) → likes (y, x)]]

• B. ∀ (x) [teacher (x) → ∃ (y) [student (y) ^ likes (y, x)]]

• C. ∃ (y) ∀ (x) [teacher (x) → [student (y) ^ likes (y, x)]]

• D. ∀ (x) [teacher (x) ^ ∃ (y) [student (y) → likes (y, x)]]

Answer: B, where ^ means conjunction. With teachers={T1,T2}, students={S1,S2}, likes(S1,T1)=true, and likes(S2,T2)=true, ∀x∃y holds although no one student likes both teachers; A can be satisfied improperly by choosing a non-student, which makes its inner implication vacuously true.

Q6. Who can fool whom, and when? (2010)

GATE 2010 solved question.

Suppose F(x,y,t) means that person x can fool person y at time t. Which statement best expresses ∀x∃y∃t(¬F(x,y,t))?

• A. Everyone can fool some person at some time

• B. No one can fool everyone all the time

• C. Everyone cannot fool some person all the time

• D. No one can fool some person at some time

Answer: B. ¬∃x∀y∀t F(x,y,t) ≡ ∀x∃y∃t¬F(x,y,t), so nobody fools every person at every time. Each x needs one failed person-time pair, not one permanently unfoolable person.

Q7. "Everyone has exactly one mother" (2025)

GATE 2025 solved question.

Which predicate-logic formulae correctly represent “Everyone has exactly one mother”? Here mother(y,x) means that y is the mother of x, and noteq(x,y) means that x and y are not equal.

• A. $∀𝑥∃𝑦∃𝑧(𝑚𝑜𝑡ℎ𝑒𝑟(𝑦, 𝑥) ∧ ¬𝑚𝑜𝑡ℎ𝑒𝑟(𝑧, 𝑥))$

• B. $∀𝑥∃𝑦[𝑚𝑜𝑡ℎ𝑒𝑟(𝑦, 𝑥) ∧ ∀𝑧(𝑛𝑜𝑡𝑒𝑞(𝑧, 𝑦) → ¬𝑚𝑜𝑡ℎ𝑒𝑟(𝑧, 𝑥))]$

• C. $∀𝑥∀𝑦[𝑚𝑜𝑡ℎ𝑒𝑟(𝑦, 𝑥) → ∃𝑧(𝑚𝑜𝑡ℎ𝑒𝑟(𝑧, 𝑥) ∧ ¬𝑛𝑜𝑡𝑒𝑞(𝑧, 𝑦))]$

• D. $∀𝑥∃𝑦[𝑚𝑜𝑡ℎ𝑒𝑟(𝑦, 𝑥) ∧ ¬∃𝑧(𝑛𝑜𝑡𝑒𝑞(𝑧, 𝑦) ∧ 𝑚𝑜𝑡ℎ𝑒𝑟(𝑧, 𝑥))]$

Answers: B and D. For fixed x=A and witness y=M, both formulas assert that M is a mother and exclude every different z from also being a mother. A only finds a mother and a non-mother; C neither guarantees a mother for every x nor excludes another one.

4. Validity and countermodel MCQs

Q8. Distribution of a universal quantifier over disjunction (2016)

GATE 2016 solved question.

Which one of the following well-formed formulae in predicate calculus is NOT valid?

• A. $(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x¬p(x) ∨ ∀x q(x))$

• B. $(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))$

• C. $∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))$

• D. $∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))$

Answer: D. On {a,b}, set p(a)=true, p(b)=false, q(a)=false, and q(b)=true; then p(x)∨q(x) holds for both objects, but ∀x p(x) and ∀x q(x) are both false. The implication therefore has a true antecedent and false consequent.

Q9. An implication that is always true (2005)

GATE 2005 solved question.

Let P(x) and Q(x) be arbitrary predicates. Which of the following statements is always TRUE?

• A. ((∀x(P(x)∨Q(x))))⟹((∀xP(x))∨(∀xQ(x)))

• B. (∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))

• C. (∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))

• D. (∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))

Answer: B. Assume ∀x(P(x)→Q(x)) and ∀xP(x); for arbitrary c, modus ponens gives Q(c), hence ∀xQ(x). The split assignment from Q8 makes A, C, and D false, so none of them is always true.

Q10. Satisfiable versus valid for a symmetric relation (2006)

GATE 2006 solved question.

Consider the following first order logic formula in which R is a binary relation symbol. ∀x∀y (R(x, y) => R(y, x)) The formula is

• A. satisfiable and valid

• B. satisfiable and so is its negation

• C. unsatisfiable but its negation is valid

• D. satisfiable but its negation is unsatisfiable

Answer: B. On {a,b}, R=∅ satisfies the formula because every implication has a false antecedent. On the same domain, R={(a,b)} satisfies its negation: R(a,b) is true while R(b,a) is false, so both sides have a model.

5. Moving quantifiers through implications and negations

Q11. A logically valid formula with a closed β (2008)

GATE 2008 solved question.

Which of the following first order formula is logically valid? Here α(x) is a first order formula with x as a free variable, and β is a first order formula with no free variable.

• A. [β→(∃x,α(x))]→[∀x,β→α(x)]

• B. [∃x,β→α(x)]→[β→(∀x,α(x))]

• C. [(∃x,α(x))→β]→[∀x,α(x)→β]

• D. [(∀x,α(x))→β]→[∀x,α(x)→β]

Answer: C. Under the antecedent, take arbitrary c:

α(c) ⇒ ∃xα(x)

∃xα(x) ⇒ β

Therefore, α(c) ⇒ β

Therefore, ∀x(α(x) ⇒ β)

The movement is legitimate because β has no free x.

Q12. Equivalent form of an existential-universal statement (2004)

GATE 2004 solved question.

Let a(x,y), b(x,y), and c(x,y) be statements over a common universe. Consider (∃x)(∀y)[(a(x,y) ∧ b(x,y)) ∧ ¬c(x,y)]. Which option is equivalent to it?

• A. (∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)]

• B. (∃x)(∀y)[(a(x, y) ∨ b(x, y)) ∧¬ c(x, y)]

• C. ¬ (∀x)(∃y)[(a(x, y) ∧ b(x, y)) → c(x, y)]

• D. ¬ (∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)]

Answer: C. Let P(x,y)=[(a(x,y)∧b(x,y))∧¬c(x,y)]:

∃x∀yP ≡ ¬∀x∃y¬P

¬P ≡ ¬[(a∧b)∧¬c]

≡ ¬(a∧b)∨c

≡ (a∧b)→c

Substitution gives exactly option C.

6. Quantifier traps: a one-page revision table

For validity, reuse the finite-model check from Q8 and Q10. Two objects {a,b} plus one careful predicate or relation assignment often refute a universal claim. Use the broader Propositional and Predicate Logic MCQ set for mixed practice.

7. How quantifier questions are tested, and the next practice step

English-to-formula translation appears in Q1 to Q5; order and uniqueness appear in Q6 to Q7; countermodel construction appears in Q8 to Q10; equivalence transformations appear in Q11 to Q12. When reviewing, record why each tempting option fails, not only the correct letter.

The short version: implication restricts universal claims, conjunction builds existential witnesses, negation swaps ∀ with ∃, and validity needs a proof or one countermodel. For the complete subject sequence, use GATE Guidance by Sanchit Sir. If the theory is clear and you need timed practice, move to the GATE Test Series.