First-Order Predicate Logic MCQs: 10 Solved Questions with Explanations

First-order logic questions can look like vocabulary tests, but the topic spans syntax, semantics, English translation, quantifier range, validity and sound inference. Recognising ∀ and ∃ is not enough if a well-formed formula says the wrong thing. The common trap is manipulating familiar symbols without checking what each name, variable and connective contributes to the claim. Attempt each of the 10 questions before reading its explanation. Then use the Introduction to First Order Predicate Logic question hub for neighbouring questions.

First-order logic in one finite model

A domain contains the objects being discussed. Constants name objects, variables range over them, functions map objects to objects, and predicates return truth values. Connectives combine formulae, while ∀ and ∃ quantify variables. Syntax asks whether an expression is well formed. Semantics asks whether it is true under an interpretation.

Take D = {1,2}. Let a = 1, f(1)=2, f(2)=1, Even(1)=false and Even(2)=true. Also let only Less(1,2) be true, with Less(1,1), Less(2,1) and Less(2,2) false.

The term f(a) denotes 2, so Even(f(a)) is true, while Less(f(a),a) is false. Both formulae are well formed; the interpretation decides their truth. This syntax-semantics distinction determines whether a formula is well formed or true under an interpretation.

Now ∃x Even(x) is true because x=2 is a witness. In contrast, ∀x∃y Less(x,y) is false. The case x=1 has witness y=2, but x=2 has no witness in D. One witness proves an existential; every case must pass for a universal. Use Quantifiers and Predicate Logic MCQs: 10 Practice Problems Explained for distribution laws, uniqueness, English exceptions and graph reachability. Terms, predicates, Prolog, expressive boundaries and syntax need the separate checks used in Questions 1-6.

First-Order Predicate Logic MCQs 1-2: components and quantifier range

Question 1: what a first-order formula is built from (Bihar STET 2025)

What does a first order predicate logic contain?

• A. Predicate and a subject

• B. Predicate and a Preposition

• C. Subject and an object

• D. None of the above

Answer: Predicate and a subject. A predicate such as Even(x) states a property of the term's object; Even(2) is true in our model. Formally, first-order logic contains terms, predicates, connectives and quantifiers. Grammatical labels are not its building blocks, so A is only the closest option.

Question 2: the calculus that quantifies over objects (UGC NET 2014)

________predicate calculus allows quantified variables to refer to objects in the domain of discourse and not to predicates or functions.

• A. Zero-order

• B. First-order

• C. Second-order

• D. High-order

Answer: First-order. In ∀x Even(x), x ranges over objects 1 and 2, not predicates such as Even or Less. Second-order logic can quantify over predicates or relations. Zero-order propositional logic has no object variables.

First-Order Predicate Logic MCQs 3-4: foundations and expressive boundaries

Question 3: the logic behind PROLOG (UPPSC Polytechnic Lecturer 2022)

Which formal system provides the semantic foundation for PROLOG?

• A. Predicate Calculus

• B. Lambda Calculus

• C. Hoare Logic

• D. Propositional Logic

Answer: Predicate Calculus. A fact such as parent(asha, ravi) and rule grandparent(X,Z) :- parent(X,Y), parent(Y,Z) use predicates, variables and implication patterns. Lambda calculus handles function abstraction, Hoare logic reasons about program correctness, and propositional logic cannot directly represent these objects and relations.

Question 4: referential opacity in propositional attitudes (UGC NET 2025, Paper 2, December)

Which logic is designed to address the problem of referential opacity in propositional attitudes?

Regular Logic

Modal Logic

First-order logic

Predicate logic

• A. 1

• B. 2

• C. 3

• D. 4

Answer: 2, which denotes Modal Logic. Suppose MaskedCoder = Anil. Riya believes that MaskedCoder solved P need not imply that she believes Anil solved P, because she may not know the identity. Substitution is unsafe inside that context. Modal and related intensional logics provide operators for necessity, possibility, belief or knowledge, so item 2 is intended. Basic modal logic does not model every propositional attitude.

First-Order Predicate Logic MCQs 5-6: connectives and syntax

Question 5: translating “both” with the right connective

Let C(x) mean “x is a Ronaldo fan” and M(x) mean “x is a Messi fan”. Which expression correctly represents: “x is both a Ronaldo fan and a Messi fan”?

• A. C(M(x))

• B. C(x) ∨ M(x)

• C. C(x) ∧ M(x)

• D. Both (B) and (C)

Answer: C(x) ∧ M(x). Let C(Arun)=true, M(Arun)=true, C(Bina)=true and M(Bina)=false. For Arun, conjunction and disjunction are true. For Bina, only disjunction is true. “Both” therefore requires ∧. Also, C(M(x)) wrongly treats the truth value of M(x) as an object term.

Question 6: which symbols are syntactic elements (UGC NET 2025, Paper 2, January)

The kind of symbols for basic syntactic elements of first-order logic are

A. Constant

B. Domain

C. Predicate

D. Temporal

E. Function

Choose the correct answer from the options given below:

• A. B, D only

• B. A, B, C only

• C. A, C, E only

• D. C, D only

Answer: A, C, E only. In Likes(a,f(a)), a is a constant, f a function symbol and Likes a predicate symbol. The domain {1,2} is semantic data, not a syntactic symbol. Temporal belongs to temporal logic.

First-Order Predicate Logic MCQs 7-8: translating English precisely

Question 7: “Agra and Gwalior are both in India” (UGC NET 2018, July)

Consider the following English sentence:

"Agra and Gwalior are both in India".

A student has written a logical sentence for the above English sentence in First-Order Logic using predicate IN(x, y), which means x is in y, as follows.

In(Agra, India) ∨ In(Gwalior, India)

Which one of the following is correct with respect to the above logical sentence?

• A. It is syntactically valid but does not express the meaning of the English sentence

• B. It is syntactically valid and expresses the meaning of the English sentence also

• C. It is syntactically invalid but expresses the meaning of the English sentence

• D. It is syntactically invalid and does not express the meaning of the English sentence

Answer: It is syntactically valid but does not express the meaning of the English sentence. The formula is well formed. With In(Agra,India)=true and In(Gwalior,India)=false, it stays true although “both” is false. The intended formula is In(Agra,India) ∧ In(Gwalior,India).

Question 8: every event except marriage

Choose the correct conversion of the below statement to the First order logic

“Milind loves every event except marriage“

Note: If love(x,e) means that person x loves event e.

I) ∀x (¬ x = Marriage → love (Milind, x))

II) ∀x (x ≠ Marriage → love (Milind, x))

• A. Only I is correct

• B. Only II is correct

• C. both I, II are not correct

• D. both I, II are correct

Answer: both I, II are correct. Under the intended event domain, ¬(x=Marriage) and x≠Marriage are equivalent. On {Conference, Concert, Marriage}, the antecedent is true for the first two items and false for Marriage. Thus both love facts are required, but no dislike is asserted. For an all-object domain, use ∀x((Event(x) ∧ x≠Marriage) → love(Milind,x)).

First-Order Predicate Logic MCQ 9: validity by proof and countermodel

Question 9: which first-order formula is valid (GATE 2007, CS)

Which one of these first-order logic formula is valid?

• A. ∀x(P(x) => Q(x)) => (∀xP(x) => ∀xQ(x))

• B. ∃x(P(x) ∨ Q(x)) => (∃xP(x) => ∃xQ(x))

• C. ∃x(P(x) ∧ Q(x)) <=> (∃xP(x) ∧ ∃xQ(x))

• D. ∀x∃y P(x, y) => ∃y∀x P(x, y)

Answer: ∀x(P(x) => Q(x)) => (∀xP(x) => ∀xQ(x)). For A, take any object a. The two premises give P(a)→Q(a) and P(a), hence Q(a). Since a was arbitrary, ∀xQ(x) follows.

Countermodels reject the rest. For B, use {a} with P(a)=true, Q(a)=false: its antecedent is true and consequent false. For C, use {a,b} with only P(a) and Q(b) true. The right side has separate witnesses, but the left side has none. For D, use {1,2} and make P true exactly for (1,2) and (2,1). Each x has a witness, but no single y works for both. Reattempt the original GATE 2007 solved question after changing the relation.

First-Order Predicate Logic MCQ 10: inference chain and next step

Question 10: sound inference from two conditionals (GATE 2015, CS, Set 2)

Consider the following two statements.

𝑆1: If a candidate is known to be corrupt, then he will not be elected

𝑆2: If a candidate is kind, he will be elected

Which one of the following statements follows from 𝑆1 and 𝑆2 as per sound inference rules of logic?

• A. If a person is known to be corrupt, he is kind

• B. If a person is not known to be corrupt, he is not kind

• C. If a person is kind, he is not known to be corrupt

• D. If a person is not kind, he is not known to be corrupt

Answer: If a person is kind, he is not known to be corrupt. Let C, E and K mean known corrupt, elected and kind. The premises are C→¬E and K→E. The first has contrapositive E→¬C; chaining gives K→¬C. For Arun, K=true, E=true, C=false satisfies both premises and the conclusion. The other choices misuse implication.

In Propositional and Predicate Logic MCQs: 12 Solved, the same GATE 2007 validity item closes mixed practice on connectives, translation and validity. For Question 9 above, prove A and build explicit countermodels for B-D. For integer and real-domain evaluations, function properties, model-size bounds and divisibility, continue with Predicate Logic and Quantifiers MCQs: 10 Solved Questions with Explanations. Keep the components, syntax, Prolog, modal boundaries and English-translation checks from Questions 1-8 alongside it. Use the GATE Test Series when you want timed practice.

The short version is simple: determine the domain, separate syntax from meaning, translate one connective at a time, and test universal claims with a countermodel.