## NPTEL Fundamentals of Artificial intelligence Week 6 Assignment Answers 2024

1. Under which of the following conditions, a clause {Li} subsumes a clause {Mi}, if there exists a substitution `s’?

Condition 1: {Li}s is a subset of {Mi}

Condition 2: {Mi}s is a subset of {Li}

A. No clause ever subsumes another clause.

B. Condition 1 only

C. Both Condition 1 and Condition 2

D. Condition 2 only

Answer :-For Answers Click Here

2. Rule of inference, called __________________ when added to resolution principle, guarantees refutation completeness, even when involving equality.

A. Demodulation

B. Paramodulation

C. Modulation

D. Equimodulation

Answer :-For Answers Click Here

For Q3 to Q6

Consider the following representation in First Order Logic:

Animal(x) : x is an animal.

Dog(x) : x is a dog.

AnimalLover(x) : x is an animal lover.

Kills(x,y) : x kills y.

Owns(x,y) : x owns y.

3. In first-order logic, each variable refers to some object in a set called the _____________.

A. domain of discourse

B. knowledge base

C. interpretation

D. semantics

Answer :-For Answers Click Here

4. Translate the following English sentence to First Order Logic statement.

Jack owns a dog.

A. ∃x( Dog(x) → Owns(Jack,x))

B. ∀x(Dog (x) ∧ Owns(Jack,x))

C. ∃x( Dog(x) ∧ Owns(Jack,x))

D. ∀x(Dog (x) → Owns(Jack,x))

Answer :-

5. Translate the following English sentence to First Order Logic statement.

Every dog owner is an animal lover.

A. ∀x (∀y Dog (y) ∧ Owns(x,y)) → AnimalLover(x)

B. ∀x (∃y Dog (y) ∧ Owns(x,y)) → AnimalLover(x)

C. ∀x∃x (Dog (x) ∧ Owns(x,y)) ∧ AnimalLover(x))

D. ∀y (∃x Dog (y) ∧ Owns(x,y)) → AnimalLover(x)

Answer :-

6. Translate the following English sentence to First Order Logic statement.

No animal lover kills an animal.

A. ∀x AnimalLover(x) → (∀y Animal (y) → ¬ Kills(x,y))

B. ∀x∃y (Animal (y) ∧ AnimalLover(x) ∧ ¬ Kills(x,y))

C. ∃x ∃y (Animal (x) ∧ AnimalLover (y) ∧ ¬Kills(x,y))

D. ∀x ∀y (AnimalLover(x) ∧ Animal (y)) → ¬ Kills(x,y)

Answer :-For Answers Click Here

For Q7 to Q10

Consider the following representation in First Order Logic:

```
P(x): x is registered for today’s race.
Q(x): x is a thoroughbred.
R(x): x has won a race this year.
U = all horses
```

Axioms: A1. A horse that is registered for today’s race is not a thoroughbred.

A2. Every horse registered for today’s race has won a race this year.

Theorem: Th1. Therefore a horse that has won a race this year is not a thoroughbred.

7. Translate the following English sentence to FOL.

A horse that is registered for today’s race is not a thoroughbred.

A. ∀x [P(x) ∨ Q(x)]

B. ∀x[P(x) → ¬ Q(x)]

C. ∃x[P(x) ∧ ¬ Q(x)]

D. ∀x [P(x) → Q(x)]

Answer :-

8. Translate the following English sentence to First Order Logic.

Every horse registered for today’s race has won a race this year.

A. ∃x[P(x) ∧ R(x)]

B. ∀x [P(x) → R(x)]

C. ∃x[P(x) ∧ R(x)]

D. ∀x R(x)

Answer :-

9. Translate the following English sentence to FOL.

A horse that has won a race this year is not a thoroughbred.

A. ∀x [R(x) ∨ Q(x)]

B. ∀x[R(x) → ¬ Q(x)]

C. ∃x[R(x) ∧ ¬ Q(x)]

D. ∀x [R(x) → Q(x)]

Answer :-

10. P(c) for some element c

∃x P(x)

We will be using this to prove statements of the form ∃x P(x). This is _____________.

A. Universal generalization

B. Universal Instantiation

C. Existential generalization

D. Existential instantiation

Answer :-For Answers Click Here