# NPTEL Fundamentals of Artificial intelligence Week 6 Assignment Answers 2024

By Sanket

## 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

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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

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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

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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)

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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

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