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 horsesAxioms: 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
