All questions of Propositional Logic for Computer Science Engineering (CSE) Exam
Which of the following is/are the type(s) of normal forms to which reduction can be performed?- a)Disjunctive Normal Form
- b)Conjunctive Normal Form
- c)Both A and B
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
Which of the following is/are the type(s) of normal forms to which reduction can be performed?
a)
Disjunctive Normal Form
b)
Conjunctive Normal Form
c)
Both A and B
d)
None of the above
|
Sanvi Kapoor answered |
The following are the types of normal forms to which reduction can be performed -
- Disjunctive Normal Form
- Conjunctive Normal Form
p is the proposition “ Indian army moves back”
q is the proposition “ Chinese army moves back”
r is the proposition “ There is no war”
Then what is the contrapositive of the statement “ If Indian army moves back and Chinese army moves back, then there is no war”- a)~r→(~p→~q)
- b)r→(p∧q)
- c)~(p∧q)→~r
- d)~r→~(p∧q)
Correct answer is option 'D'. Can you explain this answer?
p is the proposition “ Indian army moves back”
q is the proposition “ Chinese army moves back”
r is the proposition “ There is no war”
Then what is the contrapositive of the statement “ If Indian army moves back and Chinese army moves back, then there is no war”
q is the proposition “ Chinese army moves back”
r is the proposition “ There is no war”
Then what is the contrapositive of the statement “ If Indian army moves back and Chinese army moves back, then there is no war”
a)
~r→(~p→~q)
b)
r→(p∧q)
c)
~(p∧q)→~r
d)
~r→~(p∧q)
|
Kavya Mehta answered |
It seems that the statement you provided is incomplete. Could you please provide the complete proposition?
Let p, q, r, s represents the following propositions.
p: x ∈ {8, 9, 10, 11, 12}
q: x is a composite number
r: x is a perfect square
s: x is a prime number
The integer x ≥ 2 which satisfies ¬ ((p ⇒ q) ∧ (¬ r ∨ ¬ s)) is __________
Correct answer is '11'. Can you explain this answer?
Let p, q, r, s represents the following propositions.
p: x ∈ {8, 9, 10, 11, 12}
q: x is a composite number
r: x is a perfect square
s: x is a prime number
The integer x ≥ 2 which satisfies ¬ ((p ⇒ q) ∧ (¬ r ∨ ¬ s)) is __________
p: x ∈ {8, 9, 10, 11, 12}
q: x is a composite number
r: x is a perfect square
s: x is a prime number
The integer x ≥ 2 which satisfies ¬ ((p ⇒ q) ∧ (¬ r ∨ ¬ s)) is __________
|
Nitya Sharma answered |
Understanding the Propositions
To solve the expression ¬((p → q) ∧ (¬r ∨ ¬s), let's first clarify the propositions:
- p: x ∈ {8, 9, 10, 11, 12}
- q: x is a composite number
- r: x is a perfect square
- s: x is a prime number
Evaluating Each Proposition for x = 11
- For x = 11:
- p: True (11 is in the set)
- q: False (11 is prime, not composite)
- r: False (11 is not a perfect square)
- s: True (11 is prime)
Breaking Down the Expression
1. Evaluate p → q:
- This translates to "If p is true, then q is true."
- Since p is true and q is false, p → q is false.
2. Evaluate ¬r:
- Since r is false, ¬r is true.
3. Evaluate ¬s:
- Since s is true, ¬s is false.
4. Evaluate ¬r ∨ ¬s:
- This is true (true ∨ false = true).
5. Combine Results:
- Now substituting back: ¬((p → q) ∧ (¬r ∨ ¬s)) becomes ¬(false ∧ true).
- This simplifies to ¬(false), which is true.
Conclusion
Since the expression is true for x = 11, it satisfies the condition. Hence, the integer x ≥ 2 that satisfies the original expression is:
11
To solve the expression ¬((p → q) ∧ (¬r ∨ ¬s), let's first clarify the propositions:
- p: x ∈ {8, 9, 10, 11, 12}
- q: x is a composite number
- r: x is a perfect square
- s: x is a prime number
Evaluating Each Proposition for x = 11
- For x = 11:
- p: True (11 is in the set)
- q: False (11 is prime, not composite)
- r: False (11 is not a perfect square)
- s: True (11 is prime)
Breaking Down the Expression
1. Evaluate p → q:
- This translates to "If p is true, then q is true."
- Since p is true and q is false, p → q is false.
2. Evaluate ¬r:
- Since r is false, ¬r is true.
3. Evaluate ¬s:
- Since s is true, ¬s is false.
4. Evaluate ¬r ∨ ¬s:
- This is true (true ∨ false = true).
5. Combine Results:
- Now substituting back: ¬((p → q) ∧ (¬r ∨ ¬s)) becomes ¬(false ∧ true).
- This simplifies to ¬(false), which is true.
Conclusion
Since the expression is true for x = 11, it satisfies the condition. Hence, the integer x ≥ 2 that satisfies the original expression is:
11
Full form of DNF is -?- a)Disjoining Normal Form
- b)Disjunctive Normal Form
- c)Divisional Normal Form
- d)Dividend Normal Form
Correct answer is option 'B'. Can you explain this answer?
Full form of DNF is -?
a)
Disjoining Normal Form
b)
Disjunctive Normal Form
c)
Divisional Normal Form
d)
Dividend Normal Form
|
|
Poulomi Patel answered |
Understanding DNF: Disjunctive Normal Form
Disjunctive Normal Form (DNF) is a standard way of structuring logical expressions in Boolean algebra. It is essential in various fields, including computer science, digital logic design, and civil engineering when analyzing logical systems.
Definition of DNF
- DNF is a canonical form of a logical formula.
- It is defined as a disjunction (OR operation) of one or more conjunctions (AND operations) of literals.
Components of DNF
- **Literals**: These are the basic variables or their negations.
- **Conjunctions**: Groups of literals combined using the AND operator.
- **Disjunctions**: The overall expression combines multiple conjunctions using the OR operator.
Characteristics of DNF
- **Clarity**: DNF allows for a clear and systematic representation of logical expressions.
- **Simplicity**: It simplifies the process of evaluating logical statements.
- **Versatility**: DNF can represent any Boolean function, making it a universal format.
Example of DNF
Consider the logical expression:
- A AND B OR C AND D.
This expression is in DNF because it is a sum of products, where:
- (A AND B) and (C AND D) are conjunctions, and they are combined by the OR operator.
Applications in Civil Engineering
In civil engineering, DNF can be used in:
- **Decision Analysis**: Analyzing various project outcomes based on different conditions.
- **Logic Circuits**: Designing digital circuits that require specific logical functions.
In summary, Disjunctive Normal Form is a fundamental concept in logic and engineering disciplines, providing a clear and structured way to handle logical expressions.
Disjunctive Normal Form (DNF) is a standard way of structuring logical expressions in Boolean algebra. It is essential in various fields, including computer science, digital logic design, and civil engineering when analyzing logical systems.
Definition of DNF
- DNF is a canonical form of a logical formula.
- It is defined as a disjunction (OR operation) of one or more conjunctions (AND operations) of literals.
Components of DNF
- **Literals**: These are the basic variables or their negations.
- **Conjunctions**: Groups of literals combined using the AND operator.
- **Disjunctions**: The overall expression combines multiple conjunctions using the OR operator.
Characteristics of DNF
- **Clarity**: DNF allows for a clear and systematic representation of logical expressions.
- **Simplicity**: It simplifies the process of evaluating logical statements.
- **Versatility**: DNF can represent any Boolean function, making it a universal format.
Example of DNF
Consider the logical expression:
- A AND B OR C AND D.
This expression is in DNF because it is a sum of products, where:
- (A AND B) and (C AND D) are conjunctions, and they are combined by the OR operator.
Applications in Civil Engineering
In civil engineering, DNF can be used in:
- **Decision Analysis**: Analyzing various project outcomes based on different conditions.
- **Logic Circuits**: Designing digital circuits that require specific logical functions.
In summary, Disjunctive Normal Form is a fundamental concept in logic and engineering disciplines, providing a clear and structured way to handle logical expressions.
Let Q(x) be the statement “x < 5.” What is the truth value of the quantification ∀xQ(x), having domains as real numbers.- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
Let Q(x) be the statement “x < 5.” What is the truth value of the quantification ∀xQ(x), having domains as real numbers.
a)
True
b)
False
|
|
Sanya Agarwal answered |
Clarification: Q(x) is not true for every real number x, because, for instance, Q(6) is false. That is, x = 6 is a counterexample for the statement ∀xQ(x). This is false.
Given the following two statements:
S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF.
S2: AB → C, D → E, E → C is a minimal cover for the set of functional dependencies AB → C, D → E, AB → E, E → C.
Which one of the following is CORRECT?- a)S1 is TRUE and S2 is FALSE
- b)Both S1 and S2 are TRUE.
- c)S1 is FALSE and S2 is TRUE
- d)Both S1 and S2 are FALSE
Correct answer is option 'A'. Can you explain this answer?
Given the following two statements:
S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF.
S2: AB → C, D → E, E → C is a minimal cover for the set of functional dependencies AB → C, D → E, AB → E, E → C.
Which one of the following is CORRECT?
S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF.
S2: AB → C, D → E, E → C is a minimal cover for the set of functional dependencies AB → C, D → E, AB → E, E → C.
Which one of the following is CORRECT?
a)
S1 is TRUE and S2 is FALSE
b)
Both S1 and S2 are TRUE.
c)
S1 is FALSE and S2 is TRUE
d)
Both S1 and S2 are FALSE
|
Moumita Dasgupta answered |
Explanation:
S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF
- This statement is FALSE.
- A table with two single-valued attributes may not necessarily be in 3NF or BCNF.
- For a table to be in 3NF, it should be in 1NF and 2NF first, and for a table to be in BCNF, it should be in 3NF.
- Therefore, S1 is incorrect.
S2: AB → C, D → E, E → C is a minimal cover for the set of functional dependencies AB → C, D → E, AB → E, E → C
- This statement is TRUE.
- A minimal cover for a set of functional dependencies is a set of dependencies that is both minimal and logically equivalent to the original set.
- In this case, the given set of functional dependencies AB → C, D → E, E → C can be simplified to AB → C, D → E, E → C.
- Therefore, S2 is correct.
Therefore, the correct answer is option 'A' where S1 is FALSE and S2 is TRUE.
S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF
- This statement is FALSE.
- A table with two single-valued attributes may not necessarily be in 3NF or BCNF.
- For a table to be in 3NF, it should be in 1NF and 2NF first, and for a table to be in BCNF, it should be in 3NF.
- Therefore, S1 is incorrect.
S2: AB → C, D → E, E → C is a minimal cover for the set of functional dependencies AB → C, D → E, AB → E, E → C
- This statement is TRUE.
- A minimal cover for a set of functional dependencies is a set of dependencies that is both minimal and logically equivalent to the original set.
- In this case, the given set of functional dependencies AB → C, D → E, E → C can be simplified to AB → C, D → E, E → C.
- Therefore, S2 is correct.
Therefore, the correct answer is option 'A' where S1 is FALSE and S2 is TRUE.
Given a relation schema R(ABCDEFGH) in first normal form. For the set of dependencies
F= { A → B, A → C, CG → H, B → H, G → F}, which dependency is logically implied?
AC → H- a)AC → H
- b)C → H
- c)G → H
- d)A → H
Correct answer is option 'D'. Can you explain this answer?
Given a relation schema R(ABCDEFGH) in first normal form. For the set of dependencies
F= { A → B, A → C, CG → H, B → H, G → F}, which dependency is logically implied?
AC → H
F= { A → B, A → C, CG → H, B → H, G → F}, which dependency is logically implied?
AC → H
a)
AC → H
b)
C → H
c)
G → H
d)
A → H
|
|
Sanya Agarwal answered |
Concept:
If A → B and B → C then A → C is logically implied to A → B and B → C FDs this is nothing but transitivity rule.
If A → B and B → C then A → C is logically implied to A → B and B → C FDs this is nothing but transitivity rule.
Explanation:
A → B
B → H
Then we can say that A->H which is logically implied by above both FDs
So option 4 is the correct answer.
A → B
B → H
Then we can say that A->H which is logically implied by above both FDs
So option 4 is the correct answer.
Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).
Consider the two statements given below.
I. Both Y and Z are in BCNF
II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?- a)Both I and II
- b)I only
- c)II only
- d)Neither I nor II
Correct answer is option 'C'. Can you explain this answer?
Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).
Consider the two statements given below.
I. Both Y and Z are in BCNF
II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?
Consider the two statements given below.
I. Both Y and Z are in BCNF
II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?
a)
Both I and II
b)
I only
c)
II only
d)
Neither I nor II
|
|
Sanya Agarwal answered |
X = (PQRS)
Set of functional dependencies
F = {QR → S, R → P, S → Q}
QR → {Q, R, S, P}
SR → {S, R, P, Q}
QR and SR are keys of Relation schema X
R → P …
part of key (R) → non key(P)
It is partial dependency and hence not in 2nd normal form and therefore not in BCNF (Given)
X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS)
For Y = (PR)
R → P
Y is in BCNF because binary attribute (R is a key).
For Z = (QRS)
QR → S
S → Q
SR → {R, S, Q}
QR → {R, Q, S}
QR and SR are keys of Relation schema X
Z is not in BCNF because
S → Q and S is not Super key.
∴ statement I is incorrect
Dependency Preserving:
R → P is in Y
QR → S in Z
S → Q is in Z
Hence, it is dependency preserving.
Lossless:
Y ∩ Z = R which is key of Y.
Therefore, it is Lossless
Statement II is correct.
Hence, option 3 is the answer
What is logically equivalent to the following statements? "I pass only if you pass" (Note that fail is equivalent to not pass.)- a)You pass only if I pass
- b)If you fail then I fail
- c)If you pass then I pass
- d)You fail if I pass.
Correct answer is option 'B'. Can you explain this answer?
What is logically equivalent to the following statements? "I pass only if you pass" (Note that fail is equivalent to not pass.)
a)
You pass only if I pass
b)
If you fail then I fail
c)
If you pass then I pass
d)
You fail if I pass.
|
Mira Mishra answered |
Without knowing the specific statements, it is not possible to determine what is logically equivalent to them. Please provide the statements you are referring to.
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
Correct answer is option 'C'. Can you explain this answer?
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?
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
|
|
Pallabi Bajaj answered |
Given 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
To find:
Which one of the following statements follows from S1 and S2 as per sound inference rules of logic?
Answer:
The given statements can be represented as follows:
S1: Corrupt ⟹ ¬Elected (If a candidate is known to be corrupt, then he will not be elected)
S2: Kind ⟹ Elected (If a candidate is kind, he will be elected)
We need to find a statement that follows from both S1 and S2.
Explanation:
To find the statement that follows from both S1 and S2, we need to combine the given statements using logical operators.
If we take the contrapositive of S1, we get:
¬¬Elected ⟹ ¬Corrupt (If a candidate is elected, then he is not known to be corrupt)
Combining the contrapositive of S1 with S2, we get:
If a candidate is elected, then he is not known to be corrupt and is kind.
This statement can be written as:
Elected ⟹ (¬Corrupt ∧ Kind)
So, the correct answer is option C:
If a person is kind, he is not known to be corrupt.
This can be further explained by considering the following cases:
- If a person is corrupt, according to S1, he will not be elected. So, he cannot be kind.
- If a person is kind, according to S2, he will be elected. So, he cannot be known to be corrupt.
Hence, option C is the correct answer.
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
To find:
Which one of the following statements follows from S1 and S2 as per sound inference rules of logic?
Answer:
The given statements can be represented as follows:
S1: Corrupt ⟹ ¬Elected (If a candidate is known to be corrupt, then he will not be elected)
S2: Kind ⟹ Elected (If a candidate is kind, he will be elected)
We need to find a statement that follows from both S1 and S2.
Explanation:
To find the statement that follows from both S1 and S2, we need to combine the given statements using logical operators.
If we take the contrapositive of S1, we get:
¬¬Elected ⟹ ¬Corrupt (If a candidate is elected, then he is not known to be corrupt)
Combining the contrapositive of S1 with S2, we get:
If a candidate is elected, then he is not known to be corrupt and is kind.
This statement can be written as:
Elected ⟹ (¬Corrupt ∧ Kind)
So, the correct answer is option C:
If a person is kind, he is not known to be corrupt.
This can be further explained by considering the following cases:
- If a person is corrupt, according to S1, he will not be elected. So, he cannot be kind.
- If a person is kind, according to S2, he will be elected. So, he cannot be known to be corrupt.
Hence, option C is the correct answer.
If p = a number from {8, 9, 10, 11, 12}
q = not a composite number
r = a square number
s = a prime number
then what is the value of ~((p → ~q)∧(~r ∨~S) )- a)8, 9, 10, 11, 12
- b)8, 9, 10
- c)11, 12
- d)11
Correct answer is option 'D'. Can you explain this answer?
If p = a number from {8, 9, 10, 11, 12}
q = not a composite number
r = a square number
s = a prime number
then what is the value of ~((p → ~q)∧(~r ∨~S) )
q = not a composite number
r = a square number
s = a prime number
then what is the value of ~((p → ~q)∧(~r ∨~S) )
a)
8, 9, 10, 11, 12
b)
8, 9, 10
c)
11, 12
d)
11
|
|
Sanya Agarwal answered |
Given that, If p = a number from {8, 9, 10, 11, 12}
q = not a composite number
r = a square number
s = a prime number
~((p→~ q)∧(~r ∨~S) )
¬((p→¬q)∧(¬r∨¬S))
¬(¬p∨¬q)∨¬(¬r∨¬S)
(p∧q)∨(r∧s)
p∧ q = {8, 9, 10, 11, 12} ∧ {11 } = {11}
(r ∧ s) = { a square number }∧{ a prime number}
(r ∧ s) = {9} ∧ {11} = { }
(p∧ q) ∨ (r ∧ s)= {11} ∨ { } = {11}
Hence the correct answer is 11.
q = not a composite number
r = a square number
s = a prime number
~((p→~ q)∧(~r ∨~S) )
¬((p→¬q)∧(¬r∨¬S))
¬(¬p∨¬q)∨¬(¬r∨¬S)
(p∧q)∨(r∧s)
p∧ q = {8, 9, 10, 11, 12} ∧ {11 } = {11}
(r ∧ s) = { a square number }∧{ a prime number}
(r ∧ s) = {9} ∧ {11} = { }
(p∧ q) ∨ (r ∧ s)= {11} ∨ { } = {11}
Hence the correct answer is 11.
Which of the following is correct?
F1 : (p ↔ q)∧(¬ p ↔ q)
F2 : (p∨¬q)∧(¬ p∨q) ∧(¬ p∨¬q)- a)F1 is satisfiable, F2 is valid
- b)F1 is unsatisfiable, F2 is satisfiable
- c)F1 is unsatisfiable, F2 is valid
- d)F1 and F2 both are unsatisfiable
Correct answer is option 'B'. Can you explain this answer?
Which of the following is correct?
F1 : (p ↔ q)∧(¬ p ↔ q)
F2 : (p∨¬q)∧(¬ p∨q) ∧(¬ p∨¬q)
F1 : (p ↔ q)∧(¬ p ↔ q)
F2 : (p∨¬q)∧(¬ p∨q) ∧(¬ p∨¬q)
a)
F1 is satisfiable, F2 is valid
b)
F1 is unsatisfiable, F2 is satisfiable
c)
F1 is unsatisfiable, F2 is valid
d)
F1 and F2 both are unsatisfiable
|
Asha Nambiar answered |
Explanation:
Unsatisfiability of F1:
- F1 can be simplified as (p ↔ q) ∧ (¬ p ↔ q), which can be further simplified as (¬ p ∨ q) ∧ (p ∨ q) (since p ↔ q ≡ (p ∧ q) ∨ (¬ p ∧ ¬ q))
- The simplified form becomes a contradiction as both (¬ p ∨ q) and (p ∨ q) cannot be true at the same time.
- Therefore, F1 is unsatisfiable.
Satisfiability of F2:
- F2 can be simplified as (p ∨ ¬ q) ∧ (¬ p ∨ q) ∧ (¬ p ∨ ¬ q).
- This form is satisfiable as it is possible for at least one of the clauses to be true.
- Therefore, F2 is satisfiable.
Therefore, the correct answer is option 'b) F1 is unsatisfiable, F2 is satisfiable'.
Unsatisfiability of F1:
- F1 can be simplified as (p ↔ q) ∧ (¬ p ↔ q), which can be further simplified as (¬ p ∨ q) ∧ (p ∨ q) (since p ↔ q ≡ (p ∧ q) ∨ (¬ p ∧ ¬ q))
- The simplified form becomes a contradiction as both (¬ p ∨ q) and (p ∨ q) cannot be true at the same time.
- Therefore, F1 is unsatisfiable.
Satisfiability of F2:
- F2 can be simplified as (p ∨ ¬ q) ∧ (¬ p ∨ q) ∧ (¬ p ∨ ¬ q).
- This form is satisfiable as it is possible for at least one of the clauses to be true.
- Therefore, F2 is satisfiable.
Therefore, the correct answer is option 'b) F1 is unsatisfiable, F2 is satisfiable'.
Let P (x) denote the statement “x >7.” Which of these have truth value true?- a)P (0)
- b)P (4)
- c)P (6)
- d)P (9)
Correct answer is option 'D'. Can you explain this answer?
Let P (x) denote the statement “x >7.” Which of these have truth value true?
a)
P (0)
b)
P (4)
c)
P (6)
d)
P (9)
|
|
Prerna Menon answered |
It seems that the statement you mentioned is incomplete. Could you please provide the full statement P(x)?
Consider a relational table R that is in 3NF, but not in BCNF, Which one of the following statements is TRUE?- a)R has a nontrivial functional dependency X → A, where X is not a superkey and A is a prime attribute.
- b)R has a nontrivial functional dependency X → A, where X is not a superkey and A is a non-prime attribute and X is not a proper subset of any key.
- c)R has a nontrivial functional dependency X → A, where X is not a superkey and A is a non-prime attribute and X is a proper subset of some key.
- d)A cell in R holds a set instead of an atomic value.
Correct answer is option 'A'. Can you explain this answer?
Consider a relational table R that is in 3NF, but not in BCNF, Which one of the following statements is TRUE?
a)
R has a nontrivial functional dependency X → A, where X is not a superkey and A is a prime attribute.
b)
R has a nontrivial functional dependency X → A, where X is not a superkey and A is a non-prime attribute and X is not a proper subset of any key.
c)
R has a nontrivial functional dependency X → A, where X is not a superkey and A is a non-prime attribute and X is a proper subset of some key.
d)
A cell in R holds a set instead of an atomic value.
|
Subham Unni answered |
B)R has a candidate key that is a proper subset of some other candidate key.
c)R has a transitive dependency that is not in BCNF.
d)R has a nontrivial functional dependency that is not in BCNF.
The correct answer is c) R has a transitive dependency that is not in BCNF.
3NF guarantees that there are no nontrivial transitive dependencies in the table, but it does not guarantee that there are no nontrivial functional dependencies. BCNF, on the other hand, guarantees that there are no nontrivial functional dependencies. So, if a table is in 3NF but not in BCNF, it must have a transitive dependency that is not in BCNF.
c)R has a transitive dependency that is not in BCNF.
d)R has a nontrivial functional dependency that is not in BCNF.
The correct answer is c) R has a transitive dependency that is not in BCNF.
3NF guarantees that there are no nontrivial transitive dependencies in the table, but it does not guarantee that there are no nontrivial functional dependencies. BCNF, on the other hand, guarantees that there are no nontrivial functional dependencies. So, if a table is in 3NF but not in BCNF, it must have a transitive dependency that is not in BCNF.
Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is ______.- a)in 1NF but not in 2NF
- b)in 2NF but not in 3NF
- c)in 3NF but not in BCNF
- d)in BCNF
Correct answer is option 'A'. Can you explain this answer?
Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is ______.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is ______.
a)
in 1NF but not in 2NF
b)
in 2NF but not in 3NF
c)
in 3NF but not in BCNF
d)
in BCNF
|
|
Sanvi Kapoor answered |
Since attribute D is not a part of any FD, it must be a part of the candidate key.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+ = {ABCDEFGH}
CD+, GD+, and HD+ do not all the attributes. So, they can not be candidate keys.
Candidate keys are AD, BD, ED, and FD.
A → BC, B → CFH, and F → EG are partial dependencies. So, the relation is not in 2NF.
Hence the relation is in 1NF but not in 2NF.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+ = {ABCDEFGH}
CD+, GD+, and HD+ do not all the attributes. So, they can not be candidate keys.
Candidate keys are AD, BD, ED, and FD.
A → BC, B → CFH, and F → EG are partial dependencies. So, the relation is not in 2NF.
Hence the relation is in 1NF but not in 2NF.
The statement, “At least one of your friends is perfect”. Let P (x) be “x is perfect” and let F (x) be “x is your friend” and let the domain be all people.- a)∀x (F (x) → P (x))
- b)∀x (F (x) ∧ P (x))
- c)∃x (F (x) ∧ P (x))
- d)∃x (F (x) → P (x))
Correct answer is option 'C'. Can you explain this answer?
The statement, “At least one of your friends is perfect”. Let P (x) be “x is perfect” and let F (x) be “x is your friend” and let the domain be all people.
a)
∀x (F (x) → P (x))
b)
∀x (F (x) ∧ P (x))
c)
∃x (F (x) ∧ P (x))
d)
∃x (F (x) → P (x))
|
Gauri Roy answered |
Understanding the Statement
The statement "At least one of your friends is perfect" implies that there exists at least one person within the set of your friends who is considered perfect.
Translation of the Statement
- Let P(x) represent "x is perfect."
- Let F(x) represent "x is your friend."
- The domain is all people.
Logical Interpretation of Options
- a) ∀x (F(x) → P(x)): This means every friend is perfect, which is too strong for the statement.
- b) ∀x (F(x) ∧ P(x)): This suggests all friends are perfect, which again does not align with the original statement.
- c) ∃x (F(x) ∧ P(x)): This indicates that there exists at least one person who is both a friend and perfect. This aligns perfectly with the original statement.
- d) ∃x (F(x) → P(x)): This suggests that there exists at least one person such that if they are a friend, then they are perfect. This does not guarantee that any friend is perfect.
Correct Answer Explanation
The correct answer is option c:
- It captures the essence of the statement, confirming that there is at least one friend who is perfect.
- This is a direct match to the meaning of "at least one," making it the most accurate representation of the original statement.
Conclusion
Understanding logical statements often requires careful consideration of the quantifiers and the relationships between the predicates. In this case, option c effectively conveys the intended meaning of the original statement.
The statement "At least one of your friends is perfect" implies that there exists at least one person within the set of your friends who is considered perfect.
Translation of the Statement
- Let P(x) represent "x is perfect."
- Let F(x) represent "x is your friend."
- The domain is all people.
Logical Interpretation of Options
- a) ∀x (F(x) → P(x)): This means every friend is perfect, which is too strong for the statement.
- b) ∀x (F(x) ∧ P(x)): This suggests all friends are perfect, which again does not align with the original statement.
- c) ∃x (F(x) ∧ P(x)): This indicates that there exists at least one person who is both a friend and perfect. This aligns perfectly with the original statement.
- d) ∃x (F(x) → P(x)): This suggests that there exists at least one person such that if they are a friend, then they are perfect. This does not guarantee that any friend is perfect.
Correct Answer Explanation
The correct answer is option c:
- It captures the essence of the statement, confirming that there is at least one friend who is perfect.
- This is a direct match to the meaning of "at least one," making it the most accurate representation of the original statement.
Conclusion
Understanding logical statements often requires careful consideration of the quantifiers and the relationships between the predicates. In this case, option c effectively conveys the intended meaning of the original statement.
Let P(x) denote the statement “x = x + 7.” What is the truth value of the quantification ∃xP(x), where the domain consists of all real numbers?- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
Let P(x) denote the statement “x = x + 7.” What is the truth value of the quantification ∃xP(x), where the domain consists of all real numbers?
a)
True
b)
False
|
Rhea Dasgupta answered |
I apologize, but you did not provide the complete statement. Could you please provide the complete statement for P(x)?
Consider the relation schema: Singer(singerName, songName). What is the highest normal form satisfied by the "Singer" relation schema?- a)1NF
- b)2NF
- c)BCNF
- d)3NF
Correct answer is option 'C'. Can you explain this answer?
Consider the relation schema: Singer(singerName, songName). What is the highest normal form satisfied by the "Singer" relation schema?
a)
1NF
b)
2NF
c)
BCNF
d)
3NF
|
|
Sanya Agarwal answered |
Concept:
Normalization: Normalization is a database design technique that reduces data redundancy and eliminates undesirable characteristics like Insertion, Update and Deletion Anomalies.
Normalization: Normalization is a database design technique that reduces data redundancy and eliminates undesirable characteristics like Insertion, Update and Deletion Anomalies.
1NF (First Normal Form):
- Each table cell should contain a single value.
- Each record needs to be unique.
2NF (Second Normal Form):
- It should be in 1NF.
- Single Column Primary Key that does not functionally dependant on any subset of candidate key relation.
3NF (Third Normal Form):
- It should be in 2NF.
- It has no transitive functional dependencies.
Boyce-Codd Normal Form (BCNF):
- A relation R is in BCNF if R is in Third Normal Form and for every FD, LHS is super key.
- A relation is in BCNF iff in every non-trivial functional dependency X –> Y, X is a super key.
The given relation schema:
Singer(singerName, songName).
Singer(singerName, songName).
- Every Binary Relation ( a Relation with only 2 attributes ) is always in BCNF. If a relation is BCNF then it should be in 3NF, 2NF, 1NF.
Hence Singer(singerName, songName) is Boyce-Codd Normal Form (BCNF).
Hence the correct answer is BCNF.
Hence the correct answer is BCNF.
“If X, then Y unless Z” is represented by which of the following formulae in propositional logic?
(“¬” is negation “^” is conjunction, and “→” is implication)- a) (X ^ ¬ Z) → Y
- b) (X ^ Y) → ¬ Z
- c) (X → (Y ^ ¬ Z)
- d) (X → Y(^ ¬ Z)
Correct answer is option 'A'. Can you explain this answer?
“If X, then Y unless Z” is represented by which of the following formulae in propositional logic?
(“¬” is negation “^” is conjunction, and “→” is implication)
(“¬” is negation “^” is conjunction, and “→” is implication)
a)
(X ^ ¬ Z) → Y
b)
(X ^ Y) → ¬ Z
c)
(X → (Y ^ ¬ Z)
d)
(X → Y(^ ¬ Z)
|
|
Sanya Agarwal answered |
The statement “If X then Y unless Z” means, if Z doesn’t occur, X implies Y i.e. ¬Z→(X→Y), which is equivalent to Z ∨ (X→Y)
(since P→Q ≡ ¬P ∨ Q), which is then equivalent to Z ∨ (¬X ∨ Y). Now we can look into options which one matches with this.
So option (a) is (X∧¬Z)→Y = ¬( (X∧¬Z) ) ∨ Y = (¬X∨Z) ∨ Y, which matches our expression. So option (A) is correct.
P: The population of Hyderabad is more than Delhi.
Q: Last year, the number of months with 31 days are 7.
choose the correct option based on the truth value of the statements?- a)P→Q is True
- b)(P→Q)→ Q is False
- c)Q→ (P→Q )is False
- d)¬(P→Q) is True
Correct answer is option 'A'. Can you explain this answer?
P: The population of Hyderabad is more than Delhi.
Q: Last year, the number of months with 31 days are 7.
choose the correct option based on the truth value of the statements?
Q: Last year, the number of months with 31 days are 7.
choose the correct option based on the truth value of the statements?
a)
P→Q is True
b)
(P→Q)→ Q is False
c)
Q→ (P→Q )is False
d)
¬(P→Q) is True
|
|
Kavya Mehta answered |
The correct option based on the truth value of the statements is:
a) P is true, Q is false.
a) P is true, Q is false.
Consider the following logical inferences.
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.I2: If it rains then the cricket match will not be played.
It did not rain
Inference: The cricket match was played.Which of the following is TRUE?- a)Both I1 and I2 are correct inferences
- b)I1 is correct but I2 is not a correct inference
- c)I1 is not correct but I2 is a correct inference
- d)Both I1 and I2 are not correct inferences
Correct answer is option 'B'. Can you explain this answer?
Consider the following logical inferences.
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Inference: There was no rain.
I2: If it rains then the cricket match will not be played.
It did not rain
Inference: The cricket match was played.
It did not rain
Inference: The cricket match was played.
Which of the following is TRUE?
a)
Both I1 and I2 are correct inferences
b)
I1 is correct but I2 is not a correct inference
c)
I1 is not correct but I2 is a correct inference
d)
Both I1 and I2 are not correct inferences
|
|
Aniket Pillai answered |
Explanation:
I1:
- The given statement is: "If it rains then the cricket match will not be played."
- Inference I1 states: "The cricket match was played."
- From the given statement, we can infer that if it rains, the cricket match will not be played. However, the statement does not provide any information about what happens if it does not rain. Therefore, we cannot definitively conclude that the cricket match was played just because it did not rain. There could be other reasons for the match not being played.
I2:
- The given statement is: "If it rains then the cricket match will not be played."
- Inference I2 states: "It did not rain."
- This inference is correct because the given statement explicitly states that if it rains, the cricket match will not be played. Since it did not rain, we can logically conclude that the cricket match was played.
Therefore, the correct answer is option B. The first inference (I1) is not a correct inference, while the second inference (I2) is a correct inference based on the given statement.
Let domain of m includes all students, P (m) be the statement “m spends more than 2 hours in playing polo”. Express ∀m ¬P (m) quantification in English.- a)A student is there who spends more than 2 hours in playing polo
- b)There is a student who does not spend more than 2 hours in playing polo
- c)All students spends more than 2 hours in playing polo
- d)No student spends more than 2 hours in playing polo
Correct answer is option 'D'. Can you explain this answer?
Let domain of m includes all students, P (m) be the statement “m spends more than 2 hours in playing polo”. Express ∀m ¬P (m) quantification in English.
a)
A student is there who spends more than 2 hours in playing polo
b)
There is a student who does not spend more than 2 hours in playing polo
c)
All students spends more than 2 hours in playing polo
d)
No student spends more than 2 hours in playing polo
|
|
Sanvi Kapoor answered |
Clarification: There is no student who spends more than 2 hours in playing polo.
Determine the truth value of ∀n(n + 1 > n) if the domain consists of all real numbers.- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Determine the truth value of ∀n(n + 1 > n) if the domain consists of all real numbers.
a)
True
b)
False
|
|
Sanya Agarwal answered |
Clarification: There are no elements in the domain for which the statement is false.
Let p, q, and r be propositions and the expression (p → q) → r be a contradiction. Then the expression (r → p) → q is- a)A tautology
- b)A contradiction.
- c)Always TRUE when p is FALSE
- d)Always TRUE when q is TRUE
Correct answer is option 'D'. Can you explain this answer?
Let p, q, and r be propositions and the expression (p → q) → r be a contradiction. Then the expression (r → p) → q is
a)
A tautology
b)
A contradiction.
c)
Always TRUE when p is FALSE
d)
Always TRUE when q is TRUE
|
|
Sanya Agarwal answered |
(p → q) → r is a contradiction which is possible only when r is false and (p → q) is true.
Now, from here we can clearly say that option 4 is correct as (r → p) → q means ¬ (r → p) ∨ q.
Since r is false, (r → p) is true and ¬ (r → p) becomes false.
So, it becomes (false ∨ q). Now it totally depends on q. Whenever q is true, this value will always be true.
Alternate Method:
Let X ≡ p → q, Y ≡ (p → q) → r ≡ X → r,
Z ≡ r → p and W ≡ (r → p) → q ≡ Z → p
Using Truth table
So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.
Now, from here we can clearly say that option 4 is correct as (r → p) → q means ¬ (r → p) ∨ q.
Since r is false, (r → p) is true and ¬ (r → p) becomes false.
So, it becomes (false ∨ q). Now it totally depends on q. Whenever q is true, this value will always be true.
Alternate Method:
Let X ≡ p → q, Y ≡ (p → q) → r ≡ X → r,
Z ≡ r → p and W ≡ (r → p) → q ≡ Z → p
Using Truth table
So, from truth table also, it is clear that (r → p) → q is always true when q is true, and Y is false.
Let R (x) denote the statement “x > 2.” What is the truth value of the quantification ∃xR(x), having domain as real numbers?- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Let R (x) denote the statement “x > 2.” What is the truth value of the quantification ∃xR(x), having domain as real numbers?
a)
True
b)
False
|
|
Sanya Agarwal answered |
Clarification: Because “x > 2” is sometimes true—for instance, when x = 3–the existential quantification of R(x), which is ∃xR(x), is true.
How many types of normal forms are there to which reduction can be performed?- a)2
- b)3
- c)4
- d)5
Correct answer is option 'A'. Can you explain this answer?
How many types of normal forms are there to which reduction can be performed?
a)
2
b)
3
c)
4
d)
5
|
|
Sanvi Kapoor answered |
There are 2 types of normal forms in which reduction can be performed.
Determine the truth value of statement ∃n (4n = 3n) if the domain consists of all integers.- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
Determine the truth value of statement ∃n (4n = 3n) if the domain consists of all integers.
a)
True
b)
False
|
|
Sanvi Kapoor answered |
Clarification: For n = 0, 4n = 3n hence, it is true.
The statement,” Every comedian is funny” where C(x) is “x is a comedian” and F (x) is “x is funny” and the domain consists of all people.- a)∃x(C(x) ∧ F (x))
- b)∀x(C(x) ∧ F (x))
- c)∃x(C(x) → F (x))
- d)∀x(C(x) → F (x))
Correct answer is option 'D'. Can you explain this answer?
The statement,” Every comedian is funny” where C(x) is “x is a comedian” and F (x) is “x is funny” and the domain consists of all people.
a)
∃x(C(x) ∧ F (x))
b)
∀x(C(x) ∧ F (x))
c)
∃x(C(x) → F (x))
d)
∀x(C(x) → F (x))
|
|
Sanya Agarwal answered |
Clarification: For every person x, if comedian then x is funny.
”Everyone wants to learn cosmology.” This argument may be true for which domains?- a)All students in your cosmology class
- b)All the cosmology learning students in the world
- c)Both of the mentioned
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
”Everyone wants to learn cosmology.” This argument may be true for which domains?
a)
All students in your cosmology class
b)
All the cosmology learning students in the world
c)
Both of the mentioned
d)
None of the mentioned
|
|
Sanvi Kapoor answered |
Clarification: Domain may be limited to your class or may be whole world both are good as it satisfies universal quantifier.
If the truth table cannot be created, we consider the ____ to normal forms as an alternative?- a)Addition
- b)Subtraction
- c)Reduction
- d)Division
Correct answer is option 'C'. Can you explain this answer?
If the truth table cannot be created, we consider the ____ to normal forms as an alternative?
a)
Addition
b)
Subtraction
c)
Reduction
d)
Division
|
|
Sanya Agarwal answered |
If the truth table cannot be created, we consider the reduction to normal forms as an alternative.
Chapter doubts & questions for Propositional Logic - Engineering Mathematics for Computer Science Engineering 2025 is part of Computer Science Engineering (CSE) exam preparation. The chapters have been prepared according to the Computer Science Engineering (CSE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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