Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > Which of the following is correct?F1: (p &har...
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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
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Which of the following is correct?F1: (p ↔ q)∧(¬ p ↔...
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'.
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Community Answer
Which of the following is correct?F1: (p ↔ q)∧(¬ p ↔...
Satisfiable: A compound proposition that is not a contradiction is said to be satisfiable.
Tautology (Valid): A compound proposition that is always true is called tautology (valid).
Contradiction: A compound proposition that is always false is called a contradiction.
F1 : (p ↔ q)∧(¬ p ↔ q)
We know that ¬ (p ↔ q)= (¬ p ↔ q)
So, if (p ↔ q) is assumed of A.
Then A∧¬A = 0 or False
Means (p ↔ q)∧(¬ p ↔ q) is unsatisfiable.
F2 : (p∨¬q)∧(¬ p∨q) ∧(¬ p∨¬q)
= (p+q¯)(p¯+q)(p¯+q¯)
= (p+q¯)(p¯+qq¯)
= (p+q¯)(p¯)
= p¯q¯
Which is not valid but satisfiable.
So, F1 is unsatisfiable but F2 is satisfiable.
Hence the correct answer is F1 is unsatisfiable, F2 is satisfiable.
Tautology (Valid): A compound proposition that is always true is called tautology (valid).
Contradiction: A compound proposition that is always false is called a contradiction.
F1 : (p ↔ q)∧(¬ p ↔ q)
We know that ¬ (p ↔ q)= (¬ p ↔ q)
So, if (p ↔ q) is assumed of A.
Then A∧¬A = 0 or False
Means (p ↔ q)∧(¬ p ↔ q) is unsatisfiable.
F2 : (p∨¬q)∧(¬ p∨q) ∧(¬ p∨¬q)
= (p+q¯)(p¯+q)(p¯+q¯)
= (p+q¯)(p¯+qq¯)
= (p+q¯)(p¯)
= p¯q¯
Which is not valid but satisfiable.
So, F1 is unsatisfiable but F2 is satisfiable.
Hence the correct answer is F1 is unsatisfiable, F2 is satisfiable.
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Which of the following is correct?F1: (p ↔ q)∧(¬ p ↔ q)F2: (p∨¬q)∧(¬ p∨q)∧(¬ p∨¬q)a)F1is satisfiable, F2is validb)F1 is unsatisfiable, F2 is satisfiablec)F1 is unsatisfiable, F2 is validd)F1 and F2 both are unsatisfiableCorrect answer is option 'B'. Can you explain this answer?
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