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GATE | GATE-CS-2003 | Question 72

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The following resolution rule is used in logic programming.

Derive clause (P ∨ Q) from clauses (P ∨ R), (Q ∨ ¬R) 

Which of the following statements related to this rule is FALSE?

(A)

((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is logically valid

(B)

(P ∨ Q) ⇒ ((P ∨ R)) ∧ (Q ∨ ¬R)) is logically valid

(C)

(P ∨ Q) is satisfiable if and only if (P ∨ R) ∨ (Q ∨ ¬R) is satisfiable

(D)

(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable



Answer: (B)

Explanation:

Option A: According to the resolution principle, (P+R)(Q+R’) -> (P+Q), hence this is correct.

Option B: (P+Q) -> (P+R)(Q+R’) = P’Q’ + PQ + P’R’ + QR. This is the converse of the Resolution Principle and the converse function does not work in Resolution Principle.
Hence, this is false, and therefore Option B is correct.

Option C: Any formula can be satisfiable only if at least one true exists in its truth table. 
Consider here:
α = P + Q and β = (P + R)(Q + R’)
For α to be satisfiable, we must have P = 1, and Q = 1, then in that case β is also satisfiable.
For β to be satisfiable, we must have P = 1, Q = 0, and R = 0, then in that case, α is also satisfiable.
Therefore, the statement in option C is also valid.

Option D: Consider here,
Let us take α = (P+Q) as false and β: both P and Q are unsatisfiable.
α is true when only when P = false and Q = false, otherwise false in every case.
β is true only when P = false and Q = false and otherwise false in every case.
This implies that α ≣ β. 
Hence, the statement in Option D is also correct.

Conclusion: As the question is asking which option is not correct, Hence Option B is the correct answer.
 


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Last Updated : 21 Aug, 2018
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