@Verma Ashish I think your explanation will be correct if the option was **(P∨Q) if and only if (P∨R)∧(Q∨¬R) is satisfiable **but the option is **(P∨Q) is satisfiable if and only if (P∨R)∧(Q∨¬R) is satisfiable.**

Please correct me if I am wrong.

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45 votes

The following resolution rule is used in logic programming.

Derive clause $(P \vee Q)$ from clauses $(P\vee R),(Q \vee ¬R)$

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

- $((P ∨ R)∧(Q ∨ ¬R))⇒(P ∨ Q)$ is logically valid
- $(P ∨ Q)⇒((P ∨ R)∧(Q ∨ ¬R))$ is logically valid
- $(P ∨ Q)$ is satisfiable if and only if $(P ∨ R)∧(Q ∨ ¬R)$ is satisfiable
- $(P ∨ Q)⇒ \text{FALSE}$ if and only if both $P$ and $Q$ are unsatisfiable

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Option A: $(P + R)(Q + R’) \rightarrow (P + Q)$ is **valid**, since this is the resolution principle

**Option B:** $(P + Q) \rightarrow (P + R)(Q + R') = P’Q’ + PQ + PR’ + QR$ is **NOT valid** (resolution principle does not work in the converse case)

Option C:

Let, $\alpha: P + Q$ and $\beta: (P + R)(Q + R')$

**A formula is satisfiable if there exists atleast one true in its truth table.**

$A:$ If, $\alpha$ is satisfiable $(P=1, Q=1)$, then $\beta$ is also satisfiable.

$B:$ If, $\beta$ is satisfiable $(P=1, Q=0, R=0)$, then $\alpha$ is also satisfiable.

i.e., the statement is **valid.**

Option D:

Let, $\alpha : (P + Q) \rightarrow false$

and, $\beta :$ both $P$ and $Q$ are unsatisfiable.

$\alpha$ is true when $P=false$ and $Q=false$, and false otherwise

$\beta:$ is true when $P=false$ and $Q=false$, and false otherwise

$\therefore \alpha \equiv \beta$

i.e., the statement is **valid.**

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23 votes

Best answer

Taking option $(A)$

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

$((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q)$

$\equiv \neg((P \vee R) \wedge (Q \vee ¬R)) \vee (P \vee Q)$

$\equiv ((\neg P\wedge \neg R) \vee (\neg Q \wedge R)) \vee (P \vee Q)$

$\equiv P \vee \neg R \vee Q \vee R$

$\equiv 1 \vee P \vee Q$

$\equiv 1.$

So, Tautology (Logically VALID). True

Option $B$

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

$(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))$

$\equiv \neg(P \vee Q) \vee ((P\vee R) \wedge (Q \vee ¬R))$

$\equiv (\neg P\wedge \neg Q) \vee (P\wedge Q) \vee (P \wedge \neg R) \vee (Q \wedge R)$

which is a contingency $($can be TRUE or FALSE depending on values of $P,Q$ and $R)$ and hence not logically valid.

Answer B.

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

$((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q)$

$\equiv \neg((P \vee R) \wedge (Q \vee ¬R)) \vee (P \vee Q)$

$\equiv ((\neg P\wedge \neg R) \vee (\neg Q \wedge R)) \vee (P \vee Q)$

$\equiv P \vee \neg R \vee Q \vee R$

$\equiv 1 \vee P \vee Q$

$\equiv 1.$

So, Tautology (Logically VALID). True

Option $B$

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

$(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))$

$\equiv \neg(P \vee Q) \vee ((P\vee R) \wedge (Q \vee ¬R))$

$\equiv (\neg P\wedge \neg Q) \vee (P\wedge Q) \vee (P \wedge \neg R) \vee (Q \wedge R)$

which is a contingency $($can be TRUE or FALSE depending on values of $P,Q$ and $R)$ and hence not logically valid.

Answer B.

20 votes

Let us Consider the options one by one :

(a) **((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) **This option is clearly true as this is well known rule for Resolution.

Consider option (c)

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

which means

(P ⋁ Q) ↔ (P V R) $\wedge$ (Q V ~R)

So it means (P V Q) is true if and only is (P V R) ^ (Q V ~R) is true

Let's suppose assume

(P V R) $\wedge$ (Q V ~R) is true

So Either of R or ~R can be true at a time.

Say, R is true so

(P V R) term becomes true

Now for the (P V R) $\wedge$ (Q V ~R) to become true

Q must be true

and If Q becomes true, then the LHS of bi-implication i.e. P V Q will become true hence making option (C) as true.

Now consider option (d)

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

Yes this is very obvious as when both P and Q are FALSE, P V Q will result in FALSE, making the above implication true.

Being P and Q unsatisfiable means they are FALSE or a Contradiction.

So, this option is also TRUE.

Remaining option is Option (b)

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

For the above to be logically valid, it must be a tautology.

Let's check

(P+Q) -> ((P+R).(Q +~R))

~(P+Q) + (PQ + P~R + QR)

~P~Q + PQ + P~R + QR

and this is definitely not a tautology.

Hence, **answer option (B)**

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@GateOverflow04 Iff (if and only if) means both side it must be true. LHS → RHS and RHS → LHS both.

It’s also called Bi-implication $(<->)$

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7 votes

answer = **option B**

option A : follows from the correct definition of Resolution principle, hence this option is true.**option B** : does not follow from resolution principle, + it's not logically valid. It can also be verified using truth table. Hence, this option is false.

option C : this entire statement becomes true as both $P \vee Q$ and $(P \vee R) \wedge (Q \vee \neg R)$ are individually satisfiable.

option D : is always true as LHS remains false.

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