edited by
13,931 views
57 votes
57 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?

  1. $((P ∨ R)∧(Q ∨ ¬R))⇒(P ∨ Q)$ is logically valid
  2. $(P ∨ Q)⇒((P ∨ R)∧(Q ∨ ¬R))$ is logically valid
  3. $(P ∨ Q)$ is satisfiable if and only if $(P ∨ R)∧(Q ∨ ¬R)$ is satisfiable
  4. $(P ∨ Q)⇒ \text{FALSE}$ if and only if both $P$ and $Q$ are unsatisfiable
edited by

6 Answers

Best answer
27 votes
27 votes
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.
edited by
23 votes
23 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)

edited by
7 votes
7 votes
Option B is false...
   ((P ∨ R)) ∧ (Q ∨ ¬R))
=  (P∧ ¬R) ∨ (Q ∧ R)
(P∨Q) doesn't imply (P∧ ¬R) ∨ (Q ∧ R)
edited by
7 votes
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.

Answer:

Related questions