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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

@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.

edited
IN gateoverflow book
In option B there is a mistake in brackets

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.

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.

If and only if means bi-implication.

In solution first it says

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

and then concludes option B is contingency

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.

@Abhrajyoti00

thank you.

What does if and only if means here:

(Here what I understood finally):→

whenever (P v R) and ( Q V R`) is true then  (P v Q) must be true but vise versa might not be true.

@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 $(<->)$

@Abhrajyoti00

It’s mean if  on any one particular truth table value if both side satisfy then it is satisfiable.

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

B is false.

when P is true , Q false and R is true.

True -> False = false.

valid means it should be true for every value of truth assignment.
in option c can we apply resolution principle?? I mean here arguments are satisfiable and not valid. Isnt that resolution can be applied only when we have valid arguments??
In option c:-

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

If P=1,Q=0,R=1

LHS is 1 and RHS is 0

So how this is true?

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.

in (c) it is given that " (P ∨ Q) is satisfiable IF AND ONLY IF (P ∨ R) ∧ (Q ∨ ¬R) is satisfiable. "

but here they are individually satisfiable..how come this statement is true?
In option D ,if both p and q are unsatisfiable,then LHS becomes false,but RHS must be true,but here they have given RHS as false,how is this valid?