in Mathematical Logic edited by
44 votes
44 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
in Mathematical Logic edited by


@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 by
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.


6 Answers

2 votes
2 votes

please sir correct me if i am wrong any were


Ur handwriting .... very hard to understand ...
Sorry @Puja mishra
It is understandable!
can I please know why option B is not valid ?
2 votes
2 votes

Let us take each option one by one and examine them,
(Representing in a Boolean Logic for clear understanding)

Taking Option (A),

((P ∨ R) ∧ (Q ∨ ~R)) ⇒ (P ∨ Q)
≡ ((P + R)(Q + R’))’ + (P + Q)
≡ (PQ + PR’ + QR)’ + P + Q
≡ (P’ + Q’)(P’ + R)(Q’ + R’) + P + Q
≡ (P’ + P’R + P’Q’ + Q’R)(Q’ + R’) + P + Q
≡ (P’ + Q’R)(Q’ + R’) + P + Q
≡ P’Q’ + P’R’ + Q’R + P + Q
≡ P + (Q’ + Q) + (R’ + R)
≡ 1

So, This statement is Tautology i.e., Logically Valid.

Taking Option (B),

(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ~R)
(P + Q)’ + ((P + R).(Q + R’)
≡ P’Q’ + PQ + PR’ + QR

This statement is neither True nor False (can be True or False depending on the values of P, Q and R). So, this is Contingency.
Hence, definately not a Tautology i.e., Logically Invalid.

Taking Option (C),

(P ∨ Q) is satisfiable if and only if (P ∨ R) ∨ (Q ∨ ~R) is Satisfiable
this statement means: (P ∨ Q) ⇔ (P ∨ R) ∨ (Q ∨ ~R)
Now, here both R and R’ can’t be True at the same time. Either one of them may be True. 
Let’s assume, R is True. So, R’ must be False.
So, (P ∨ R) is True. 
Now, (Q ∨ R’) to become True, Q must be True.
and If Q becomes true, then the L.H.S of the bi-implication i.e., (P ∨ Q) will become true and hence option (C) is True.

Taking Option (D),

(P ∨ Q) ⇒ FALSE if and only if both P and Q are Unsatisfiable
P and Q are unsatisfiable means both P and Q are False i.e., Contradiction. So, it is very much clear that if both P = 0 and Q = 0, then (P ∨ Q) = 0 i.e., Contradiction which is Unsatisfiable.
Hence option (D) is True.

Finally, Answer is Option (B).


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