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
I am able to prove a and d are true. For proving $b$ as false$:- P=1, Q=0, R=1$ Now $LHS$ becomes true,but $RHS$ is $0$, hence invalid implication.
But i can also prove c is not valid as:- $(P ∨ Q)$ is satisfiable $\iff (P ∨ R) ∧ (Q ∨ ¬R)$ is satisfiable
Here if I take, $P=1,\,Q=0,\,R=1$ then LHS becomes true,but RHS is false. So it is also false.
But my book has given b as the answer.Can someone tell where i am wrong in proving that c is also false?