The following resolution rule is used in logic programming. Derive clause (P ∨…

Answer and summary:

The statement that is FALSE is: (P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) is logically valid.

• Why the resolution direction ((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is valid:

  Assume both (P ∨ R) and (Q ∨ ¬R) hold. If P holds then P ∨ Q holds. If P does not hold then R must hold. If Q holds then P ∨ Q holds. If Q does not hold then ¬R must hold. R and ¬R cannot both be true, so it is impossible that both P and Q are false under these premises; thus at least one of P or Q is true, so P ∨ Q follows.

• Why the implication (P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) fails (counterexample):

  Take an assignment with P = true, Q = false, R = true. Then P ∨ Q is true, but (P ∨ R) is true and (Q ∨ ¬R) is false, so their conjunction is false. Therefore the implication does not hold in general.

• About satisfiability of (P ∨ Q) and (P ∨ R) ∨ (Q ∨ ¬R):

  The expression (P ∨ R) ∨ (Q ∨ ¬R) simplifies to P ∨ Q ∨ (R ∨ ¬R). Since (R ∨ ¬R) is a tautology, the whole formula is always true and therefore satisfiable. P ∨ Q is satisfiable as well (for example by setting P = true). Hence both formulas are satisfiable; note the difference between being satisfiable and being a tautology.

• About (P ∨ Q) ⇒ FALSE and unsatisfiability:

  P ∨ Q is unsatisfiable exactly when both P and Q are unsatisfiable (each false in every assignment). Thus (P ∨ Q) ⇒ FALSE holds if and only if both P and Q are unsatisfiable.

Conclusion: The only false statement among the choices is the claim that (P ∨ Q) implies the conjunction (P ∨ R) ∧ (Q ∨ ¬R); all other listed statements are correct under the usual propositional interpretation.