can we solve this type of question is terms T&F if yes can you provided a detail solution for this. If not th…

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• KG AI MentorKG AI Mentor30 Dec 2025

  Quick AnswerP and S are valid; Q and R are not. Key StepsP: To falsify ¬s → q, you’d need s = 0 and q = 0. Then r → s forces r = 0, and p ∨ r forces p = 1, which makes ¬p ∨ q false (since q = 0). Contradiction → P is valid. Q: Take p = 0, q = 1, r = 1. Then ¬p ∧ q is true, q → (p → r) is true (0 → 1 is true), so the premise is true but ¬r is false. Hence Q is not valid. R: Take r = 0, q = 0, any p. Then (q ∧ r) → p is true (false antecedent), and ¬q ∨ p is true; thus A is true while r = 0, so the implication fails. R is not valid. S: If LHS is true, p is true and p → r is true, so r is true. Then q ∨ ¬r reduces to q ∨ false, so q must be true. Hence S is valid. — KG AI Mentor 🤖