Correct Option
The problem provides two initial statements:
- If P is true, then both Q and S are true (symbolically: P → (Q ∧ S)).
- If R and S are true, then T is false (symbolically: (R ∧ S) → ¬T).
Let's evaluate Conclusion 1: "If T is true, then at least one of P and R must be false."
- From statement 2: (R ∧ S) → ¬T.
- The contrapositive of this statement is T → ¬(R ∧ S).
- By De Morgan's laws, ¬(R ∧ S) is equivalent to (¬R ∨ ¬S).
- Therefore, if T is true, it implies (¬R ∨ ¬S), meaning either R is false or S is false (or both).
- Now, we consider the implications of (¬R ∨ ¬S) on P and R:
- If S is false (¬S is true): From statement 1 (P → (Q ∧ S)), if P were true, S would necessarily be true. Since S is false, P cannot be true. Thus, P must be false (¬P). If P is false, then the condition "at least one of P and R must be false" (¬P ∨ ¬R) is satisfied.
- If R is false (¬R is true): In this case, the condition "at least one of P and R must be false" (¬P ∨ ¬R) is directly satisfied because R is false.
- Since both possibilities arising from T being true (¬R ∨ ¬S) lead to "at least one of P and R must be false," Conclusion 1 is valid.
Incorrect Options
Let's evaluate Conclusion 2: "If Q is true, then P is true."
- Statement 1 is P → (Q ∧ S), which logically implies P → Q.
- Conclusion 2 proposes Q → P. This is the converse of P → Q.
- The converse of a conditional statement is not necessarily true. The truth of P → Q does not guarantee the truth of Q → P.
- Consider a scenario where P is false, Q is true, and S is true. In this case, statement 1 (If P is true, then both Q and S are true) holds true because the premise 'P is true' is false, making the entire implication true regardless of Q and S. However, Conclusion 2 (If Q is true, then P is true) would be false in this scenario, as Q is true but P is false.
- Therefore, we cannot definitively conclude that if Q is true, then P is true. Conclusion 2 is not valid.
Based on the analysis, only Conclusion 1 is valid. Thus, option 1 is the correct answer.