Question

Let $a : \sim(p \wedge \sim r) \vee (\sim q \vee s)$ and $b : (p \vee s) \leftrightarrow (q \wedge r)$. If the truth values of $p$ and $q$ are true and that of $r$ and $s$ are false, then the truth values of $a$ and $b$ are respectively

• A. T, F
• B. T, T
• C. F, F
• D. F, T

Hint:

To speed up evaluation, look for shortcuts: in statement $a$, the second block $(\sim q \vee s) \equiv (F \vee F) \equiv F$. This means the final value of $a$ depends entirely on the first block. Evaluating $\sim(T \wedge T) \equiv F$ gives $F \vee F = F$, which instantly eliminates options (A) and (B)!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Evaluate truth values of compound statements a and b given p≡T, q≡T, r≡F, s≡F.

Step 2: Key Formula or Approach:
Substitute truth values and evaluate using standard truth tables for ∧, ∨, ∼, and ↔ step-by-step.

Step 3: Detailed Explanation:
a: ∼(T∧∼F) ∨ (∼T∨F) = ∼(T∧T) ∨ (F∨F) = ∼T ∨ F = F ∨ F = F. b: (T∨F) ↔ (T∧F) = T ↔ F = F. Both are F, giving (F, F).

Step 4: Final Answer:
Truth values are F, F, matching option (C).