1. **Problem Statement:** Given the logical expression $q \lor \sim[(\sim p \to q) \land \sim p]$, we need to analyze it. 2. **Understanding the expression:** - $\sim$ means NOT. - $\land$ means AND. - $\lor$ means OR. - $\to$ means implication. 3. **Rewrite implication:** Recall that $a \to b$ is equivalent to $\sim a \lor b$. So, $$\sim p \to q \equiv \sim (\sim p) \lor q = p \lor q$$ 4. **Substitute back:** The expression becomes: $$q \lor \sim[(p \lor q) \land \sim p]$$ 5. **Construct the truth table:** We have variables $p$, $q$. Calculate intermediate columns: - $\sim p$ - $p \lor q$ - $(p \lor q) \land \sim p$ - $\sim[(p \lor q) \land \sim p]$ - Final expression: $q \lor \sim[(p \lor q) \land \sim p]$ | $p$ | $q$ | $\sim p$ | $p \lor q$ | $(p \lor q) \land \sim p$ | $\sim[(p \lor q) \land \sim p]$ | $q \lor \sim[(p \lor q) \land \sim p]$ | |-----|-----|----------|------------|----------------------------|---------------------------------|------------------------------------| | 0 | 0 | 1 | 0 | 0 | 1 | 0 \lor 1 = 1 | | 0 | 1 | 1 | 1 | 1 | 0 | 1 \lor 0 = 1 | | 1 | 0 | 0 | 1 | 0 | 1 | 0 \lor 1 = 1 | | 1 | 1 | 0 | 1 | 0 | 1 | 1 \lor 1 = 1 | 6. **Interpretation:** The final column is always 1 (true) for all combinations of $p$ and $q$. 7. **Conclusion:** The compound statement is a **tautology** because it is true for all truth values of $p$ and $q$. --- **Note:** Part (a) asks for a circuit diagram which cannot be represented in this text format.