1. **Problem 1:** Use De Morgan’s Laws to simplify and describe the solution set for: $$\sim(x < -2 \text{ or } x \geq 5)$$ **Step 1:** Recall De Morgan’s Laws: $$\sim (A \text{ or } B) = \sim A \text{ and } \sim B$$ **Step 2:** Apply the law: $$\sim(x < -2 \text{ or } x \geq 5) = \sim(x < -2) \text{ and } \sim(x \geq 5)$$ **Step 3:** Negate each inequality: $$\sim(x < -2) = x \geq -2$$ $$\sim(x \geq 5) = x < 5$$ **Step 4:** Combine: $$x \geq -2 \text{ and } x < 5$$ **Solution set:** All $x$ such that $-2 \leq x < 5$. 2. **Problem 2:** Form a compound statement that represents a contradiction using: $p$: "The shop is open." $q$: "The shop is closed." **Step 1:** A contradiction is a statement that is always false. **Step 2:** The compound statement: $$p \text{ and } q$$ because the shop cannot be both open and closed simultaneously. **Step 3:** Truth table: | $p$ | $q$ | $p \land q$ | |---|---|---| | T | T | F | | T | F | F | | F | T | F | | F | F | F | Since $p \land q$ is always false, it is a contradiction. 3. **Problem 3:** Conditional statement about discounts. **Step 1:** Symbolic form: Let $c$: "Customer buys more than 10 items." Let $d$: "Customer gets a discount." Conditional statement: $$c \to d$$ **Step 2:** Inverse: "If a customer does not buy more than 10 items, then they will not get a discount." Symbolically: $$\sim c \to \sim d$$ **Step 3:** Converse: "If a customer gets a discount, then they bought more than 10 items." Symbolically: $$d \to c$$ 4. **Problem 4:** Security system logic. (a) The alarm rings if the door or window (or both) is open. This is an OR gate. (b) Boolean expression: Let $A$: door is open. Let $B$: window is open. Alarm $= A \lor B$ Truth table: | $A$ | $B$ | $A \lor B$ | |---|---|---| | 0 | 0 | 0 | | 0 | 1 | 1 | | 1 | 0 | 1 | | 1 | 1 | 1 | **Final answers:** Problem 1 solution set: $-2 \leq x < 5$ Problem 2 contradiction: $p \land q$ Problem 3 symbolic: $c \to d$, inverse: $\sim c \to \sim d$, converse: $d \to c$ Problem 4 logic gate: OR, Boolean expression: $A \lor B$