1. **Problem statement:** Determine the truth value of each proposition from part II of Exercise 1. 2. **Recall:** The propositions involve quantifiers over real numbers $\mathbb{R}$ and inequalities. 3. **Proposition 1:** $\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : 2x + y > 0$. - For any fixed $x$, choose $y = -2x + 1$. - Then $2x + y = 2x + (-2x + 1) = 1 > 0$. - So proposition 1 is **true**. 4. **Proposition 2:** $\exists x \in \mathbb{R}, \forall y \in \mathbb{R} : 2x + y > 0$. - For fixed $x$, consider $y = -2x - 1$. - Then $2x + y = 2x + (-2x - 1) = -1 \not> 0$. - So no such $x$ exists. - Proposition 2 is **false**. 5. **Proposition 3:** $\forall x \in \mathbb{R}, \forall y \in \mathbb{R} : 2x + y > 0$. - Counterexample: $x=0, y=-1$ gives $2(0) + (-1) = -1 \not> 0$. - Proposition 3 is **false**. 6. **Proposition 4:** $\exists x \in \mathbb{R}, \exists y \in \mathbb{R} : 2x + y > 0$. - Choose $x=0, y=1$. - Then $2(0) + 1 = 1 > 0$. - Proposition 4 is **true**. 7. **Proposition 5:** $\exists x \in \mathbb{R}, \forall y \in \mathbb{R} : y^2 > x$. - For fixed $x$, if $x < 0$, then $y^2 \geq 0 > x$ for all $y$. - So choose $x = -1$. - Proposition 5 is **true**. 8. **Proposition 6:** $\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : (2x + y > 0 \lor 2x + y = 0)$. - For any $x$, choose $y = -2x$. - Then $2x + y = 0$. - So proposition 6 is **true**. 9. **Proposition 7:** $\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : (2x + y > 0 \land 2x + y = 0)$. - The conjunction $A \land \neg A$ is always false. - So no $y$ satisfies both. - Proposition 7 is **false**. **Final answers:** 1. True 2. False 3. False 4. True 5. True 6. True 7. False