Question 2

Given $F: \forall x (\exists y R(x, y))$. This asserts that for every element $x$ in the domain, there is at least one element $y$ such that the relation $R(x, y)$ holds.

Statement I: $\exists y (\exists x R(x, y))$. Since the logical domain is non-empty, we can pick any element $x_0$. By sentence $F$, there exists some $y_0$ such that $R(x_0, y_0)$ holds. Therefore, there exists $y$ and exists $x$ such that $R(x, y)$, so statement I is necessarily true and implied by $F$.

Statement II: $\exists y (\forall x R(x, y))$. This asserts that there is a single element $y$ that is related to every element $x$. In general, $\forall x \exists y$ does not imply $\exists y \forall x$. For example, over the set of integers $\mathbb{Z}$ with $R(x, y): y > x$, for every integer $x$ there is a larger integer $y$, but there is no single integer $y$ that is greater than all integers. Hence, II is not implied.

Statement III: $\forall y (\exists x R(x, y))$. This asserts that every element $y$ is related to by some element $x$. Sentence $F$ does not guarantee that every $y$ in the domain is chosen by some $x$. Hence, III is not implied.

Statement IV: $\neg \exists x (\forall y \neg R(x, y))$. Applying quantifier duality (De Morgan's laws for quantifiers):

This is identical to sentence $F$. Thus, statement IV is logically equivalent to $F$, and therefore implied by $F$.

Thus, statements I and IV are implied by $F$.