Consider the first-order logic sentence \(F:\forall x(\exists yR(x,y))\).…

Key insight: the premise is ∀x ∃y R(x,y).

• Why ∃y(∃x R(x,y)) is implied: because the domain is non-empty, pick any element a. From ∀x ∃y R(x,y) we have ∃y R(a,y), so there exists at least one pair (x,y) with R(x,y). Existential quantifiers can be reordered, so ∃y(∃x R(x,y)) holds.

• Why ¬∃x(∀y ¬R(x,y)) is implied: this sentence is logically equivalent to ∀x ∃y R(x,y). Indeed, ¬∃x(∀y ¬R(x,y)) ⇔ ∀x ¬(∀y ¬R(x,y)) ⇔ ∀x ∃y R(x,y). So it follows directly.

• Why the other two are not implied: neither requires exactly what the premise gives. ∃y(∀x R(x,y)) demands one y that works for all x, and ∀y(∃x R(x,y)) demands every y has some x mapping to it. Both can fail while ∀x ∃y R(x,y) holds (e.g., different x's may require different y's, or all x's may map to a single y leaving some y without preimage).