Show $\forall x \exists y F(x,y)$ does not imply $\exists y \forall x F(x,y)$ Show:
$\exists y \forall x R(x,y) \rightarrow \forall x \exists y R(x,y)$
$\forall x \exists y F(x,y)$ does not imply $\exists y \forall x F(x,y)$
How do proofs of this nature usually work? When I try to prove the first one by saying:
$\mathcal{M} \models R(n,m)$ for all $n$ and some $m$ in the domain, so $\mathcal{M} \models \forall x \exists y R(x,y)$, I don't see why the same can't be applied for the second problem with $F(x,y)$. Can someone help me to intuitively understand these operations?

"does not imply" is almost always proved by a counter-example. Let be $\mathbb{Q}\times\mathbb{Q}$ domain of F(x,y) defined by F(x,y) if only if x=2y.Then, ∀x∃y such that x=2y (obviously) i.e., F(x,y), but there is no y in $\mathbb{Q}$ such that x=2y for all x in $\mathbb{Q}$
