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?

This is similar to Shaun's answer, however, it uses different software.
One can generate a tree proof for the first one that closes as follows:  

However, for the second one the tree proof does not close allowing a countermodel to be constructed showing it is invalid:

The countermodel specifies a domain with two members, $0$ and $1$. The predicate $F$ is true for the pairs $(0,0)$ and $(1,1)$. It is false for all other pairs, $(0,1)$ and $(1,0)$. That means for all $x$ there exists $y$ such that $Fxy$ is true. So the antecedent is true. However, it is not true that there exists a $y$ such that all $x$ $Fxy$ true. So the consequent is false.
All one has to do to prove invalidity is to find a countermodel showing the antecedent is true while the consequent is false.

Tree Proof Generator. https://www.umsu.de/trees/
