Show that $\exists v_{0} (\phi \rightarrow \forall v_{0} \phi)$ is valid.
So what I have done is that I start with the assumption that it is not valid and trying to get a inconsistency.
$s \not\in \text{Sat}_{M} (\exists v_{0} (\phi \rightarrow \forall v_{0} \phi)) $
$s \not\in \text{Sat}_{M} (\neg \forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$
$s \not\in {^NM} \setminus \text{Sat}_{M} (\forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$
$s \in \text{Sat}_{M} (\forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$
$s \in A_{0}(\text{Sat}_{M} (\neg(\phi \rightarrow \forall v_{0} \phi)))$
$s(a/0) \in \text{Sat}_{M} (\neg(\phi \rightarrow \forall v_{0} \phi))$ for all $a\in M$
$s(a/0) \in {^NM} \setminus \text{Sat}_{M} (\phi \rightarrow \forall v_{0} \phi)$ for all $a\in M$
$s(a/0) \not\in \text{Sat}_{M} (\phi \rightarrow \forall v_{0} \phi)$ for all $a\in M$
$s(a/0) \not\in ({^NM} \setminus \text{Sat}_{M} (\phi)) \cup (\text{Sat}_{M}(\forall v_{0} \phi))$ for all $a\in M$
From here I get two conditions (not sure if the right word).
1.
$s(a/0) \in \text{Sat}_{M} (\phi)$ for all $a\in M$
and
2.
$s(a/0) \not\in \text{Sat}_{M}(\forall v_{0} \phi)$ for all $a\in M$
$s(a/0) \not\in A_{0}(\text{Sat}_{M}(\phi))$ for all $a\in M$
$s(a/0)s(b/0) \not\in \text{Sat}_{M}(\phi)$ for all $a\in M$ and for some $b \in M$
Now since we have that for all $a \in M$, $s(a/0) \in \text{Sat}_{M} (\phi)$ and $s(a/0) \not\in \text{Sat}_{M} (\phi)$ we have a inconsistency.
This is what I've done but I have a feeling in the back of my head that there is something wrong in my solution.