Is formula $\mathtt{(∃x)(∀y)ϕ(x,y)}$ provable or refutable from $\mathtt{T}$ (in a Sound & Complete Proof System using the Axioms of $\mathtt{T}$)?

Question

Is the formula $\mathtt{(∃x)(∀y)ϕ(x, y)}$ provable or refutable from $\mathtt{T}$ (in a Sound and Complete Proof System using the Axioms of $\mathtt{T}$)

where $$\mathtt{T} = \{\mathtt{(∀x)¬E(x, x), (∀x)(∀y)(E(x, y) → E(y, x)), (∀x)(∃y)ϕ(x, y)}\}$$ over the language $\mathtt{L}$ where $\mathtt{L}=\mathtt{(E)}$.
$(\mathtt{E(x, y)}$ means in a graph that “the vertices $\mathtt{x}$ and $\mathtt{y}$ are adjacent” or “the vertices $\mathtt{x}$ and $\mathtt{y}$ are neighbors”$)$

Full Question

The reference for $Q2$ is here. I am interested in Guidance for $Q3$.

Answer 1

Since the poof system is sound and complete,

$\psi:=(\exists x)(\forall y) \varphi(x,y)$ is provable from $T$ iff every model $\mathcal{G}$ of $T$ satisfies $\psi$

Notice that a model of $T$ is neccesarily a graph (for the first two formulas involving $E$) which also satisfies that for all node $x$ in the graph there is a node $y$ in the graph such that $y$ has exactly two common neighbors with $x$.

So, the question is: Is true that a graph with all the previous properties satisfies that there is a node $z$ having exactly two common neighbors with all the nodes in the graph?

The answer is no because we can consider the graph $\Diamond=(V,E)$ where $V=\{1,2,3,4\}$ and $E=\{(1,2),(2,3),(3,4),(4,1),(2,1),(3,2),(4,3),(1,4)\}$.

For $\neg\psi = (\forall x)(\exists y)\neg\varphi(x,y)$ you can check that the complete graph with $4$ vertices is a $T$-model where it doesn't hold $\neg\psi$.

Therefore, $\psi$ is neither provable or refutable from $T$.