Existentially closed model

Question

Let $T$ be the complete first-order arithmetic, with signature $0, 1, + , \dot{}$. Show that the natural numbers form an exsitentially closed model of T.

I am having some trouble proving this directly, I tried contradiction: Consider a structure $\mathcal B \supseteq \mathbb N $ and an existential formula $\phi(\bar x)=\exists y $ $ \psi(\bar x,y) $ such that $\mathcal B \vDash \phi(\bar x) $ but $\mathbb N \nvDash \phi(\bar x) $. Equivalently $$ \mathbb N \vDash \forall y\neg \psi(\bar x,y)$$ Or in other words, there is a system of equations which has solution in $\mathcal B$ but not in $\mathbb N$. Then I fail to reach a contradiction. How is that not possible? The book I'm taking this from (Hodges) states this as a "freak" example. Is it trivial maybe? Thanks for any help. (Sorry if some of my syntax is not too acurate).

Answer 1

Hint: If there were no parameters $\bar{x}$, then $\phi$ would be a sentence and so it would be true in $\mathcal{B}$ iff it is true in $\mathbb{N}$. Can you find a way to get rid of the parameters so you just have a sentence?

A full solution is hidden below.

Every element $n\in \mathbb{N}$ can be represented by a closed term, by just adding together $n$ copies of $1$. So replace each parameter appearing in $\phi$ by a closed term representing the same natural number and you get an equivalent formula $\phi'$ which now has no parameters. This $\phi'$ is a sentence and thus if it is true in $\mathcal{B}$, it must also be true in $\mathbb{N}$.