By transfer principle, is the set of hypernatural the set of naturals? In Jech, we learn that $x=\mathbb{N}$ is a $\Delta_0$-formula. Can you tell me what is wrong with the following reasoning ?
Let $\phi(x)$ be the formula $x=\mathbb{N}$
The tranfer principle tells us that $\forall y\in\mathbf{S},\phi^{\mathbf{S}}(y)\iff \phi^{\mathbf{I}}(^*y)$ where $\mathbf{S}$ and $\mathbf{I}$ are the class of standard sets and the class of internal sets respectively.
But, as $\mathbf{S}$ is transitive, and $\phi$ is $\Delta_0$, $\forall y\in\mathbf{S},\phi(y)\iff\phi^{\mathbf{S}}(y)$ and in particular $\phi(x)\iff \phi^{\mathbf{S}}(x)$.
Similarly, as $\mathbf{I}$ is transitive, and $\phi$ is $\Delta_0$, $\forall y\in\mathbf{I},\phi(y)\iff\phi^{\mathbf{I}}(y)$. But, $y=^*x$ is internal so  $\phi(^*x)\iff\phi^{\mathbf{I}}(^*x)$.
Using the last two facts, the transfer principle gives $\phi(x)\iff \phi(^*x)$. Taking $x$ equal to $\mathbb{N}$, then $\mathbb{N}=^*\mathbb{N}$.
EDIT : My framework is *ZFC, see the paper here.

Jech proves that "$x=\mathbb{N}$" can be expressed by a $\Delta_0$ formula in ZFC.  That is, there is some $\Delta_0$ formula $\phi(x)$ such that ZFC proves there is exactly one set satisfying $\phi(x)$ (and that set is what we think of intuitively as $\mathbb{N}$).
However, you are not working in ZFC.  You are working in *ZFC, which does not include all the axioms of ZFC: it is missing the axiom of regularity.  In fact, the axiom of regularity is crucial to the proof that the $\Delta_0$ formula $\phi(x)$ defines a unique set.  So *ZFC actually cannot prove that $\phi(x)$ defines a unique set, and so you cannot conclude that ${}^*\mathbb{N}=\mathbb{N}$.  In fact, your argument shows that in *ZFC, there is no $\Delta_0$ formula that defines $\mathbb{N}$.
