A question about collections that agree with logical statements in $ZF$ Working in $ZF + AR$ or $Z + AR$, where $AR$ stands for the axiom of regularity.
Terminology:
Following the terminology of Théorie des ensembles — Jean-Louis Krivine:
Consider a collection $X$ and a logical statement $E(x_1, \dots, x_n)$ with $n$ variables whose parameters are objects of X. The statement $E$ restricted to $X$ is denoted as $E^X(x_1, \dots, x_n)$ and is defined recursively as:


*

*$E$ is one of the forms: $x \in y, x = y, x \in a, x = a, a \in x, a \in b, a = b$, where $a$ and $b$ are objects of $X$; the restricted statement $E^X$ is $E$ itself.

*$E$ is $\text{not}\; F$; then $E^X$ is $\text{not}\; F^X$

*$E$ is $F \; \text{or} \; G$; then $E^X$ is $F^X \,\text{or} \; G^X$.

*$E$ is $\exists x F(x, x_1, \dots, x_n)$; then $E^X$ is $\exists x[X(x) \land F^X(x, x_1, \dots, x_n)]$.


We say that a collection $X$ agrees with $E$ if
$$\forall x_1 \dots \forall x_k[X(x_1) \land \dots \land X(x_k) \implies (E(x_1, \dots, x_k) \Leftrightarrow E^X(x_1, \dots, x_k))].$$
We can see that if the statement $E$ is without quantifiers; $E^X = E$; therefore, any collection agrees with $E$.
If $E$ doesn't have a free variable then "$V_\beta$ agrees with $E$" means that "$E$ is true in the universe iff it is true in $V_\beta$."
Problem
From my understanding for all statements $E(x_1, \dots, x_n)$, whose parameters are in an arbitrary collection $X$, it should trivially follow that $X$ agrees with $E(x_1, \dots, x_n)$. Is there any statement $E(x_1, \dots, x_n)$ whose parameters are in $X$, for some collection $X$, such that $X$ doesn't agree with $E$?
Also, when dealing with statements without free variables. Is
$$E = \forall x\exists y[P(x) \in y]$$ a statement that doesn't agree with, say, $V_3$, where $V_\alpha$ is the von Neumann stage of order $\alpha$, but agrees with, say, $V_\omega$?
 A: 
From my understanding for all statements $E(x_1, \dots, x_n)$, whose parameters are in an arbitrary collection $X$, it should trivially follow that $X$ agrees with $E(x_1, \dots, x_n)$.

No, not at all.  I think you perhaps have a misunderstanding: using parameters doesn't mean you can't also have other variables that are quantified.  For a very simple example, let $a$ be any set and $X=\{a\}$.  Let $E(x)$ be the statement $\exists y[x\in y]$.  Then $E(a)$ is true (take $y=\{a\}$) but $E^X(a)$ is false (since $a\not\in a$).
Note also that any statement without free variables is also a "statement with parameters in $X$".  Its number $n$ of parameters just happens to be $0$.  Or, you can consider it as a statement with any number of parameters, except that its parameters don't actually occur anywhere in it (so their value does not affect the truth of the statement).  The operative part of the phrase "with parameters in $X$" is "in $X$": you are not allowed to make parameters take values that are not in $X$.

Is
  $$E = \forall x\exists y[P(x) \in y]$$ a statement that doesn't agree with, say, $V_3$, where $V_\alpha$ is the von Neumann stage of order $\alpha$, but agrees with, say, $V_\omega$?

This one is actually a bit subtle.  The problem is that "$P(x)$" is not actually a valid term in the language of set theory.  Rather, the expression $z=P(x)$ is an abbreviation for $$\forall w[w\in z\Leftrightarrow\forall v[v\in w\Rightarrow v\in x]].$$  If you want to express your statement $E$ in the language of set theory, there are a few different ways you can do it, which are equivalent in the universe but not necessarily when relativized to a set $X$.  For instance, you might write $E$ as $$\forall x\exists y\forall z[z=P(x)\Rightarrow z\in y]$$ (where $z=P(x)$ is an abbreviation as above).  Or you might write it as $$\forall x\exists y\exists z[z=P(x)\wedge z\in y].$$  These two statements are equivalent in the universe, since for any $x$, there is a unique $z$ which satisfies $z=P(x)$.  However, in a set like $X=\{a\}$, they are not equivalent: the first is true, since taking $x=a$, there does not exist any $z\in X$ such that $(z=P(a))^X$ is true, so $\forall z[z=P(a)\Rightarrow z\in y]$ is vacuously true (and we can set $y=a$).  But for the exact same reason, the second is false.
As for your question, it turns out that actually both versions are true in $V_\omega$ and false in $V_3$, so both statements agree with $V_\omega$ but not with $V_3$.  Even the first version fails in $V_3$ because if you set $x=\{\emptyset\}$, then $z=P(x)$ is an element of $V_3$ but it is not an element of any element of $V_3$.
