So the subject I'm stumbling around a bit is elementary equivalence and elementary substructures. So maybe you could help me sort things out.
So the definitions as I know them.
Structures $M, N$ over signature $S$ are called elementary equivalent if for every formula $\phi$ over $S$, $M \vDash \phi \iff N \vDash \phi$. $M, N$ being isomorphic implies they are elementary equivalent (but not vice versa).
A substructure $N$ of structure $M$ is called an elementary substructure of $M$, if for every formula $\phi$, and for every $b_1, \cdots, b_n \in N: N \vDash \phi (b_1, \cdots, b_n) \iff M \vDash \phi(b_1, \cdots, b_n) $
All is cool up to now. My problem is understanding the examples given to show that a substructure might be elementary equivalent with its structure, but is not an elementary substructure. To my understanding, in order to prove a substructure isn't elementary, you need to find a formula $\phi(v_1 \cdots v_n)$ so that for every $b_1, \cdots, b_n \in N$, so that $M$ won't satisfy $\phi (b_1, \cdots ,b_n)$, and $N$ will, or vice versa.
So an example: $M=(\mathbb Z; +,0)$, and its substructure $N=(2\mathbb Z;+,0)$, and we need to show that $N$ is not an elementary substructure of $M$. So before that, we can see that they are elementary equivalent, as they are isomorphic.
So, for the rebuttal: Lets take $\phi:= \forall x \exists y: y+y=x$. But this formula is 'independent' of choosing variables (there are not free variables), so by being vacuously truth it seems that for every $b_1, \cdots, b_n \in N$, we have $N \vDash \phi(b_1, \cdots, b_n)$ and $M \not\vDash \phi(b_1, \cdots, b_n)$ so $N$ is not elementary equivalent.
My question is: Is using this type of formula correct for refutal? Or should I only use formulas with free variables, such as $\phi := \exists y: y+y=x$, and then $M \vDash \phi(6)$, but $N$ doesn't.
The second, smaller question: Doesn't $\phi:= \forall x \exists y: y+y=x$ show they are not actually elementary equivalent?
I deeply apologize for the wall of text. Any help will be greatly appreciated! Thank you.
The sentence $\forall x\exists y(y+y=x)$ is true in neither $\mathbb{Z}$ nor $2\mathbb{Z}$. So the sentence does not refute elementary equivalence.
But there is no $y$ in $2\mathbb{Z}$ such that $y+y=2$, while there is in $\mathbb{Z}$, so $2\mathbb{Z}$ is not an elementary substructure of $\mathbb{Z}$. There is no real difference between using formulas with free variables, for which we substitute elements of $N$, and sentences in an extended language with symbols for elements of $N$.