Q. If for every sentence $\phi$ of a first-order language $\mathcal{L}$ is s.t. $\mathcal{A} \vDash \phi$ iff $\mathcal{A'} \vDash \phi $, then there is an isomorphism between two interpretations/structures/models $\mathcal{A}$ and $\mathcal{A'}$ of a first-order language $\mathcal{L}$. How do I show that this is false?
$\mathcal{A'}$ and $\mathcal{A'}$ are structures of $\mathcal{L}$ that make the same sentences $\phi$ of $\mathcal{L}$ true.
Yet they don't have to be isomorphic for this to be the case.
I'm not sure how to prove this.
Maybe we could use the fact that it's converse is a theorem, which is the following the theorem: If there is an isomorphism between two interpretations $\mathcal{A}$ and $\mathcal{A'}$ of a first-order language $\mathcal{L}$, then for every sentence $\phi$ of a first-order language $\mathcal{L}$ is s.t. $\mathcal{A} \vDash \phi$ iff $\mathcal{A'} \vDash \phi$.
Many thanks in advance!
The simplest counterexample would be to take $\mathcal{L}$ the language with only the equality symbol, and take $\mathcal{A}$ and $\mathcal{A'}$ to be infinite sets of different cardinalities. They will both satisfy the same first-order $\mathcal{L}$-sentences, but can never be isomorphic because of their different cardinality.