Does elementary equivalence imply L-equivalence for structure L?

Question

Does elementary equivalence imply L-equivalence for structure L? In the textbook "A Shorter Model Theory" by Wilfrid Hodges, page 39 defines both of these terms but does not tie them together. I was wondering if for A, B (2 L-structures), is A L-equivalent to B when A and B are elementarily equivalent.

Answer 1

$L$-equivalence is defined for a language $L$. $A$ is $L$-equivalent to $B$ when $A\models\varphi$ iff $B\models \varphi$ for all sentences $\varphi$ in the language $L$.

As far as I can tell, Hodges never gives a formal definition of "language", but he does give lots of examples of languages. Given a signature $\Sigma$ (consisting of constant, function, and relation symbols), we can build from $\Sigma$, for example, the first-order language $L_{\omega,\omega}$, the infinitary language $L_{\infty,\omega}$, and the quantifier-free first-order language $L_{\omega,0}$.

Each of these languages has a different set of sentences, so they have different notions of $L$-equivalence. We say that $A$ and $B$ are elementarily equivalent when $A\models \varphi$ iff $B\models\varphi$ for all first-order sentences $\varphi$. So elementary equivalence is the same as $L_{\omega,\omega}$-equivalence. But $L_{\omega,0}\subsetneq L_{\omega,\omega} \subsetneq L_{\infty,\omega}$, so $L_{\omega,0}$-equivalence is strictly weaker than elementary equivalence, and $L_{\infty,\omega}$-equivalence is strictly stronger.