Is a sentence true for two substructures also true for their intersection? [closed]

Question

Let $L$ be a first order language and $M$ an interpretation of $L$. If $A$ and $B$ are two substructures of $M$ and their intersection $C=A\cap B\ne \emptyset$, then is it the case that every sentence (in language $L$) true for both $A$ and $B$ is also true for $C$?

By "intersection" of $A$ and $B$, I mean the smaller substructure whose domain is the intersection of the domain of $A$ and $B$.

I came to this conjecture when I learned that there was a smallest standard submodel in a universe of some set theories (such as $ZF$) if there was a standard submodel in it. The smallest standard submodel is the intersection of all the standard submodels. But is the intersection of any two standard submodels still a standard submodel? If so, considered every standard submodel is nothing more than a substructure of the universe for which a group of sentences, i.e. the axioms of some set theory, are true, can we further generalize it into the conjecture I mentioned above?

I haven't found a proof or disproof for this so far (actually I am not very good at it). Can anyone prove it or give a counterexample?

-----update----

The question in the title is trivial. Thanks for answers from Andrej Bauer and 喻良. But whether the intersection of two standard submodels of $ZF$ is still a stanadard submodel? (which is my earliest concern and let's only focus on $ZF$ or $ZFC$ here)

Answer 1

If $\mathcal C$ is any substructure of a structure $\mathcal A$, then $\mathcal C$ is the intersection of $\mathcal A$ and an isomorphic copy of it, $\mathcal A'$, inside a larger structure $\mathcal M$. (Take an isomorphic copy of $\mathcal A$ and combine it with $\mathcal A$, identifying corresponding elements of $\mathcal C$.) So a sentence is preserved by intersections if and only if it's preserved by substructures. And that's well-known to be the case only for sentences equivalent to universal sentences.

Answer 2

Of course not. For example, consider the empty language. A model is just a set. Take $M = \{1, 2, 3\}$, $A = \{1, 3\}$ and $B = \{2, 3\}$. Both $A$ and $B$ satisfy $\exists x y . x \neq y$, but $A \cap B$ does not.

Supplemental: The OP comments that the simple answer makes the original question look stupid. I disagree, no question is stupid. In fact, very often a seemingly stupid question can lead to interesting things.

So we have discovered that the answer to the question is negative. But now, rather than hiding in shame, we should drill some more, by asking trickier questions. For instance, can we put some further restrictions, either on the language, the statements, or the substructures, so that the answer becomes positive?

For example, suppose our language only has function symbols (on relation symbols, the only primitive relation is $=$). If the statement $\phi$ under consideration contains no $\exists$ and $\lor$, is its validity preserved by substructures? What if we have a statement with only $\forall$, $\land$ and $=$ (so we drop implications as well)?

Answer 3

Let $L$ be the language of multiplication, let $M=\mathbf{R},\ A=[0,1), B=(0,1]$. Then $\exists x(xx=x)$ is true in $A$ and in $B$ but not in $A \cap B$.

Answer 4

Of course not. For example, if $g$ and $h$ are mutually Cohen reals, then $L[g]\cap L[h]=L$. But both $L[g]$ and $L[h]$ satisfy $V\neq L$.