# Horn sentences and power structures

It is well-known that Horn sentences are "preserved" under products (see for instance Show that the direct product of structures satisfies a Horn sentence).

I was wondering what happens with the converse. I.e. if a Horn sentence is true in the product is it true in the components? What if all components are equal (i.e. we have a power structure)?

No. For example, "There are at least four elements" is expressible as a Horn sentence but does not "descend to factors" (even in power structures) for obvious reasons.

To see that "There are at least four elements" is indeed expressible as a Horn sentence, we proceed as follows:

• For each pair of distinct variables $$i,j\in\{1,2,3,4\}$$, the expression $$x_i\not=x_j$$ is a negated atomic formula and so is a Horn formula. (Clause $$2$$ with $$n=1$$ in the linked post)

• The arbitrary conjunction of Horn formulas is again a Horn formula, so $$\bigwedge_{1\le i is a Horn formula. (Clause $$3$$, repeatedly)

• Finally, we can existentially generalize, so that $$\exists x_1,x_2,x_3,x_4\left(\bigwedge_{1\le i is a Horn sentence as claimed. (Clause 5, repeatedly)

EDIT: Note that while the above shows that it is easy to express "There are at least $$k$$ elements" for each $$k$$ as a Horn sentence, things are murkier if we want to say "There are not exactly $$k$$ elements" - while the preservation theorem says that this can't be done for composite $$k$$, it turns out to be doable-but-tricky for prime $$k$$. See this MO thread.

• Noah: quick question about the edit: is it that the preservation theorem shows that it can't be done for any composite $k$, or that there exists a composite $k$ for which it can't be done? (A clue about the reasoning involved would be nice too $\ddot{\smile}$.) Commented Aug 8 at 20:17
• @RobArthan Suppose $k=ab$. Let $\varphi$ be a Horn sentence saying "There are not exactly $k$ elements" and let $\mathfrak{A},\mathfrak{B}$ be structures with cardinalities $a,b$ respectively. If $a\not=1\not=b$ then $\mathfrak{A}\models\varphi$ and $\mathfrak{B}\models\varphi$ but $\mathfrak{A}\times\mathfrak{B}\not\models\varphi$, contradicting the preservation theorem. Commented Aug 8 at 20:55