# Consistent full Horn theories of two structures

Suppose that two structures $$A$$ and $$B$$ whose cardinality is greater than 1 (added in a revision) have the same positive primitive theory. Does it follow that the union of the full Horn theory of $$A$$ and that of $$B$$ is consistent? (Here I allow possibly empty finite conjunctions in the antecedent of a basic Horn formula and $$\bot$$ in the consequent; also a Horn sentence can have an arbitrarily many quantifiers.)

One way of having a contradiction from two Horn theories is having a sentence in one and its negation in the other. Such a sentence is positive primitive (or negations thereof). Hence the question in the beginning. In fact, it is true of any concrete structures that I can think of.

• There is still something wrong with the question. You might want to specify exactly what form of sentence you mean by "p.p." or "Horn". Otherwise, please explain why a sentence from one Horn theory whose negation belongs to another Horn theory must be p.p. May 17 at 16:21
• Doesn't $(\mathbb{Z},<)$ and $(\mathbb{N},<)$ give a counterexample? May 17 at 16:22
• Take $A$ to be the $1$-element unstructured set and $B$ to be the $2$-element unstructured set. There are homomorphisms in both directions, so they satisfy the same p.p. sentences. The Horn sentence that separates them is $(\exists x)(\forall y)(x=y)$. May 17 at 16:36
• OK, then take $A$ to be a complete irreflexive graph on an infinite set and take $B$ to equal $A$ union a single isolated point. $A$ and $B$ are mutually embeddable, so the satisfy the same p.p sentences. The Horn sentence that separates them is $(\forall x)(\exists y)(E(x,y))$. May 17 at 16:40
• @Pteromys: Regarding Alex's suggestion, $(\mathbb Z,<)$ and $(\mathbb N,<)$ have the same finitely generated substructures up to isomorphism so they satisfy the same p.p. sentences. Now you need an example of a Horn sentence that is true in $(\mathbb Z,<)$ and false in $(\mathbb N,<)$. May 17 at 17:50

Secondly, here is an example involving total orders in the language $$\{\le\}$$. Consider again naturals and integers. The chain of natural numbers satisfy the Horn sentence $$(\exists x)(\forall y)[y \le x \to y = x]$$, where as the chain of integers satisfy the Horn sentence $$(\forall x)(\exists y)[y \not \ge x]$$. Now the theory of partial orders is Horn, so it suffices to show that no partial order satisfies those two Horn sentences. This is easy, as the minimum element of a partial order, if it exists, cannot have an element strictly below it or an element incomparable with it.