In first-order logic without equality, is the theory of partial orders the same as preorders? This is related to my previous question on equality and equivalence relations. In first-order logic without equality with a single binary relation $R$, is the theory of reflexive partial orders the same as the theory of preorders?
 A: Per a couple of my previous answers (1, 2), there is a standard technique for answering this sort of question. Namely, we have the following (phrased more generally here than in the above-linked answers):

Suppose $\mathfrak{A},\mathfrak{B}$ are structures in the same relational language and $R\subseteq\mathfrak{A}\times\mathfrak{B}$ is a total surjective relation which preserves and reflects all the relations in that language. Then $\mathfrak{A}\equiv_{\mathsf{FOLw/o=}}\mathfrak{B}$; in fact, $\mathfrak{A}$ and $\mathfrak{B}$ have the same equality-free $\mathcal{L}_{\infty,\infty}$-theories.

The proof is by a standard induction on formula complexity; note that preserving and reflecting equality amounts to bijectivity, so this does actually match the usual notion of isomorphism if equality is treated as atomic. (The restriction to relational languages is convenient but not truly necessary.)
This lets us prove results about classes of structures. Specifically, say that $\mathfrak{A}$ and $\mathfrak{B}$ are quasi-isomorphic if there is some $R$ with the properties above. Then we have:

Suppose $\mathbb{K}_0\subseteq\mathbb{K}_1$ are classes of structures in the same relational language, and every structure in $\mathbb{K}_1$ is quasi-isomorphic to a structure in $\mathbb{K}_0$. Then $\mathbb{K}_1$ and $\mathbb{K}_0$ have the same equality-free-first-order theories: the equality-free-first-order sentences true in every structure in $\mathbb{K}_0$ are also true in every structure in $\mathbb{K}_1$ (since $\mathbb{K}_0\subseteq\mathbb{K}_1$ the other direction is trivial).

In this case, take $\mathbb{K}_1$ to be the class of preorders and $\mathbb{K}_0$ to be the class of partial orders (with "partialization" giving the desired $R$s).
