# Is the equality-free theory of linear orders the same as the equality free-theory of linear preorders?

This is a natural follow-up to my question, here:In first-order logic without equality, is the theory of partial orders the same as preorders?. My current question is, consider first-order logic without equality with a single binary relation $$R$$. Is the equality-free theory of reflexive linear orders the same as the equality-free theory of linear preorders? By the way, I define a linear preorder to be a preorder such that for all $$x$$ and $$y$$ in the preorder, either $$x \leq y$$ or $$y \leq x$$. Also, if the answer is no, what is an explicit axiomatization of the equality-free theory of reflexive linear orders?

Specifically, given a linear preorder $$L$$, let $$L'$$ be the quotient of $$L$$ by the relation $$x\sim y\iff x\le y\wedge y\le x$$. Then $$L'$$ is clearly a linear order. Meanwhile, the quotient map $$L\rightarrow L'$$ preserves and reflects the linear order $$\le$$, so we get $$L\equiv_{\mathsf{FOL_{w/o=}}}L'$$. This shows that the $$\mathsf{FOL_{w/o=}}$$-theory of linear preorders contains the $$\mathsf{FOL_{w/o=}}$$-theory of linear orders; since the converse inclusion is trivial, we're done.
I strongly recommend that you carefully read the general result in the third-linked answer above (relink); I think mastering it will help you answer questions of this type yourself much more quickly. In general, always look for "quasi-isomorphism" relations when trying to show equivalence of $$\mathsf{FOL_{w/o=}}$$-theories.