For an algebraic signature (= set of function symbols) $\Sigma$, say that an *ordered $\Sigma$-algebra* is a pair $\mathfrak{A}=(\mathcal{A};\le)$ where $\mathcal{A}$ is a $\Sigma$-algebra in the sense of universal algebra and $\le$ is a **total** ordering of (the underlying set of) $\mathcal{A}$. Note that no compatibility conditions are placed on $\le$ and the structure of $\mathcal{A}$. Let $\mathit{OA}(\Sigma)$ be the class of orderd $\Sigma$-algebras. Say that a **weak $\Sigma$-equation** is an expression of the form $t=s$ or $t\le s$ for $t,s$ terms in $\Sigma\sqcup\{\max,\min\}$, and define satisfaction of weak equations in ordered algebras in the obvious way. Finally, let $\mathit{Th_{weq}}(\mathfrak{A})$ (resp. $\mathit{Th_{weq}}(\mathbb{K})$) be the set of weak $\Sigma$-equations satisfied in the ordered $\Sigma$-algebra $\mathfrak{A}$ (resp. each $\mathfrak{A}\in\mathbb{K}$). My question is: > Given a class of of ordered $\Sigma$-algebras $\mathbb{K}$, is there an "operations-based" description of the class $$\mathbb{K}':=\{\mathfrak{A}\in\mathit{OA}(\Sigma): \mathfrak{A}\models \mathit{Th}_{\mathit{weq}}(\mathbb{K})\}?$$ Here, by "operations-based" I mean that it should look like "$\mathbb{K}'=\mathsf{Z}_1\mathsf{Z}_2...\mathsf{Z}_n(\mathbb{K})$" for some reasonably-nice monotonically-increasing operations on classes of ordered algebras $\mathsf{Z}_1,\mathsf{Z}_2,...,\mathsf{Z}_n$ whose definitions don't refer to weak equations per se. Basically, I'm asking whether there an "(totally) ordered algebra" version of the HSP theorem. The key obstacle I see here is the poor behavior of products: the obvious definition of $\mathfrak{A}\times\mathfrak{B}$ (let alone $\prod_{i\in I}\mathfrak{A}_i$) gives a *partially* ordered algebra. We could look at maximal linearly ordered subalgebras of the full product, but this doesn't seem to help much: for example, if we let $\mathfrak{S}$ be the two-element $\{f\}$-algebra where $f$ acts as a nontrivial involution, then $\mathit{Th_{weq}}(\mathfrak{S})$ has lots of models but no power of $\mathfrak{S}$ has an ordered subalgebra of size $>2$.