TITLE: Strict and non-strict orderings
QUESTION [7 upvotes]: Consider a set $A$ equipped with two binary relations $\le$ and $<$, related in the appropriate ways for the strict and non-strict version of an ordering.  One might make different choices about exactly what axioms to impose, but some reasonable ones to choose from are:

$\le$ is a preorder (or maybe a partial order).
$<$ is irreflexive.
$<$ is transitive.
$x\le y$ and $y<x$ cannot both hold.
If $x\le y$ and $y<z$, then $x<z$.  Dually, if $x<y$ and $y\le z$, then $x<z$.
If $x<y$, then $x\le y$.
If $x<z$, then either $x<y$ or $y<z$.

In classical mathematics, strict and non-strict orders are usually interdefinable.  (Edit: As Joel pointed out, this is only really true in the partial-order case.)   Since this generally fails in constructive mathematics, the above abstract structure seems more likely to be interesting there.  But even in classical mathematics there are interesting examples where $\le$ and $<$ don't determine each other in the naive way, such as the collection of transitive sets in ZFC with the relations $\subseteq$ and $\in$.
Has anyone defined an abstract structure like this?  If so, what is it called?

REPLY [2 votes]: An answer in the opposite direction (helpfully meant, for completing the picture):
In 'old-fashioned' constructive mathematics, the set $A$ already has an apartness $\#$.
Then we have: $x<y\leftrightarrow (x\leq y\wedge x\# y)$ and $x\leq y\leftrightarrow \forall z\, [\,z<x\rightarrow z<y\,]$.
So there is always a sensible way to express $\leq$ in terms of $<$, and the necessary information to express $<$ in terms of $\leq$ is precisely the apartness $\#$.
[Addition for completeness: if we have $<$, and it is a 'total' relation in the sense that $(\neg(x<y)\wedge\neg(y<x))\rightarrow x=y$, then we can define $x\# y\leftrightarrow(x<y\vee y<x)$, and the above also still holds.]  

[Update to incorporate some exchange in the comments below:]
In constructive mathematics usually the symbol $<$ is used for a relation where $x<y$ indicates constructive existence of numerical/finitary evidence that $x$ and $y$ are apart in some sense, and order-comparable. The symbol $\leq$ usually is derived from $<$, for total orders most often I believe by defining $x\leq y\leftrightarrow \neg(y<x)$. This usually amounts, quantifier-wise, to for all.
The alternative $x\leq y\leftrightarrow \forall z\, [\,z<x\rightarrow z<y\,]$ mentioned above also uses for all, not surprisingly.
In intuitionistic mathematics, for expressing $<$ in terms of $\leq$ without using an existing apartness, perhaps we could turn to the strong intuitionistic interpretation of the logical connective $\vee$. Then, without using an existing apartness, we can define:
$$(*)\ \ \ \    x<y\leftrightarrow \,\forall z\,[\,(z\leq y\wedge \neg(y\leq z))\vee (x\leq z\wedge \neg(z\leq x))\,]$$  
I'm moderately optimistic that this definition will satisfy the specified axioms in many situations, like the real numbers for example. 
One might think the quantifier $\forall$ in contrast to what is stated earlier, but it helps to read $(*)$ as $ x<y\leftrightarrow \,\forall z\,\exists i\in\{0,1\}[\,(i=0\wedge z\leq y\wedge \neg(y\leq z))\vee (i=1\wedge x\leq z\wedge \neg(z\leq x))\,]$