# Is there a preference between proving a total order (strict vs partial)?

I know that proving a relation $\mathcal{R}$ to be a strict total order (asymmetric, transitive,and total ) implies that the relation $S$ defined as $X\mathcal{S} Y \longleftrightarrow (X\mathcal{R}Y\vee X=Y)$ is a total order (antisymmetric, transitive, reflexive, and complete).

Conversely, if a relation $\mathcal{R}$ is a total order then the relation $S$ defined as $X\mathcal{S} Y \longleftrightarrow (X\mathcal{R}Y\wedge X\neq Y)$ is a strict total order.

My perception is that people use to prove a relation to be a total order over proving that a relation is a strict total order. If proving a relation to be a total order automatically implies the other to hold why people don't do it by considering strict total orders?

For example in the definition of $\mathbb{N}$ I think it should be the same (even easier) if we prove that it's a totally strict ordered set and then automatically it's a totally order set, but books prefer the other way around. They prove that it's a totally ordered set and then the other option follows automatically. What am I missing? is it just matter of custom, style, or something?

For example, the ubiquitous subset relation $X \subseteq Y$ and its strict counterpart $X \subsetneq Y$. We express $X \subseteq Y$ by $$\forall x: x\in X \to x \in Y$$ This is a lot shorter than $X \subsetneq Y$: $$\forall x(x \in X \to x \in Y) \land \exists y( y \in Y \land \neg (y \in X))$$ which can of course be shortened by using properties of $\subseteq$, but in principle the expression above is what it comes down to.
On the other hand, the membership relation $X \in Y$ itself (considered on transitive sets such as ordinals) is (almost) never discussed in the inclusive form, because (usually) we don't want to consider sets as elements of themselves.