What does associativity mean for orders? I'm watching the class Category Theory for Programmers and it's said that an order (preorder, partial order, or total order) constitutes a category, and one of the conditions for this is that the relation is associative.
I understand how and why $(x+y)+z = x+(y+z)$, in the sense that the order of applying the $+$ operator doesn't matter - after doing $x+y$ I get a number $w$, then I can do $w+z$.
However I don't really understand what $(a\leq b)\leq c$ means. $(a\leq b)$ doesn't produce a number like $(x+y)$; it's just a true/false value of whether or not $a$ is less than or equal to $b$.
What does it mean to ask "is $a$ is smaller than $b$ smaller than $c$"?
What's a better interpretation of asking whether an order is associative?
 A: I think part of your confusion is just the notation being used. To make a poset into a category, you have precisely one morphism from $a$ to $b$ exactly if $a \leq b$, but $a \leq b$ is not (usually) the "name" of that morphism. For the purposes of this answer, we'll call such a morphism (if it exists) $l_{a, b} \in \hom(a, b)$.
The key idea that will affect everything below is that elements of $\hom(a, b)$ are unique if they exist (for this kind of category - this is not true for all categories). For any morphisms $f, g \in \hom(a, b)$, $f = g$ because both $f$ and $g$ are equal to $l_{a, b}$.
The axioms of a category then require that there is an identity morphism $\mathrm{id}_a : a \to a$, indicating that the relation $\leq$ needs to be reflexive ($a \leq a$). Since there's exactly one morphism between two objects if it exists, $\mathrm{id}_a$ must be $l_{a, a}$.
Next, we require composition: a map $\hom(b, c) \times \hom(a, b) \to \hom(a, c)$. Remembering that $\hom(a, b)$ is inhabited (by $l_{a, b}$) exactly if $a \leq b$, this translates to transitivity of the relation: $ b \leq c$ and $a \leq b$ implies $a \leq c$.
But what is $l_{b, c} \circ l_{a, b}$? We know that it's an element of $\hom(a, c)$, but that set has at most one element, and if it has any element, it's $l_{a, c}$. So $l_{b, c} \circ l_{a, b} = l_{a, c}$.
The other conditions talk about equality of morphisms. But remember that any morphisms with the same domain and codomain are equal, so these equalities are all trivial.
Left identity is $\mathrm{id}_b \circ f = f$ for all morphisms $f \in \hom(a, b)$. Both sides of the equality are morphisms from $a$ to $b$, so they're automatically equal. Similarly, right identity, $f \circ \mathrm{id}_a = f$ is trivially true.
Associativity says that for morphisms $f \in \hom(c, d)$, $g \in \hom(b, c)$ and $h \in \hom(a, b)$, $(f \circ g) \circ h = f \circ (g \circ h)$. Both sides of the equality are morphisms from $a$ to $d$, so they are trivially equal once again. Both sides would equal $l_{a, d}$ in this case.
