Linear algebra with discrete, ordinal, but non-numeric values? I have a series of statements relating various discrete, ordinal, but non-numeric values.  I'd like to reason about those statements intelligently.  Looking at the relationships I want to say that it is a linear algebra problem but I'm not sure you can use linear algebra rules with non-numeric coefficients.
vastly oversimplified example case
I have the following ordinal relationships
a < b < c
d < e

and the following equations
{a, d} = f
{c, e} = g   

in this case, there is no relationship between, say, a and e.  It's not that the relationship isn't known, it doesn't exist.
I want a formal way of analyzing those equations to say that f < g given that all the components of f are < all the comparable components of g.  Is linear algebra applicable?  Is there a better toolset I should be looking at?
 A: I don't see how linear algebra would be relevant here; the "linear" in "linear algebra" doesn't refer to an ordering but rather to linearity in the sense of a function satisfying $f(ax)=af(x)$ and $f(x+y)=f(x)+f(y)$ for $x,y$ vectors and $a$ a scalar. Rather, what you're considering here are partial orders (or posets), and the relevant subject is order theory - although partial orders are so ubiquitous in mathematics that you won't need to specifically look at order theory to see them.
Incidentally, your proposed ordering on subsets of a given poset - namely, $A\trianglelefteq B$ iff for each $a\in A$ and $b\in B$, either $a$ and $b$ are incomparable or $a<b$ - isn't very well-behaved. Specifically:


*

*It's not antisymmetric: if $a$ and $b$ are incomparable then $\{a\}\trianglelefteq \{b\}$ and $\{b\}\trianglelefteq \{a\}$.

*More importantly, it's not transitive: suppose we have a partial order with elements $a,b,c$ where $a<b$ and $c$ is incomparable with both $a$ and $b$. Then $\{b\}\trianglelefteq\{c\}\trianglelefteq\{a\}$ but $\{b\}\not\trianglelefteq\{a\}$.
A better notion is to compare all elements at once: say $A\preccurlyeq B$ if for every $a\in A$ and every $b\in B$ we have $a<b$. This still isn't entirely great (if $a<b<c$ then consider $\{a,c\}$ versus $\{a,b,c\}$) but at least it's transitive. 
