Constructing a metric over a lattice Consider a lattice $({\cal L}, \wedge, \vee)$ with an antimonotonic function $f: {\cal L} \rightarrow {\mathbb R}$ defined on it (i.e $x \preceq y \implies f(x) \ge f(y)$). 
$f$ is said to be submodular if for all $x,y \in {\cal L}$, $$f(x) + f(y) \ge f(x \wedge y) + f(x \vee y)$$ and supermodular if the inequality is flipped (again for all $x,y$). 
It's generally known (there's an easy proof), that a submodular $f$ induces a metric on ${\cal L}$ via the defn $$ d_s(x,y) = 2f(x \wedge y) - f(x) - f(y)$$. If $f$ is supermodular, then the construction $$d^s(x,y) = f(x) + f(y) - 2f(x \vee y)$$ yields a metric. 
Question I'm dealing with an $f$ that is nether sub- nor supermodular. I can define the "distance" $$ d(x,y) = \min ( d^s(x,y), d_s(x,y))$$
Conjecture: $d(x,y)$ is a metric. 
I have very little sound mathematical intuition for why this conjecture should be true, and bucketloads of empirical evidence (from a lattice I'm actually working with). This seems like the kind of thing that if true, would be reasonably well known to experts, and if false, might have a clear counterexample. So this is a plea for help. 
Since it might make a difference, I should mention that the lattice I'm working with is nondistributive in general, but it has distributive sublattices where I'm still unable to prove the conjecture. 
 A: First failed attempt (this poset is not a lattice). Triangle inequality fails for the three .5 nodes in the middle:

Second attempt Our lattice consists of sets, with intersection and union.  I show them by Venn diagrams here... The same three middle sets fail the triangle inequality for the same reasons as before.  But now it is surely a lattice, right?

A: Don't you need some strict inequality somewhere in your definitions?  For example, a constant function meets your definitions of antimonotonic, submodular, and supermodular, but does not induce a metric (assuming your lattice has more than one element) since $d^s$ and $d_s$ would then always evaluate to zero.
A: Probably to define a distance starting with a general antimonotonic $f$ we should do this:
if $a\lt b$ then $d(a,b) = d(b,a) = f(a)-f(b)$, and for general $a,b$ let $d(a,b)$ be $$\inf \sum_{n=1}^n d(x_i,x_{i-1}),$$ where the infimum is over all sequences $a=x_0, x_1, \dots, x_n=b$ such that adjacent terms are comparable (call such sequences paths from $a$ to $b$).  In your original proposal we used only two-step paths from $a$ to $b$, but to get the triangle inequality we need to allow longer paths as well.  Presumably submodular implies that a certain two-step path is shortest, and supermodular that a different two-step path is shortest.  Plus, this will apply to a poset that is not a lattice.
