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.

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