Let $D(X)$ denote the free bounded distributive lattice on a set $X$. Let $x_0 \in X$ be a chosen generator.
In the paper Recursive formulas on free distributive lattices (1968) by N. M. Reviere it is claimed without proof that every element $u \in D(X)$ has the decomposition $$u = \bigl(x_0 \wedge l(u)\bigr) \vee k(u) \quad (\ast)$$ where the two lattice homomorphisms $l,k : D(X) \to D(X \setminus x_0)$ are defined by $l(x)=k(x)=x$ for $x \in X \setminus x_0$ and $l(x_0)=1$, $k(x_0)=0$. The notation in the paper is a bit different, but it's the same meaning. It follows that every element has a unique decomposition $(x_0 \wedge b) \vee a$ with $a,b \in D(X \setminus x_0)$ and $a \leq b$.
I don't know how to prove this. Of course $(\ast)$ is true when $u \in D(X \setminus x_0)$. To prove it in general, I tried to write $u = u_1 \wedge u_2$ (likewise, $u_1 \vee u_2$) and assume that the claim is true for $u_1,u_2$. But when I calculcate both sides of $(\ast)$, it just doesn't fit. Abstractly, this is because a priori the right hand side of $(\ast)$ is not homomorphic in $u$. I am sure there is an easy way to fix this, but I just do not see it.
I have already verified the claim by hand for the free distributive lattice on three generators $x_0,x_1,x_2$ (which has $20$ elements, see here), also because I couldn't really believe that such a simple decomposition is valid for all elements*. A nice example here is $$u := (x_0 \wedge x_1) \vee (x_0 \wedge x_2) \vee (x_1 \wedge x_2),$$ for which we find $l(u) = x_1 \vee x_2$, $k(u) = x_1 \wedge x_2$, and in fact $$u = (x_0 \wedge (x_1 \vee x_2)) \vee (x_1 \wedge x_2)$$ holds because of the distributive law applied to the first two components of $u$.
*Also, because the number of elements of $D([9])$ has only been computed last year in 2023 (see here), and the number of elements of $D([10])$ is still unknown.
