Let $L$ be a lattice and $S \subseteq L$ be such that $\bigvee S$ exists in $L$. Then $\bigvee S$ is called an irredundant join if $\bigvee(S\setminus\{s\}) \neq \bigvee S$ for all $s \in S$.
Let $J(L)$ be the set of join-irreducible elements of $L$.
In this question (which I've answered myself):
Irredundant join of join-irreducibles
I asked to prove that if $L$ is such that ${\downarrow} a$ has no infinite chains, for $a \in L$, then every element of $L$ has a representation as an irredundant join of a finite subset of $J(L)$.
The question I'm now trying to solve is:
If additionally $L$ is distributive, then each of these irredundant representations is unique, that is, each element has one and only one.
A hint. If $a \in L$, and $S,T\subseteq J(L)$ are that $a = \bigvee S = \bigvee T$ are irredundant, we know that under these conditions $S$ and $T$ are finite.
Now consider $t\wedge \bigvee S$ for $t \in T$ and $s \wedge \bigvee T$, for $s \in S$.
For each $t_0 \in T$, $$t_0 = t_0 \wedge \bigvee T = t_0 \wedge \bigvee S = \bigvee\{t_0 \wedge s : s \in S\}.$$ Since $t_0 \in J(L)$, it follows that $t_0 = t_0 \wedge s_0$, for some $s_0 \in S$, whence $t_0 \leq s_0$.
Now apply the same reasoning to $s_0$ and conclude that $s_0 \leq t_1$, for some $t_1 \in T$, whence $t_0 \leq t_1$.
As $T$ is irredundant, if follows that $t_0 = t_1$, and so $t_0 = s_0$. Thus, $T \subseteq S$.
This reasoning is symmetric, and so $T = S$.