Here's an answer to a related question which involves much more standard terminology.
Say $A \subset L$ is an antichain iff $\forall x, y \in A(x \neq y \rightarrow x \wedge y = 0)$. If $A_1$ and $A_2$ are two antichains, then $A_1 \wedge A_2$ is yet another antichain which we say "refines" both $A_1$ and $A_2$.
Proof: If $x_i, y_i \in A_i$ with $x_i \wedge y_i = 0$, then clearly $(x_1 \wedge y_1) \wedge (x_2 \wedge y_2) = 0. \\ \\ \square $
The notion of antichain makes sense even if the joins $\bigvee (S - \{ x \})$ aren't defined. If $L$ satisfies $\forall x \in L\ \forall S \subset L\ (x \wedge \bigvee S = \bigvee (\{ x \} \wedge S)$ then the notions of "antichain" and "mutually disjoint" coincide.