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 $A_2$.  

<u>Proof</u>: If $x_i, y_i \in A_i$ with $x_i \wedge y_i = 0$, then clearly $(x_1 \wedge x_2) \wedge (y_1 \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.