Disjoint product of lattices / the lattice of disjoint pairs of open sets is not a lattice, but is nearly so

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Consider a lattice $L=(|L|,\wedge,\vee,\top,\bot)$ and define its disjoint product $|L\otimes L|$ to be a set as follows: $$ |L\otimes L| = \{(l,l')\in L\times L \mid l\wedge l'=\bot\} . $$ We see that $|L\otimes L|$ does not immediately inherit a componentwise lattice structure $$ (l_1,l_1')\circ (l_2,l_2') = (l_1\circ l_2,l_1'\circ l_2') \quad \text{for}\quad \circ\in\{\wedge,\vee\} . $$ This is because (for example) from $l_1\wedge l_1'=\bot$ and $l_2\wedge l_2'=\bot$ it does not necessarily follow that $(l_1\vee l_2)\wedge (l_1'\vee l_2')=\bot$, so we lose joins.

But we clearly have some meets and joins; notably $(l_1,l')\circ (l_2,l')$ still makes sense for $\circ\in\{\wedge,\vee\}$. In words: we have meets and joins when one component is fixed.

Has this construction, or a construction like it, been studied? Is there an algebraic theory for $|L\otimes L|$?

This question has arisen in topologies, where $L$ is the lattice of open sets and I am interested in understanding the structure of pairs of disjoint open sets.

Thank you.