I’ve proved the following lemma:

Lemma Let for every $X, Y \in S$ and $Z \in \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y)$ there is a $T \in S$ such that $T \sqsubseteq Z$.
Then for every $X_0, \ldots, X_n \in S$ and $Z \in \mathrm{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n)$ there is a $T \in S$ such that $T \sqsubseteq Z$.

I spent much time (probably a few hours) to prove it, but the found proof is really simple, almost trivial.

The proof is currently located in this PDF file.

After prayer in tongues and going down anointment of Holy Spirit I proved this conjecture about funcoids. The proof is currently located in this PDF file. Well, the proof is for special cases of distributive lattices, but more general case seems not necessary (at least now).

It seems easy to generalize it for more general lattices than the lattice of funcoids, what I hope to do a little later.

New conjecture:

Conjecture $\mathrm{up} (f \sqcap^{\mathsf{FCD}} g) \subseteq \{ F \sqcap G \mid F \in \mathrm{up}\, f, G \in \mathrm{up}\, g \}$ for all funcoids $f$, $g$ (with corresponding sources and destinations).

Looks trivial? But how to (dis)prove it?

In this draft I present some definitions and conjectures on how to generalize filter bases for more general filtrators (such as the filtrator of funcoids). This is a work-in-progress.

This seems an interesting research by itself, but I started to develop it as a way to prove this conjecture.

I claimed that I have proved the following conjecture:

Conjecture $\forall H \in \mathrm{up} (g \circ f) \exists F \in \mathrm{up}\, f, G \in \mathrm{up}\, g : H \sqsupseteq G \circ F$ for every composable funcoids $f$ and $g$.

The proof was with an error. So it remains a conjecture.

I have proved the following theorem:

Theorem $\forall H \in \mathrm{up} (g \circ f) \exists F \in \mathrm{up}\, f, G \in \mathrm{up}\, g : H \sqsupseteq G \circ F$ for every composable funcoids $f$ and $g$.

This theorem (being a conjecture at that time) was used in my alternative proof of Urysohn’s lemma. Now the proof is complete.

See my site for details.