Funcoid bases/Proving existence of funcoid through lattice Gamma

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Statement Under the conditions "1" or "2" of our conjecture, we have that $S=\operatorname{up}\bigwedge^{\mathsf{FCD}} S=\operatorname{up}\bigwedge^{\mathsf{FCD}} S'$.

Proof Denote $S' = S \cap \Gamma$. It's easy to show that $S'$ is a filter on $\Gamma$ (in my book it's proved that $\Gamma$ is a sublattice of $\mathsf{FCD}$).

Let's denote $f = \bigwedge^{\mathsf{FCD}} S$.

$\operatorname{up} f \supseteq S$ is obvious. It remains to prove $Y \in \operatorname{up} f \Rightarrow Y \in S$.

For every ultrafilter $x$ we have

$\operatorname{up} \langle Y \rangle x \subseteq \operatorname{up} \left\langle \bigwedge^{\mathsf{FCD}} S' \right\rangle x = \\ \operatorname{up} \bigwedge_{X \in S'} \langle X \rangle x = \\ \left\{ \langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x \mid i = 0, \ldots, n, X_i \in S' \right\} = \\ \left\{ \langle X \rangle x \mid X \in S' \right\}$

(taken into account that $\langle X_i\rangle x$ is a principal filter).

It follows $\langle Y \rangle x \in \left\{ \langle X \rangle x \mid X \in S' \right\}$ that is

$\langle Y \rangle x = \langle X \rangle x$ for some $X \in S'$.

The next question is whether we can find such $X \in S$ (not necessarily $X \in S'$!) that $\langle Y \rangle x = \langle X \rangle x$ for all ultrafilters $x$. Or is there a counter-example for existence of such $X$?