Open problem: co-separability of filter objects

Conjecture Let a and b are filters on a set U. Then

a\cap b = \{U\} \Rightarrow \\ \exists A,B\in\mathcal{P}U: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U). [corrected]

This conjecture can be equivalently reformulated in terms of filter objects:

Conjecture Let \mathcal{A} and \mathcal{B} are filter objects on a set U. Then

\mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A,B\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge B\subseteq\mathcal{B} \wedge A\cup B=U).

(where \mathfrak{F} is the set of filter objects on U).

It is simple to show (by applying the above conjecture twice) that it is equivalent to a yet simpler conjecture

Conjecture Let \mathcal{A} and \mathcal{B} are filter objects on a set U. Then

\mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge A\cup^{\mathfrak{F}}\mathcal{B}=U).

Put in yet simpler words, this conjecture can be formulated: the filtrator of filters on a set is with co-separable center.

Looking innocent? Indeed I currently don’t know how to attack this looking simple problem. Maybe you will help?

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