I proposed this open problem for the next polymath project. Now I will consider some its special simple cases.

First, it is quite obvious that every strong partitioning is a weak partitioning.

If (finite) meets of our lattice are distributive over arbitrary joins (or, equivalently, our lattice is brouwerian) then the reverse implication holds, that is a weak partitioning is a strong partitioning.

Proof: Let $latex S$ is a weak partitioning, let $latex A,B\in\mathscr{P}\mathfrak{A}$ and $latex A\cap B=0$. Then

$latex \bigcup^{\mathfrak{A}} A \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} B =\bigcup^{\mathfrak{A}} \left\{ \left(\bigcup^{\mathfrak{A}} A \right) \cap^{\mathfrak{A}} Y | Y \in B \right\} \subseteq\bigcup^{\mathfrak{A}} \left\{ \bigcup^{\mathfrak{A}} (S \setminus Y) \cap^{\mathfrak{A}} Y | Y \in B \right\} =\bigcup^{\mathfrak{A}} \left\{ 0 \right\} = 0$
It was taken into account that $latex S\setminus Y\supseteq A$.

So $latex S$ is a strong partitioning. QED

This does not solve the problem for me, because I want to prove it for the cases when our complete lattice is not brouwerian but is dual that is co-brouwerian. Co-brouwerian lattices appear as lattices of filter objects in my manuscript about filters and as lattices of funcoids in my texts about funcoids and reloids; it is simple to prove that for infinite set it is not brouwerian.

Further: it is trivial that every finite distributive lattice is brouwerian. From this follows that weak and strong partitioning coincide for finite distributive lattices.

A natural question to ask: Are there non-distributive finite lattices for which weak and strong partitioning do not coincide?

Solving this question (what I have not yet attempted) would involve some combinatorics. If that shall go complex we may use a computer experiment enumerating many lattices. But I hope that the solution to this tiny question is simpler.

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