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 is a weak partitioning, let and . Then
It was taken into account that .
So 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.