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Let $\mathfrak{A}$ be a bounded lattice. I call a $2$-staroid a relation $f\in\mathscr{P}(\mathfrak{A}\times\mathfrak{A})$ such that $i\sqcup j \mathrel{f} b\Leftrightarrow i\mathrel{f} b\vee j\mathrel{f} b$ and $a \mathrel{f} i\sqcup j\Leftrightarrow a\mathrel{f} i\vee a\mathrel{f} j$ and $\neg(a\mathrel{f}0)$ and $\neg(0\mathrel{f}b)$ for every $a,b,i,j\in\mathfrak{A}$.

Here I denote $\sqcup$ the join on the lattice $\mathfrak{A}$ and $0$ its least element.

Let $\mathfrak{F}$ be the lattice of all filters (including the improper filter), ordered reversely to set-theoretic inclusion of filters, on the lattice $\mathscr{P}U$ for some set $U$.

I denote $\mathfrak{P}$ the lattice of principal filters (ordered reversely to set-theoretic inclusion of filters).

Conjecture: There are at least two different $2$-staroids $f$ and $g$ on $\mathfrak{F}$ such that $f\cap(\mathfrak{P}\times\mathfrak{P}) = g\cap(\mathfrak{P}\times\mathfrak{P})$.

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I've proved it in http://www.mathematics21.org/binaries/staroids2.pdf draft article. (There I name $2$-staroids as pseudofuncoids.)

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