Multifuncoid has atomic arguments

A counter-example against this conjecture have been found.See http://www.mathematics21.org/algebraic-general-topology.htmlAlgebraic General Topology.

Prerequisites: http://www.mathematics21.org/algebraic-general-topology.htmlAlgebraic General Topology.

Conjecture. $L\in \left[f\right]\Rightarrow \left[f\right]\cap {\prod }_{i\in \mathrm{dom}𝔄}\mathrm{atoms}{L}_i\ne \mathrm{\varnothing }$ for every pre-multifuncoid $f$ of the form whose elements are atomic posets.

A weaker conjecture: It is true for forms whose elements are powersets.

The following is an attempted (partial) proof:

If $\mathrm{arity}f=0$ our theorem is trivial, so let $\mathrm{arity}f\ne 0$. Let $⊑$ is a well-ordering of $\mathrm{arity}f$ with greatest element $m$.

Let $\mathrm{\Phi }$ is a function which maps non-least elements of posets into atoms under these elements and least elements into themselves. (Note that $\mathrm{\Phi }$ is defined on least elements only for completeness, $\mathrm{\Phi }$ is never taken on a least element in the proof below.) brown [TODO: Fix the ”universal set” paradox here.]

Define a transfinite sequence $a$ by transfinite induction with the formula $a_c=\mathrm{\Phi }{⟨f⟩}_c\left({a|}_{X\left(c\right)\setminus \left\{c\right\}}\cup {L|}_{\left(\mathrm{arity}f\right)\setminus X\left(c\right)}\right)$.

Let $b_c={a|}_{X\left(c\right)\setminus \left\{c\right\}}\cup {L|}_{\left(\mathrm{arity}f\right)\setminus X\left(c\right)}$. Then $a_c=\mathrm{\Phi }{⟨f⟩}_c{b}_c$.

Let us prove by transfinite induction $a_c\in \mathrm{atoms}{L}_c.$ $a_c={\mathrm{\Phi }{⟨f⟩}_{c}L|}_{\left(\mathrm{arity}f\right)\setminus \left\{c\right\}}⊑{{⟨f⟩}_{c}L|}_{\left(\mathrm{arity}f\right)\setminus \left\{c\right\}}$. Thus $a_c⊑L_c$. [TODO: Is it true for pre-multifuncoids?]

The only thing remained to prove is that ${⟨f⟩}_c{b}_c\ne 0$

that is ${⟨f⟩}_c({a|}_{X(c)\setminus \{c\}}\cup {L|}_{(\mathrm{arity}f)\setminus X(c)})\ne 0$ that is $y\not\asymp {⟨f⟩}_c{b}_c$.