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Prerequisites: Algebraic General Topology.

Conjecture. $$L \in \mathrel{\left[ f \right]} \Rightarrow \mathrel{\left[ f \right]} \cap \prod_{i \in \operatorname{dom} \mathfrak{A}} \operatorname{atoms} L_i \neq \emptyset$$ 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 proof:

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

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

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

Let $$b_c = a|_{X \left( c \right) \setminus \left\{ c \right\}} \cup L|_{\left( \operatorname{arity} f \right) \setminus X \left( c \right)}$$. Then $$a_c = \Phi \left\langle f \right\rangle_c b_c$$.

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

The only thing remained to prove is that $$\left\langle f \right\rangle_c b_c \neq 0$$

that is $$\langle f \rangle _ c  ( a|_{ X ( c ) \setminus \{ c \} } \cup L|_{( \operatorname{arity} f ) \setminus X ( c )} ) \neq 0$$ that is $$y \not\asymp \left\langle f \right\rangle_c b_c$$.