### A surprisingly easy proof of yesterday conjecture

I have found a surprisingly easy proof of this conjecture which I proposed yesterday. Theorem Let $latex S$ be a set of binary relations. If for every $latex X, Y \in S$ we have $latex \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$ then…

Conjecture Let $latex S$ be a set of binary relations. If for every $latex X, Y \in S$ we have $latex \mathrm{up} (X \sqcap^{\mathsf{FCD}} Y) \subseteq S$ then there exists a funcoid $latex f$ such that $latex S = \mathrm{up}\, f$.
The converse of this theorem does not hold. Counterexample: Take $latex S = \mathrm{up}\, \mathrm{id}^{\mathsf{FCD}}_{\Omega}$. We know that $latex S$ is not a filter base. But it is trivial to prove that $latex S$ is a base of the funcoid $latex \mathrm{id}^{\mathsf{FCD}}_{\Omega}$.
Definition A set $latex S$ of binary relations is a base of a funcoid $latex f$ when all elements of $latex S$ are above $latex f$ and $latex \forall X \in \mathrm{up}\, f \exists T \in S : T \sqsubseteq X$. It…