Problem with understanding the graph formulation of union-closed sets conjecture. I have read
"The graph formulation of the union-closed sets conjecture" by Henning Bruhn,  Pierre Charbit, Oliver Schaudt and Jan Arne Telle,  but i don't understand the main proof.
Let's recall:
Union-closed sets conjecture: Let $\mathcal{F}$ be a finite family of finite sets, such if $A,B\in \mathcal{F}$, then $A\cup B\in \mathcal{F}$. The conjecture states that there exists an element $a\in U:=\bigcup_{A\in \mathcal{F}} A$, so that the number of sets $A\in \mathcal{F}$, where $a$ occurs is equal or greater than $\frac{|\mathcal{F}|}{2}$.
Here is the another conjecture:
Graph formulation of union-closed sets conjecture: Let $G$ be a finite bipartite graph with at least one edge. Then each of the two bipartition classes contains a vertex belonging to at most half of the maximal stable sets.
The vertex that is contianed to at most half of the maximal stable sets is called a rare vertex.
The authors of this paper claim that these two conjectures are indeed equivalent. I was reading the proof that second conjecture implies the first conjecture and i came across with following difficulties:
In this proof we construct a bipartite graph, where first bipartition class is an universum(it is $U$), and the second class is a family ($\mathcal{F}$). Further, it is written that it is sufficient to find only one rare vertex. Here is the problem:

In order to prove the conjecture do we need to find two rare vertices or just one rare vertex?

Thank you in advance for any clarification.
 A: Maybe this will make it clear. Another equivalent formulation of the graph conjecture is
Graph conjecture v2.0. Let $G$ be a finite bipartite graph with at least one edge; let $A$ and $B$ be the bipartition. Then $A$ contains a rare vertex.
This is equivalent to the union-closed sets conjecture, in just the way described in Theorem 6 of the paper. We let $A$ be the side corresponding to elements, and $B$ be the side corresponding to sets in our family.
However, using graph conjecture v2.0, we can actually prove the existence of two rare vertices, one in each part! First, we can feed in $G$ with bipartition $A \cup B$, and get a rare vertex in $A$. Second, we can feed in $G$ with bipartition $B \cup A$, and get a rare vertex in $B$. In other words, graph conjecture v2.0 is equivalent to the graph conjectures in the paper.

The following conjecture would not be good enough:
Inferior graph conjecture.  Let $G$ be a finite bipartite graph with at least one edge. Then $G$ has a rare vertex.
First of all, this is not a conjecture; we can prove it. Let $vw$ be the edge. Then either $v$ is rare or else $v$ is contained in more than half of all independent sets; in that case, $w$ cannot be in any of those independent sets, so $w$ is rare.
Second, we cannot apply the inferior graph conjecture to prove the union-closed sets conjecture. If we construct the bipartite graph with elements on one side and sets on the other, as before, the inferior graph conjecture might end up giving us a rare vertex on the "wrong side" of the graph: a rare set, rather than a rare element! This does not help us in any way.
