Lexicographically maximal edges of dual hypergraphs and Nash-solvability of tight game forms

Let $\mathcal{A} = \{A_1, \ldots, A_m\}$ and $\mathcal{B} = \{B_1, \ldots, B_n\}$ be a pair of dual multi-hypergraphs on the common ground set $O = \{o_1, \ldots, o_k\}$. Note that each of them may have embedded or equal edges. An edge is called containment minimal (or just minimal, for short) if it is not a strict superset of another edge. Yet, equal minimal edges may exist. By duality, (i) $A \cap B \neq \emptyset$ for every pair $A \in \mathcal{A}$ and $B \in \mathcal{B}$; (ii) if $A$ is minimal then for every $o \in A$ there exists a $B \in \mathcal{B}$ such that $A \cap B = \{o\}$. We will extend claim (ii) as follows. A linear order $\succ$ over $O$ defines a unique lexicographic order $\succ_L$ over the $2^O$. Let $A$ be a lexicographically maximal (lexmax) edge of $\mathcal{A}$. Then, (iii) $A$ is minimal and for every $o \in A$ there exists a minimal $B \in \mathcal{B}$ such that $A \cap B = \{o\}$ and $o \succeq o'$ for each $o' \in B$. This property has important applications in game theory implying Nash-solvability of tight game forms as shown in the old (1975 and 1989) work of the first author. Here we give a new, very short, proof of (iii). Edges $A$ and $B$ mentioned in (iii) can be found out in polynomial time. This is trivial if $\mathcal{A}$ and $\mathcal{B}$ are given explicitly. Yet, it is true even if only $\mathcal{A}$ is given, and not explicitly, but by a polynomial containment oracle, which for a subset $O_A \subseteq O$ answers in polynomial time whether $O_A$ contains an edge of $\mathcal{A}$.