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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. Anedge is called containment minimal (or just minimal, for short) if it is not astrict superset of another edge. Yet, equal minimal edges may exist. Byduality,  (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 $O$ definesa unique lexicographic order $\succ_L$ over the $2^O$. Let $A$ be alexicographically 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 $o' \inB$.  This property has important applications in game theory implyingNash-solvability of tight game forms as shown in the old (1975 and 1989) workof 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 trivialif $\mathcal{A}$ and $\mathcal{B}$ are given explicitly. Yet, it is true evenif only $\mathcal{A}$ is given, and not explicitly, but by a polynomialcontainment oracle, which for a subset $O_A \subseteq O$ answers in polynomialtime whether $O_A$ contains an edge of $\mathcal{A}$.

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