a 1FD-set of $H_n$. Since $x_1, x_n
\in N(y_1) \cap S$, it follows that $y_1 \in S$. Hence, as $x_n, y_1
\in N(x_i) \cap S$ for any $1< i < n$, we have that $X \subseteq S$. This implies also that $y_j \in S$ for any $1 <j < n$ and thus $V
\setminus \{y_n\} \subseteq S$ and therefore $|S| \ge \nH-1$.

*Case I.3. $x_1, y_n \in S$.* We divide this case in three parts.

*(i) Suppose that $y_1 \notin S$.* Then $y_j \in S$ for every $1 < j < n$, otherwise $|N(y_1) \cap S| > |N(y_j) \cap S|$ since $N(y_j) \subset N(y_1)$ and $x_1 \in N(y_1) \setminus N(y_j)$. As $N(x_i) \varsubsetneqq N(x_{i+1})$ for all $1 < i < n-1$, it follows that there is at most one index $1 < i < n$ such that $x_i
\notin S$, implying that $|S| \ge \nH-2$.

*(ii) Suppose that $y_1 \in S$ and $x_n \notin S$.* Then as $N(x_i) \subset N(x_n)$ and $y_n \in S \cap (N(x_n) \setminus
N(x_i))$ for each $1 < i < n$, it follows that $X \setminus \{x_n\}
\subset S$. Hence, as every $y_j$ has a different number of neighbors in $X \setminus \{x_n\}$, $y_j \in S$ 