of order $n$ with no isolated vertex, then ${{\rm fd}}(G) \le n - 2$. This establishes the base cases. Let $n \ge 5$ and assume that every graph $G'$ of order $n'$, where $3 \le n' < n$, with no isolated vertex satisfies ${{\rm fd}}(G') \le n' - 2$. Let $G = (V,E)$ be a graph of order $n$ with no isolated vertex. Every graph on at least two vertices has two vertices of the same degree. Let $u$ and $v$ be two vertices in $G$ with the same degree. Then, $d_G(u) = d_G(v) = k$ for some $k$, where $1 \le k \le n-1$. If $u$ and $v$ are not adjacent in $G$ or if $u$ and $v$ are adjacent in $G$ and $k \ge 2$, then $D = V
\setminus \{u,v\}$ is a FD-set, and so ${{\rm fd}}(G) \le |D| = n - 2$, as desired. Hence we may assume that $u$ and $v$ are adjacent in $G$ and $k = 1$. We now consider the graph $G' = G - \{u,v\}$ of order $n' =
n - 2 \ge 3$. Since $G$ has no isolated vertex, neither does $G'$. Applying the inductive hypothesis to $G'$, we have 