This is a question inspired by "A question about independent set in regular graphs".

Suppose that $G$ is a simple $r$-regular graph with $n$ vertices. We say $H$ is a dominating set for $T$, if for every vertex $v\in T$, we have $v\in H$ or there is a vertex $u\in H$ such that $vu\in E(G)$.

Can any one prove or disprove the following:

QuestionSuppose that $G$ is an $r$-regular graph with $r\neq 0$. Is there a maximal independent set $T$ in $G$, such that there exists an independent dominating set $H$ for $T$ such that $T \cap H =\emptyset$?

We Know the following facts:

**Fact 1.** If $G$ is a graph, $G\neq \overline{K_{n}}$ and $T_{1}$ is an independent set of $G$, then there exists $T_{2}$ such that, $T_{2}$ is an independent dominating set for $T_{1}$ and
$ \vert T_{1} \cap T_{2}\vert \leq \frac{2\Delta(G) -\delta(G) }{2\Delta(G)} \vert T_{1}\vert$.

**Fact 2.** For each $m\in\mathbb N$ there is a connected cubic graph $G=(V,E)$ of order $|V|=n=50m$, and there is a maximal independent set $T\subseteq V$, with $|T|=14m$, such that, for any independent set $H\subseteq V$ which is dominating for $T$, we have $|T\cap H|\ge2m=\dfrac1{25}n$.

**EDIT:** Fact 3. There is a counterexample for the following claim:" Every nonempty regular simple graph contains two disjoint maximal independent sets".