A new question about maximal independent sets in regular graphs

Question

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:

Question Suppose 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".

Answer 1

As Joro said, there is a counterexample for your question. In fact, C. Berge and C. Payan (independently) conjectured that any regular graph has two disjoint maximal independent sets of vertices, and this conjecture has shown to be false for some regular graphs. See the following papers:

1. C. Payan, A counter-example to the conjecture: "Every nonempty regular simple graph contains two disjoint maximal independent sets", Graph Theory Newsletter 6 (1977) 7-8.

2. C. Payan, Coverings by minimal transversals, Discrete Math. 23 (1978) 273-277.

Note that this conjecture has also positive answer in some classes of regular graphs!