Minimality of selfadjoint dilation of operator knot of dissipative operator

Let $A$ is dissipative densely defined operator in the space $\mathfrak{H}$ and $-i \in \rho\left(A\right)$.
Let denote $R = {\left(A+iI\right)}^{-1}$ and consider the defect operators
\begin{gather*} B = iR - i{R}^{*} -2{R}^{*}R,\\ \widetilde{B} = iR - i{R}^{*} - 2R{R}^{*},\\ T = I - 2iR. \end{gather*}
A set of linear bounded operators acting from an entire Hilbert space ${H}_{1}$ into a Hilbert space ${H}_{2}$ will be denoted by $L\left({H}_{1}, {H}_{2}\right)$.
Definition. The assembly of Hilbert spaces $\mathfrak{H}$, ${E}_{-}$ и ${E}_{+}$ and operators $A: \mathfrak{H} \rightarrow \mathfrak{H}, \Phi \in L\left({E}_{-}, \mathfrak{H}\right), \Psi \in L\left(\mathfrak{H}, {E}_{+}\right), K \in L\left({E}_{-}, {E}_{+}\right)$ is called the operator knot, which has been introduced in work of U. L. Kudryashov «Selfadjoint dilation of operator knot of dissipative operator» in «Dynamic systems», 3(31), №1-2, 2013, p. 45-48${}^{\left[1\right]}$.
$Q = \left(A, \Phi, K, \Psi, \mathfrak{H}, {E}_{-}, {E}_{+}\right)$, if the following relations hold:
\begin{gather*} B = {\Psi}^{*}\Psi;\\ \widetilde{B} = {\Phi}{\Phi}^{*};\\ {T}^{*}\Phi + {\Psi}^{*}K = 0;\\ T{\Psi}^{*}+{\Phi}{K}^{*} = 0;\\ 2{\Phi}^{*}{\Phi}+{K}^{*}K = I;\\ 2{\Psi}{\Psi}^{*} + K{K}^{*} = I. \end{gather*}

Selfadjoint dilation $S$ of dissipative operator $A$ is constructed using the knot $\Theta$ in [1] in the following manner.
The spaces ${H}_{-} = {L}_{2} \left( \left( -\infty, 0 \right], {E}_{-} \right)$, ${H}_{+} = {L}_{2} \left( \left[0, +\infty\right) \right)$ and $H = {H}_{-} \oplus \mathfrak{H} \oplus {H}_{+}$ are considered.
The vector $h = \left({h}_{-}, {h}_{0}, {h}_{+}\right) \in \mathfrak{D}\left(S\right)$ if and only if

• $\left\{ {h}_{\pm}, \frac{{dh}_{\pm}\left(t\right)}{dt} \right\} \subset {H}_{\pm}$;
• $\widetilde{h} = {h}_{0} + \Phi {h}_{-} \left(0\right) \in \mathfrak{D}\left(A\right)$;
• ${h}_{+}\left(0\right) = -K{h}_{-} \left(0\right) + i\Psi \left(A+iI\right) \widetilde{h}$.

Theorem. The dilation $S$ is minimal, i.e.
$$H = \overline{span\left\{ {R}_{\pm i} \left(S\right) h \mid h \in \mathfrak{H}, n \in \left\{0\right\} \cup \mathbb{N} \right\}}$$
if the spaces ${E}_{+} = \overline{\Psi \mathfrak{H}}$, ${E}_{-} = \overline{{\Phi}^{*}\mathfrak{H}}$ are separable.
The following expressions was used for the proof:
$$ {R}^{n}_{-i}\left(S\right) \begin{pmatrix} 0\\ {h}_{0}\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ {a}_{n}\\ {b}_{n} \end{pmatrix}, $$
where $n \in \mathbb{N}$, ${a}_{n} = {R}^{n} {h}_{0}$,
$${b}_{n} = {e}^{-t} \sum\limits_{k=1}^{n} \frac{{t}^{n-k}}{\left(n-k\right)! {i}^{n-k-1}} \Psi {R}^{k-1} {h}_{0}.$$

$$ {R}^{n}_{i} \left(S\right) \begin{pmatrix} 0 \\ {h}_{0} \\ 0 \end{pmatrix} = \begin{pmatrix} {c}_{n} \\ {d}_{n} \\ 0 \end{pmatrix}, $$
where ${d}_{n} = {{R}^{*}}^{n} {h}_{0}$,
$${c}_{n} = {e}^{t} \sum\limits_{k=1}^{n} \frac{{t}^{n-k}}{\left(n-k\right)! {\left(-i\right)}^{n-k-1}} {\Phi}^{*} {{R}^{*}}^{k-1} {h}_{0},$$
where ${h}_{0} \in \mathfrak{H}$.