of basis operators have the following properties. First, the operator $Y_1$ in this set are identical with ${\cal P}_{|0\rangle}$ and hence also satisfies the interesting properties described by Eq. (\[Y1property\]). Second, $$\begin{aligned}
[Y_{j},Y_{1}]&=&0,\ {\rm for}\ j=1,2,\cdots,10; \nonumber \\
\{Y_{j},Y_{1}\}_{+}&=&0, \ {\rm for}\ j=11,12,\cdots,16,\end{aligned}$$ where $[\cdot]$ represents the commutator and $\{\cdot\}_{+}$ represents an anti-commutator. Third, $$\begin{aligned}
\left[\sum_{i=1}^{10}A_{i}Y_{i},
\sum_{j=11}^{16}B_{j}Y_{j}\right]&=&\sum_{j=11}^{16}C_{j}Y_{j},
\nonumber \\ \left(\sum_{i=1}^{10}A_{i}Y_{i}\right) \left(
\sum_{j=1}^{10}B_{j}Y_{j}\right)&=&\sum_{j=1}^{10}C_{j}Y_{j},
\nonumber \\ \left(\sum_{i=11}^{16}A_{i}Y_{i}\right)
\left(\sum_{j=11}^{16}B_{j}Y_{j}\right)&=&\sum_{j=1}^{10}C_{j}Y_{j}.
\label{antic square}\end{aligned}$$ With these observations, we next split the total uncontrolled Hamiltonian into two terms, i.e., $H=H_{0}+H'$, where $$\begin{aligned}
H_{0}=W_{1}Y_{1}+W_{2}Y_{2}+\cdots +W_{10}Y_{10}, \end{aligned}$$ and $$\begin{aligned}
H'=W_{11}Y_{11}+\cdots +W_{16}Y_{16}. \end{aligned}$$ Evidently, we have the anti-commuting relation $$\begin{aligned}
\{Y_1,H'\}_{+}=0, \label{antic}\end{aligned}$$ an important fact for our proof below.

Consider now the following control Hamiltonian describing a sequence of extended UDD $\pi$-pulses $$\begin{aligned}
\label{control1}
H_{c}=\sum_{j=1}^{N}\pi \delta(t-T_j)\frac{Y_1}{2}.\end{aligned}$$ After the $N$ control pulses, the unitary evolution operator for the whole system of the two qubits plus a bath is given by ($\hbar=1$ throughout) $$\begin{aligned}
{U}(T) & = & e^{-i[H_{0}+H'](T-T_N)} (-iY_1) \nonumber \\
& &\ \times\ e^{-i[H_{0}+H'](T_N-T_{N-1})} (-iY_1) \nonumber \\
& &\ \cdots \nonumber \\
 & &\ \times\  e^{-i[H_{0}+H'](T_3-T_2)} (-iY_1) \nonumber \\
& &\ \times\ e^{-i[H_{0}+H'](T_2-T_1)} (-iY_1)\nonumber \\
& &\ \times\ \ e^{-i[H_{0}+H']T_1}.
\label{Ut}
  \end{aligned}$$ We can then take advantage of the anti-commuting 