}{{{\lambda }}},t_0,\dots,t_{\kappa-1})={\varrho }({\widetilde }{{{\lambda }}},\gamma(t_0),\dots,\gamma(t_{\kappa-1}))$$ and $$\bar{R}:\mathcal{F}(\Lambda,K) \to \mathcal{F}({\widetilde }{\Lambda},{\widetilde }{K})$$ by $$\bar{R}(f) = R(\theta\circ f).$$ Then for $f\in \mathcal{F}(\Lambda,K)$ and ${\widetilde }{{{\lambda }}}\in {\widetilde }{\Lambda}$, $$\begin{aligned}
(\bar{R}(f))({\widetilde }{{{\lambda }}})&=&(R(\theta\circ f))({\widetilde }{{{\lambda }}})
={\varrho }({\widetilde }{{{\lambda }}}, \theta\circ f(\eta_0({\widetilde }{{{\lambda }}})),\dots,\theta\circ f(\eta_{\kappa-1}({\widetilde }{{{\lambda }}})))\\
&=&\bar{{\varrho }}({\widetilde }{{{\lambda }}}, \beta\circ f(\eta_0({\widetilde }{{{\lambda }}})),\dots,\beta\circ f(\eta_{\kappa-1}({\widetilde }{{{\lambda }}}))).\end{aligned}$$ Thus, the mapping $\bar{R}$ is of the form needed to apply Corollary 1 of [@Hei01b]. Accordingly, considering ${\widetilde }{A}$ as a quantum algorithm from $\mathcal{F}({\widetilde }{\Lambda},{\widetilde }{K})$ to ${\widetilde }{G}$, we can find a quantum algorithm $A$ from $\mathcal{F}(\Lambda,K)$ to ${\widetilde }{G}$ with $n_q(A)=2\kappa n_q({\widetilde }{A})$ and $$A(f)={\widetilde }{A}(\bar{R}(f)) \quad (f\in \mathcal{F}(\Lambda,K)).$$ For ${\widetilde }{{{\lambda }}}\in {\widetilde }{\Lambda}_0$ we have $\eta_j({\widetilde }{{{\lambda }}})\in \Lambda_0$, and therefore, by assumption (\[U4\]), for $f\in F_0$, $$\begin{aligned}
(\bar{R}(f))({\widetilde }{{{\lambda }}})&=&
{\varrho }({\widetilde }{{{\lambda }}}, \theta\circ f(\eta_0({\widetilde }{{{\lambda }}})),\dots,\theta\circ f(\eta_{\kappa-1}({\widetilde }{{{\lambda }}})))\\
&=&{\varrho }({\widetilde }{{{\lambda }}}, (\Theta(f))(\eta_0({\widetilde }{{{\lambda }}})),\dots,(\Theta(f))(\eta_{\kappa-1}({\widetilde }{{{\lambda }}})))\\
&=& (R(\Theta(f)))({\widetilde }{{{\lambda }}}).\end{aligned}$$ This implies $$\label{U5}
{\widetilde }{Q}_{l,\bar{R}(f)}={\widetilde }{Q}_{l,R(\Theta(f))}   \quad (l=0,\dots,,k-1),$$ and consequently $$\label{U6}
A(f)={\widetilde }{A}(\bar{R}(f))={\widetilde }{A}(R(\Theta(f)))\quad (f\in F_0).$$ Now fix $f\in F_0$ and let $\zeta$ be a random variable with distribution $A(f)$. We have, by assumption (\[U2\]) $$\begin{aligned}
\label{U7}
\|S(f)-\Psi(\zeta)\|&\le& \|S(f)-S(\Theta(f))\|+\|S(\Theta(f))-\Psi(\zeta)\|\nonumber\\
&\le& \|S(\Theta(f))-\Psi(\zeta)\|+\delta.\end{aligned}$$ Furthermore, by (\[XG4\]), $$\begin{aligned}
