${\prod{\Phi}} = (\thinspace{\prod{\Psi}}) \mid R$. By Theorem \[T:CHC\_RSTR\_EQ\_RSTR\_CHC\], the restriction of the choice space equals the choice space of the restriction: $(\thinspace\prod\Psi) \mid R = \prod ({{\Psi}\negmedspace\mid\negmedspace{R}})$. Transitivity of equality implies ${\prod{\Phi}} = \prod ({{\Psi}\negmedspace\mid\negmedspace{R}})$. Then, by Theorem \[T:SPACE\_UNIQ\_ENSEMBLE\] (invertibility of the Cartesian product), $\Phi = {{\Psi}\negmedspace\mid\negmedspace{R}}$.

Suppose term $(i,P) \in \Phi$. Since $\Phi = {{\Psi}\negmedspace\mid\negmedspace{R}}$, then $(i,P) \in {{\Psi}\negmedspace\mid\negmedspace{R}}$. By the definition of restriction, this implies both $(i,P) \in \Psi$ and $i \in R$. Since $(i,P) \in \Phi$ implies $(i,P) \in \Psi$, we conclude $\Phi \subseteq \Psi$.

\[L:SUBSET\_RESTRICTION\] Let $\Psi$ and $\Phi$ be ensembles. If $\Phi \subseteq \Psi$ and $R = {{\operatorname{dom}{\Phi}}}$, then $\Phi = {{\Psi}\negmedspace\mid\negmedspace{R}}$.

Consider $(i,P) \in {{\Psi}\negmedspace\mid\negmedspace{R}}$. It then follows from the definition of restriction that $(i,P) \in \Psi$ and $i \in R$. But $R = {{\operatorname{dom}{\Phi}}}$, so $i \in {{\operatorname{dom}{\Phi}}}$. This implies there exists $(i,Q) \in \Phi$. Since $\Phi \subseteq \Psi$, then $(i,Q) \in \Psi$. Since $\Psi$ is a mapping, then $(i,P) \in \Psi$ and $(i,Q) \in \Psi$ implies $P = Q$. From $P = Q$ and $(i,Q) \in \Phi$, we infer that $(i,P) \in \Phi$. Thus $(i,P) \in {{\Psi}\negmedspace\mid\negmedspace{R}}$ implies $(i,P) \in \Phi$, so ${{\Psi}\negmedspace\mid\negmedspace{R}} \subseteq \Phi$.

Next suppose $(i,P) \in \Phi$. From this it follows that $i \in R = {{\operatorname{dom}{\Phi}}}$. From the premises $(i,P) \in \Phi$ and $\Phi \subseteq \Psi$ we conclude $(i,P) \in \Psi$. Together $(i,P) \in \Psi$ and $i \in R$ imply that $(i,P) \in {{\Psi}\negmedspace\mid\negmedspace{R}}$. Thus $(i,P) \in \Phi$ implies $(i,P) \in {{\Psi}\negmedspace\mid\negmedspace{R}}$, so $\Phi \subseteq {{\Psi}\negmedspace\mid\negmedspac