e{R}}$.

From ${{\Psi}\negmedspace\mid\negmedspace{R}} \subseteq \Phi$ and $\Phi \subseteq {{\Psi}\negmedspace\mid\negmedspace{R}}$ we infer $\Phi = {{\Psi}\negmedspace\mid\negmedspace{R}}$.

\[L:SUBSET\_SUBSPACE\] Let $\Psi$ and $\Phi$ be ensembles. If $\Phi \subseteq \Psi$, then ${\prod{\Phi}}$ is a subspace of ${\prod{\Psi}}$.

Set $R = {{\operatorname{dom}{\Phi}}}$. Since $\Phi \subseteq \Psi$ by hypothesis, then by applying lemma \[L:SUBSET\_RESTRICTION\] we infer $\Phi = {{\Psi}\negmedspace\mid\negmedspace{R}}$. With this equality and Theorem \[T:SPACE\_UNIQ\_ENSEMBLE\] (invertibility of the Cartesian product), we have ${\prod{\Phi}} = \prod ({{\Psi}\negmedspace\mid\negmedspace{R}})$. Theorem \[T:CHC\_RSTR\_EQ\_RSTR\_CHC\] asserts that 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 ${\prod{\Phi}} = (\thinspace\prod\Psi) \mid R$. This last equality is exactly the premise of definition \[D:SUBSPACE\]: ${\prod{\Phi}}$ is a subspace of ${\prod{\Psi}}$.

\[T:SUBSET\_IFF\_SUBSPACE\] Let $\Psi$ and $\Phi$ be ensembles. ${\prod{\Phi}}$ is a subspace of ${\prod{\Psi}}$ if and only if $\Phi \subseteq \Psi$.

Lemma \[L:SUBSPACE\_SUBSET\] asserts that if ${\prod{\Phi}}$ is a subspace of ${\prod{\Psi}}$, then $\Phi \subseteq \Psi$. Lemma \[L:SUBSET\_SUBSPACE\] asserts that if $\Phi \subseteq \Psi$, then ${\prod{\Phi}}$ is a subspace of ${\prod{\Psi}}$. This pair of converse implications establishes the biconditional.

\[L:ENSEMBLE\_PROD\_SUBSETS\] If $\Upsilon$, $\Psi$, and $\Phi$ are ensembles such that $\Upsilon = \Psi\Phi$, then $\Psi \subseteq \Upsilon$ and $\Phi \subseteq \Upsilon$.

Since $\Upsilon$ is the dyadic product of $\Psi$ and $\Phi$, then by definition \[D:DYADIC\_ENSEMBLE\_PRODUCT\], $\Psi$ and $\Phi$ are disjoint ensembles and $\Upsilon = \Psi \cup \Phi$.

Suppose $i \in {{\operatorname{dom}{\Upsilon}}} = {{\operatorname{dom}{(\Psi \cup \Phi)}}}$. Through de