ran}{\Phi}}}$.

\[D:DYADIC\_SPACE\_PRODUCT\] Let $\Theta$ and $\Phi$ be disjoint ensembles. The dyadic space product of $\prod\Theta$ and $\prod\Phi$ is $$\prod \Theta \prod \Phi = \prod \Theta\Phi$$ (that is, with $\Upsilon = \Theta\Phi$, the set of all $\Upsilon$-choices).

\[D:DYADIC\_CHOICE\_PROD\] Let $\Theta$ and $\Phi$ be disjoint ensembles generating choice spaces $\prod\Theta$ and $\prod\Phi$. Suppose $\alpha \in \prod\Theta$ and $\beta \in \prod\Phi$ are choices. Their dyadic *choice* product $(\alpha,\beta) \mapsto \alpha\beta$ is $$\alpha\beta = \alpha \cup \beta.$$

\[T:DYADIC\_CHOICE\_PROD\] Let $\Theta$ and $\Phi$ be disjoint ensembles. The dyadic choice product is a bijection $$\prod\Theta \times \prod\Phi \leftrightarrow \prod\Theta\Phi.$$

Suppose $(\theta, \phi) \in \prod\Theta \times \prod\Phi$, whence it follows that choices $\theta \in \prod\Theta$ and $\phi \in \prod\Phi$. Consider the dyadic products $\theta\phi = \theta \cup \phi$ and $\Theta\Phi = \Theta \cup \Phi$. Since ${{\operatorname{dom}{\theta}}} = {{\operatorname{dom}{\Theta}}} = T$ and ${{\operatorname{dom}{\phi}}} = {{\operatorname{dom}{\Phi}}} = P$, then ${{\operatorname{dom}{\theta\phi}}} = T \cup P = {{\operatorname{dom}{\Theta\Phi}}}$. Let $i \in {{\operatorname{dom}{\theta\phi}}} = T \cup P$. Since $\Theta$ and $\Phi$ are disjoint, then exactly one of two cases holds: either $i \in T$ and $i \notin P$, or $i \notin T$ and $i \in P$.

In the first case, $i \in T$, $\theta\phi(i) = \theta(i)$ and $\Theta\Phi(i) = \Theta(i)$. Since $\theta$ is a $\Theta$-choice, then $\theta(i) \in \Theta(i)$. But since $\theta\phi(i) = \theta(i)$ and $\Theta\Phi(i) = \Theta(i)$, then $\theta\phi(i) \in \Theta\Phi(i)$ for any $i \in T$.

The second case is similar and leads to $\theta\phi(i) \in \Theta\Phi(i)$ for any $i \in P$. From these two cases we conclude that $\theta\phi(i) \in \Theta\Phi(i)$ for any $i \in T \cup P = {{\operatorname{dom}{\theta\phi}}}$, and that $\theta\phi \in \prod\Theta\Phi$.

The preceding establishes that the cho