Example of Boolean Algebra that satisfies distributive law but violates complete distributive law More precisely, I'm interested to know the example of Boolean Algebra $B$, such that for any $a, b, c \in B$, $a \cap (b \cup c) = (a \cap b) \cup (a \cap c)$, but there exists $\{ P_{ij}:i\in I, j \in J\} \subseteq B$, $\bigwedge_{i \in I}\vee_{j \in J}P_{ij} \neq \bigvee_{a \in J^I}\wedge_{i \in I}P_{ia(i)}$.
Added: As reminded by Hagen von Eitzen's comment, I'm also interested in whether Axiom of Choice plays a role in complete distributive law. Is it really necessary to well-order all the functions of $J^I$in the RHS of  complete distributive law to make it meaningful? In the finite case, we usually specify an order of subformulaes in order to calculte it. Is it true for infinite operations?
In our case, the problem boils down to whether $I$ and $J$ are well-ordered, since we can define a lexicographic order in $J^I$ induced by the order of $I$ and $J$..
 A: This is essentially taken from the Handbook of Boolean Algebras (vol.1) (Example 14.3, p.214).
Consider the partial order $P$ consisting of all finite partial functions $p : \omega \to \omega$, and let $B$ denote the regular-open algebra of $P$ (giving $P$ the partial-order topology).  This is a complete Boolean algebra.
Fixing $W \subseteq \omega$ of size $\omega$, for each $i \in W$ define $$\begin{gather}
b_{i,0} = \{ p \in P : i \in \mathrm{dom} (p) , p(i) = 0 \} \\
b_{i,1} = - b_{i,0} = \{ p \in P : i \in \mathrm{dom} (p), p(i) \neq 0 \}
\end{gather}$$  Clearly $b_{i,0} + b_{i,1} = 1$ for all $i$, and therefore $$\prod_{i \in W} \sum_{n < 2} b_{i,n} = 1.$$
Suppose, however, that $f : W \to 2$ is given.  If $\prod_{i \in W} b_{i , f(i)} \neq 0$ then there is a $p \in P$ such that $p \in b_{i , f(i)}$ for all $i \in W$, which is impossible as $p$ must have finite domain.  Therefore $\prod_{i \in W} b_{i , f(i)} = 0$ for all $f : W \to 2$, and thus $\sum_f \prod_{i \in W} b_{i , f(i)} = 0$.
