Product of Products over Overlapping Domains

Theorem

Let $R: \Z \to \set {\T, \F}$ and $S: \Z \to \set {\T, \F}$ be propositional functions on the set of integers.

Let $\ds \prod_{\map R i} x_i$ denote a continued product over $R$.

Then:

$\ds \prod_{\map R j} a_j \prod_{\map S j} a_j = \paren {\prod_{\map R j \mathop \lor \map S j} a_j} \paren {\prod_{\map R j \mathop \land \map S j} a_j}$

where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively.

Proof

\(\ds \prod_{\map R j} a_j \times \prod_{\map S j} a_j\)

\(=\)

\(\ds \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map R j} } \times \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map S j} }\)

Definition of Continued Product by Iverson's Convention

\(\ds \)

\(=\)

\(\ds \prod_{j \mathop \in \Z} a_j^{\paren {\sqbrk {\map R j} + \sqbrk {\map S j} } }\)

Product of Powers

Let:

$A := \set {j \in \Z: \map R j}$

$B := \set {j \in \Z: \map S j}$

The result then follows from Cardinality of Set Union.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products