Product of Products
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Theorem
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{R \paren i} x_i$ denote a continued product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
- $\ds \prod_{R \paren i} \paren {b_i c_i} = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}$
Proof
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
\(\ds \prod_{R \paren i} \paren {b_i c_i}\) | \(=\) | \(\ds \prod_{R \paren i} \paren {a_{i 1} a_{i 2} }\) | by definition | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{R \paren i} \paren {\prod_{1 \mathop \le j \mathop \le 2} a_{i j} }\) | Definition of Continued Product by Propositional Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{1 \mathop \le j \mathop \le 2} \paren {\prod_{R \paren i} a_{i j} }\) | Exchange of Order of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\prod_{R \paren i} a_{i 1} } \paren {\prod_{R \paren i} a_{i 2} }\) | Definition of Continued Product by Propositional Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}\) | by definition |
$\blacksquare$
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: Exercise $22$