Change of Index Variable of Product

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Theorem

$\ds \prod_{\map R i} a_i = \prod_{\map R j} a_j$

where $\ds \prod_{\map R i} a_i$ denotes the product over $a_i$ for all $i$ that satisfy the propositional function $\map R i$.


Proof

Let $S = \set {i \in \Z: \map R i}$.

Let $T = \set {j \in \Z: \map R j}$.

Let $i \in S$.

Then $\map R i$.

Let $j = i$.

By Leibniz's Law, $\map R j$.

Thus $i \in T$.

By definition of subset:

$S \subseteq T$


Similarly, let $j \in T$.

Then $\map R j$.

Let $i = j$.

By Leibniz's Law, $\map R i$.

Thus $j \in S$.

By definition of subset:

$T \subseteq S$


Thus by definition of set equality:

$S = T$


Thus:

\(\ds \prod_{\map R i} a_i\) \(=\) \(\ds \prod a_i \sqbrk {\map R i}\) Definition of Iverson's Convention
\(\ds \) \(=\) \(\ds \prod a_i \times \chi_S\) Definition of Characteristic Function of Set
\(\ds \) \(=\) \(\ds \prod a_j \times \chi_T\) as $S = T$
\(\ds \) \(=\) \(\ds \prod a_j \sqbrk {\map R j}\) Definition of Iverson's Convention
\(\ds \) \(=\) \(\ds \prod_{\map R j} a_j\)

$\blacksquare$


Sources

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