Summation of Summation over Divisors of Function of Two Variables

Theorem
Let $c, d, n \in \Z$.

Then:
 * $\displaystyle \sum_{d \mathop \backslash n} \sum_{c \mathop \backslash d} f \left({c, d}\right) = \sum_{c \mathop \backslash n} \sum_{d \mathop \backslash \left({n / c}\right)} f \left({c, c d}\right)$

where $c \mathrel \backslash d$ denotes that $c$ is a divisor of $d$.

Proof
From Exchange of Order of Summation with Dependency on Both Indices:

We have that:
 * $R \left({d}\right)$ is the propositional function:
 * $d \mathrel \backslash n$
 * $S \left({d, c}\right)$ is the propositional function:
 * $c \mathrel \backslash d$

Thus $R' \left({d, c}\right)$ is the propositional function:
 * Both $d \mathrel \backslash n$ and $c \mathrel \backslash d$

This is the same as:
 * $c \mathrel \backslash n$ and $\dfrac d c \mathrel \backslash \dfrac n c$

Similarly, $S' \left({c}\right)$ is the propositional function:
 * $\exists d$ such that both $d \mathrel \backslash n$ and $c \mathrel \backslash d$

This is the same as:
 * $c \mathrel \backslash n$

This gives:
 * $\displaystyle \sum_{d \mathop \backslash n} \sum_{c \mathop \backslash d} f \left({c, d}\right) = \sum_{c \mathop \backslash n} \sum_{\left({d / c}\right) \mathrel \backslash \left({n / c}\right)} f \left({c, d}\right)$

Replacing $d / c$ with $d$:
 * $\displaystyle \sum_{d \mathop \backslash n} \sum_{c \mathop \backslash d} f \left({c, d}\right) = \sum_{c \mathop \backslash n} \sum_{d \mathop \backslash \left({n / c}\right)} f \left({c, c d}\right)$