Let $(\cdot\mid\cdot) : \Bbb{N}\times\Bbb{N} \to \Bbb{Z}_2$ be the divisibility function which takes on the value $(x|y) = 1$ whenever $x$ divides $y$ and the value $(x|y) = 0$ whenever it does not divide. We define the non-divisibility function $(x\nmid y) = 1 - (x|y)$. Clearly, it equals $1$ if and only if $x$ does not divide $y$.
Define $\chi(n) = \prod_{i = 1}^{\pi(\sqrt{n+2})} (p_i \nmid n(n+2))$. Clearly $\chi: \Bbb{N} \to \Bbb{Z}_2$ and is $1$ if and only if $n(n+2)$ is coprime to all primes $\leq \sqrt{n+2}$. That is the same thing as saying $\chi(n)$ is $1$ if and only if $n$ is the first in a twin prime pair. Put $m = n(n+2)$. Let's see what happens when we expand $\chi(n) = (1 - (p_1| m))(1-(p_2|m)) \cdots (1-(p_{\pi(\sqrt{n+2})}|m))$.
For example: $$(1 - (2|m))(1-(3|m))(1-(5|m)) = \\ 1 - (2|m) - (3|m) - (5|m) + (2\cdot 3|m) + (2\cdot 5|m) + (3\cdot 5|m) - (2\cdot 3\cdot 5 |m)$$
After some becoming comfortable with such expansions, you will eventually be a natural at concluding:
$$ \chi(n) = \sum_{d \mid \sqrt{n+2}\#}(-1)^{\omega(d)}(d \mid m) $$
So now define $\phi(n,m) = \sum_{d \mid n}(-1)^{\omega(d)}(d \mid m)$. Because $((-1)^x)^2 = 1$ we can rewrite this as:
$$ \phi(n, m) = (-1)^{\omega(n)}\sum_{d\mid n}(-1)^{\omega(\frac{n}{d})}(d\mid m) $$
which is a Dirichlet convolution of two multiplicative functions: $(-1)^{\omega(\cdot)}$ and $(\cdot \mid m)$, all multiplied (pointwise) by $(-1)^{\omega(\cdot)}$, which makes $\phi(n, m)$ a multiplicative function in $n$. But more interestingly, if we define:
$$ \psi(n,m) = \mu(n)\sum_{d\mid n}\mu(\frac{n}{d})(d\mid m) $$
Then $\psi(n, m) = \phi(n,m)$ whenever $n$ is square-free by this note here at top of the article: "It is related to the Mobius...".
Now we're getting somewhere, because then by the converse of Mobius inversion, we have that $(\cdot, m)$, or the function that was convolved with $\mu$, must be equal to:
$$ (n\mid m) =\sum_{d \mid n} \mu(d) \psi(d,m) $$
for any $n, m \in \Bbb{N}$. Now we assume the opposite of the twin prime conjecture.
Suppose there were only a finite number of twin primes $n, n+2$. Then $\chi(n)$ eventually will vanish, meaning for all $n \gt N$, for some constant $N \in \Bbb{N}$, we must have $\chi(n) = 0 = \psi(\sqrt{n+2}\#, n(n+2))$.
Since $(\sqrt{n+2}\#| n(n+2)) = 0$ for sufficiently large $n$, we have that eventually:
$$ 0 = \sum_{d \mid \sqrt{n+2}\#} \mu(d)\psi(d, n(n+2))=\\ \sum_{d \mid \sqrt{n+2}\#}\sum_{c \mid d}\mu(\dfrac{d}{c})(c\mid n(n+2)) = ? \\ \tag{1} $$ But we can take out of the outter summation range $d = \sqrt{n+2}\#$ because by assumption of finite twin primes, $\psi(\sqrt{n+2}\#, n(n+2))$ vanishes eventually when $n$ is sufficiently large enough.
Question
My question is the question mark at (1). I'm not sure how to "play" with this expression. I've tried converting it back into a product to see "exactly what it means". I've tried swapping the summation ranges with the summand divisibility functions because they're each of the form $x\mid y$ in the summation range or $(x\mid y)$ in the summand area, so that you can swap them where appropriate. I've checked all the above statements (verified for small examples) using Python code, so I'm 99% sure (1) is correct. I just need help getting to the next step, if there is one.
Note that, wherever you see $\sqrt{n+2}\#$ this can be replaced with $a\#$ for any $\sqrt{n+2} \leq a \leq n-1$ by Sieve of Eratosthenes. Also note that although I tried, I was not able to convert the formula in (1) to an equivalent product expression (i.e. similar to our derivation of the summation of $\chi(n)$ but in the other direction).
The general form of the summation in quesiton is just a sum over divisors of a Dirichlet convolution at each given divisor, or:
$$ \varphi(n,m) = \sum_{d \mid n}(\mu * (\cdot | m))(d) $$