Remark 1. The following counting formula can be derived using arithmetic properties of inequalities, properties of floor / ceiling, inclusion-exclusion, sieve-of-Eratosthenes, and the inclusion-exclusion counting principle for OR-logic. To give an idea of what would be needed on the proof assistant side.
Definition 1. Define the twin prime (average) counting function to be, for any $x \in \Bbb{R}$:
$$ F(x) := T(f)(x) = \sum_{d \mid N_x\#} \mu(d)\sum_{r^2 = 1 \mod d} f(d,r,x) $$
where $N_x = \sqrt{|x| + 1}\#$ (primorial), where $T$ is the transformation of functions $f \in \{ \Bbb{N}^2 \times \Bbb{R} \to \Bbb{Z}\}$ given by the present double summation; and finally where $f$ is specifically $f(d,r,x) :=\left\lfloor\frac{x - r}{d} \right\rfloor - \delta_1(d)$. Note all the $-\delta_1(d)$ does is simply add a $-1$ into the total summation. The need for it can be found in a derivation of such counting function.
It is known for $x \geq 0$ to count the number of twin prime averages occurring in the shifting, growing interval $[p_n + 2, x]$ with $n = \pi (\sqrt{x + 1})$. The reason it doesn't count the earlier averages is because they naturally get cancelled out by the inherent Sieve-of-Eratosthenes. You could state in terms of the interval $[0, x]$ but you'd say additionally "twin prime averages $a$ such that $a \pm 1 \notin \{ p_1, \dots, p_n\}$."
Definition 2. Given our function space transformation $T: \{\Bbb{N}^2 \times \Bbb{R}\to \Bbb{Z}\} \to \{\Bbb{R} \to \Bbb{Z}\}$, and since these function spaces are $\Bbb{Z}$-modules themselves, we define the limit kernel to be just what it sounds like:
$$ \lim\ker T = \{ f \in \text{dom}(T) : \lim_{x \to \infty} T(f)(x) = 0 \} $$
That is the set of all domain functions which vanish in the limit after $T$ is applied, as $x \to \infty$.
Lemma 1. The limit kernel of $T$ is a $\Bbb{Z}$-submodule. Thus $\lim \ker T$ is closed under $+,-$ and scalar $\cdot$ .
Proof. This follows simply from $T$ being defined as a double summation, where of course this is the case. Though a more rigorous proof should be made since a limit is taken, which one usually does not see in basic linear algebra.
Lemma 2. For $x \in \Bbb{R}$ and $x \lt 0$ we have:
$-(|f(|x|)| + 2) \leq f(x) \leq |f(|x|)|$. In other words, we have almost that $f(x) = - f(|x|)$ except for an additive constant of $2$ on the lower bound side.
Code.
from sympy import *
def F(x):
S = -1
s = int(sqrt(abs(x) + 1))
if s in {0, 1}:
P = 1
else:
P = primorial(s, nth=False)
for d in divisors(P):
for r in range(0, d):
if (r ** 2 - 1) % d == 0:
S += (-1) ** primeomega(d) * floor((x - r)/d)
return S
X = 200
for x in range(0, X):
pos = F(x)
neg = F(-x)
print(x, pos, neg)
assert -abs(pos) -2 <= neg <= abs(pos)
Lemma 3. $$ \sum_{i = 0}^{md - 1} \left\lfloor\frac{x - i}{d} \right\rfloor = mx - \frac{m(m + 1)}{2} d + m $$
Code.
from sympy import *
N = 10
for d in range(1, N):
for m in range(1, N):
for x in range(-d, d):
S = 0
for i in range(0, m*d):
S += floor((x-i)/d)
print(d, x, S)
assert S == m * x - (m * d) * (m + 1) // 2 + m
Lemma 4. The twin prime conjecture is false if and only if $f \in K = \lim \ker T$.
Proof. This is because if false, the counter function as $x \to \infty$ which is $F(x) \geq 0$ must vanish. And vice versa.
Lemma 5. We have already that:
$$ G(d,r,x) =\sum_{i = 0}^{N_x -1} \left\lfloor\frac{x -i- r}{d} \right\rfloor - N_x\delta_1(d) - \sum_{i = 0}^{N_x} f(x - i)\delta_1(d) $$ sits in $K = \lim\ker T$. We simply compute $f(x-i)$ using our formula and then subtract $f(x-i)$ using the coefficient of $\delta_1(d)$. Then sum up everything.
Lemma 6. If the twin prime conjecture is false, there exists a natural number $N_0$ such that for all $|x| \geq N_0$ we have that $f(x) = 0$.
Corollary 1. This mean there exists a constant $C \in \Bbb{Z}$ and a constant $M_0 \in \Bbb{N}$ such that for all $x \geq M_0$ we have that:
$$ g(d, r,x) = \sum_{i = 0}^{N_x - 1} \left\lfloor\frac{x-i-r}{d} \right\rfloor - N_x \delta_1(d) + C $$
Lemma 7. $\lim T$ basically eats constant functions, or in other words $\lim_{x \to \infty}T( C)(x) = 0$ for any constant $C \in \Bbb{Z}$.
Proof. Can be proved by elementary means, expanding $T( C)(x)$.
Corollary 8. If the twin prime conjecture is false, then:
$$ \lim_{x \to \infty}T(g)(x) = \lim_{x \to \infty}\left(-(N_x + 1) + \sum_{d \mid N_x}\mu(d)\sum_{r^2 = 1 \mod d} \left\lfloor\frac{x - i - r}{d} \right\rfloor\right) $$
Lemma 8. If the twin prime conjecture is false, then by Lemma 3:
$$
\lim_{x \to \infty} T(g)(x) = \\ \ \\ \lim_{x \to \infty}\sum_{d \mid N_x}\mu(d)\sum_{r^2 = 1 \mod d}\left(-(N_x + 1) +\sum_{d \mid N_x}\mu(d)\sum_{r^2 = 1 \mod d}\left(\frac{N_x}{d}x - \frac{N_x(\frac{N_x}{d}+1)}{2} +\frac{N_x}{d}\right)\right) \\
\tag{1}
$$
Corollary 8. By rearranging the formula at tag (1), we have:
$$
\lim_{x \to \infty} T(g)(x) = \\
-1 +\lim_{x \to \infty} N_x\left( \sum_{d \mid N_x}\mu(d)\sum_{r^2 = 1 \mod d}\left(1 + \frac{x}{d} -\frac{N_x}{2d} -\frac{1}{2}+ \frac{1}{d}\right)\right) \\
\ = -3 + \lim_{x \to \infty} N_x\left( x - \frac{N_x}{2} +1\right) \sum_{d \mid N_x}\mu(d)\sum_{r^2 = 1 \pmod d}\frac{1}{d} \\
\tag{2}
$$
Lemma 9. The double sum formula at tag (2) can be elegantly handled via Merten's third theorem.
But thinking much simpler: as $x \to \infty$ we obviously would have $T(g)(x) \to -\infty$, but this is impossible since by Lemma 5 we showed that necessarily we have that $\lim_{x\to \infty} T(g)(x) = 0$, since the summation defining $G$ must sit in the limit kernel of $T$ when we suppose a finite number of twin primes.
Obviously it is too big of a question to ask the community to proof-check this sketch. So I have a really good question instead.
Question.
Is the above theory encodable in Coq or not yet?
I mention Coq because I just installed it, I'm learning about it in order to understand presenters at ANR coREACT seminars, and its awesome to me that it has a standalone IDE. I'm open though to both Lean4 and Isabelle. Preferring Coq right now though.
