Definitive proof of twin prime conjecture

$$When we write $\log$ in this paper, we refer to natural logarithm. Let $i$, $k$, $n$ and $r$ be positive integers, $p$ be a prime and $p_n$ be the $n$th prime. To begin with, we prove that there are at least three $p+2$ type integers $N$ that satisfy $p_n<N<2p_n$ when $n\geqq 5$ holds. We consider that $r$ is the number of $N$. According to the following Dusart’s inequality
$$n(\log n+\log (\log n)-1)<p_n<n(\log n+\log (\log n))$$
holds for $n\geqq 6$, let $f(n)$ be as below.
$f(n)=2n(\log n+\log (\log n)-1)-((n+2)(\log (n+2)+\log (\log (n+2)))+2)$
$f(9)=-2.27780067371110733430155271965419\dots$
$f(10)=-0.00935015299988545656167212393386\dots$
$f(11)=2.40516122715988664631909754841655\dots$
$f(12)=4.95069428318310030843435876234474\dots$
$f(13)=7.61490398664676903251831055964017\dots$
It is confirmed that $f(n)>0$ holds for $11\leqq n\leqq 13$. $f(n)>0$ holds for $n\geqq 11$ since the first term has a greater divergence velocity than the second term. The following inequalities hold for $5\leqq n\leqq 10$ where $p_n<p_{n+k}+2$ and $p_{n+k}+2<2p_n(0\leqq k\leqq 3～6)$ hold.
$p_5<13,15,19,21<2p_5$
$p_6<15,19,21,25<2p_6$
$p_7<19,21,25,31,33<2p_7$
$p_8<21,25,31,33<2p_8$
$p_9<25,31,33,39,43,45<2p_9$
$p_{10}<31,33,39,43,45,49,55<2p_{10}$
From the above, it is proved that $r\geqq 3$ holds for $n\geqq 5$.
$$We regard $n\geqq 5$ and $r\geqq 3$ hold. Let $M$ be an odd integer and the product of the elements $p_n+2$, $p_{n+1}+2$, …, $p_{n+r-1}+2$. We take account of $0\leqq i\leqq r-1$ holds.
$$M=\prod _{k=0}^{r-1}(p_{n+k}+2)$$
$$M\equiv 2\prod _{k=0,k\ne i}^{r-1}(p_{n+k}+2)(mod{p}_{n+i})$$
Let $a_i$ be a positive integer.
$$M=a_i{p}_{n+i}+2\prod _{k=0,k\ne i}^{r-1}(p_{n+k}+2)$$
$$\prod _{i=0}^{r-1}(M-a_i{p}_{n+i})=2^r{M}^{r-1}\dots (1)$$
$$M\prod _{k=0,k\ne i}^{r-1}(M-a_k{p}_{n+k})\equiv 2^r{M}^{r-1}(mod{p}_{n+i})$$
$$(\prod _{k=0,k\ne i}^{r-1}(M-a_k{p}_{n+k})-2^r{M}^{r-2})M\equiv 0(mod{p}_{n+i})$$
$$We assume that $M\not\equiv 0(mod{p}_{n+i})$ holds for all $i$ where $0\leqq i\leqq r-1$ holds after this. Let $b_i$ be a positive integer. We consider the following equation.
$$\prod _{k=0,k\ne i}^{r-1}(M-a_k{p}_{n+k})=b_i{p}_{n+i}+2^r{M}^{r-2}\dots (2)$$
By the equation (1),
$$(M-a_i{p}_{n+i})(b_i{p}_{n+i}+2^r{M}^{r-2})=2^r{M}^{r-1}$$
$$(M-a_i{p}_{n+i})b_i=a_i{2}^r{M}^{r-2}$$
$$(b_i-a_i{2}^r{M}^{r-3})M=a_i{b}_i{p}_{n+i}$$
holds. Let $c_i$ be a positive integer.
$$b_i=c_i{p}_{n+i}+a_i{2}^r{M}^{r-3}\dots (3)$$
Comparing the equations (2) and (3), we find that the second term on the equation (3) is $a_i/M$ times the one on the equation (2). If $M\not\equiv 0(mod{p}_{n+i})$ holds for all $i$, then it becomes a contradiction since $a_i/M$ is less than one and if this operation is repeated a finite number of times, then the terms will no longer be integers. Hence, the assumption is false. When $r\geqq 3$ holds, $M\equiv 0(mod{p}_{n+i})$ holds for at least one $i$.
$$When $M\equiv 0(mod{p}_{n+i})$ holds for some $i$, one of the elements in $M$ must be $p_{n+i}$ because $3p_{n+i}>2p_n$ holds and they cannot be three times or more odd multiples of $p_{n+i}$. From the above, it is proved that there are an infinite number of twin prime numbers since at least one of the numbers $p_n+2$, $p_{n+1}+2$, …, $p_{n+r-1}+2$ must be a prime number when $n\geqq 5$ and $r\geqq 3$ hold and the sets of numbers between $p_n$ and $2p_n$ can be taken infinitely by increasing $n$. (Q.E.D.)