Definitive proof of twin prime conjecture

$ $When we write log, 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 $n≧5$ holds. We suppose $r$ 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 $n≧6$, let $f(n)$ be as below.
$f(n)=2n(\log n+\log \log n-1)-((n+2)(\log(n+2)+\log \log(n+2))+2)$
$f(9)=-2.27780067371110733430155271965419…$
$f(10)=-0.00935015299988545656167212393386…$
$f(11)=2.40516122715988664631909754841655…$
$f(12)=4.95069428318310030843435876234474…$
$f(13)=7.61490398664676903251831055964017…$
It is confirmed that $f(n)>0$ holds for $11≦n≦13$. $f(n)>0$ holds for $n≧11$ since the first term has a greater divergence velocity than the second term. The following inequalities hold for $5≦n≦10$ where $p_n< p_{n+k}+2$ and $p_{n+k}+2<2p_n$ $(0≦k≦3～6)$ 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 $r≧3$ holds for $n≧5$.
$ $We suppose $n≧5$ and $r≧3$ hold. Let $M$ be an odd integer and the product of all the elements $p_n+2$, $p_{n+1}+2$, …, $p_{n+r-1}+2$. We suppose $0≦i≦r-1$ holds.
$$M=\prod_{k=0}^{r-1}(p_{n+k}+2)$$
$$M≡2\prod_{k=0,k≠i}^{r-1}(p_{n+k}+2)\ (mod\ p_{n+i})$$
Let $a_i$ be a positive integer.
$$M=a_ip_{n+i}+2\prod_{k=0,k≠i}^{r-1}(p_{n+k}+2)$$
$$\prod_{i=0}^{r-1}(M-a_ip_{n+i})=2^rM^{r-1}\ …(1)$$
Let $s_i$ and $t_i$ be integers. We suppose the following equation and $0≦t_i< p_{n+i}$ hold.
$$2\prod_{k=0,k≠i}^{r-1}(p_{n+k}+2)=s_i p_{n+i}+t_i$$
$$2^rM^{r-1}=\prod_{i=0}^{r-1}(s_i p_{n+i}+t_i)\ …(2)$$
By the equation (1),
$$\prod_{i=0}^{r-1}(M-a_ip_{n+i})=\prod_{i=0}^{r-1}(s_ip_{n+i}+t_i)$$
$$M\prod_{k=0,k≠i}^{r-1}(M-a_kp_{n+k})≡t_i \prod_{k=0,k≠i}^{r-1}(s_kp_{n+k}+t_k)\ (mod\ p_{n+i})$$
holds. Let $b_i$ be an integer.
$$M\prod_{k=0,k≠i}^{r-1}(M-a_kp_{n+k})=b_ip_{n+i}+t_i \prod_{k=0,k≠i}^{r-1}(s_k p_{n+k}+t_k)$$
$$(s_ip_{n+i}+t_i)M\prod_{k=0,k≠i}^{r-1}(M-a_kp_{n+k})=(s_ip_{n+i}+t_i)b_i p_{n+i}+t_i 2^rM^{r-1}$$
$$t_iM\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})≡t_i2^rM^{r-1}\ (mod\ p_{n+i})$$
$$(\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})-2^rM^{r-2})t_iM≡0\ (mod\ p_{n+i})\ …(3)$$
When $t_i≡0\ (mod\ p_{n+i})$ holds, $M≡0\ (mod\ p_{n+i})$ must hold by the equation (2) since $p_{n+i}$ is odd. We assume that $M≢0\ (mod\ p_{n+i})$ holds for all $i$ where $0≦i≦r-1$ holds after this. Let $c_i$ be a positive integer. We consider the following equation.
$$\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})=c_ip_{n+i}+2^rM^{r-2}\ …(4)$$
By the equation (1),
$$(M-a_ip_{n+i})(c_ip_{n+i}+2^rM^{r-2})=2^rM^{r-1}$$
$$(M-a_ip_{n+i})c_i=a_i2^rM^{r-2}$$
$$(c_i-a_i2^rM^{r-3})M=a_ic_ip_{n+i}$$
holds. Let $d_i$ be a positive integer.
$$c_i=d_ip_{n+i}+a_i2^rM^{r-3} …(5)$$
Comparing the equations (4) and (5), we find that the second term on the equation (5) is $a_i/M$ times the one on the equation (4). If $M≢0\ (mod\ p_{n+i})$ 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. Therefore, the assumption is false. When $r≧3$ holds, for at least one $i$, the equation (4) does not hold and $M≡0\ (mod\ p_{n+i})$ holds by the congruent expression (3).
$ $When $M≡0\ (mod\ p_{n+i})$ 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 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 $p_{n+i}$ when $n≧5$ and $r≧3$ hold and the sets of numbers between $p_n$ and $2p_n$, excluding $p_n$ and $2p_n$ can be taken infinitely by increasing $n$. (Q.E.D.)