Definitive proof of twin prime conjecture

$ $When we write $\log$ in this paper, we refer to natural logarithm. Let $e$ be Napier's constant, $i$, $k$, $n$ 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 consider that $r$ is the number of $N$. According to the following Rosser and Dusart's inequalities
$$n(\log\ n+\log(\log\ n)-1)< p_n< n(\log\ n+\log(\log\ n))$$
hold for $n≧6$, let $f(n)$ be as below. The upper bound is due to Rosser (1941) and the lower bound to Dusart (1999).
$f(n)=2n(\log\ n+\log(\log\ n)-1)-((n+2)(\log(n+2)+\log(\log(n+2)))+2)$
$f(9)=-2.27780067371110733430155271965419…$
$f(10)=-0.00935015299988545656167212393386…$
$f(11)=2.40516122715988664631909754841655…$
$f(12)=4.95069428318310030843435876234474…$
$f(13)=7.61490398664676903251831055964017…$
$$\frac{df(n)}{dn}=2\log(n)-\log(n+2)+2\log(\log\ n)-\log(\log(n+2))+2/(\log\ n)-1/\log(n+2)-1$$
Let $g_1(n)=2\log\ n-\log(n+2)$ and $g_2(n)=2\log(\log\ n)-\log(\log(n+2))$. In this case, $\frac{dg_1(n)}{dn}=(n+4)/(n^2+2n)$ and $\frac{dg_2(n)}{dn}=(2(n+2)\log(n+2)-n\log\ n)/(n(n+2)(\log\ n)\log(n+2))$ hold, $\frac{dg_1(n)}{dn}>0$ holds for $n>0$ and $\frac{dg_2(n)}{dn}>0$ holds for $n>e$. Therefore, $\frac{df(n)}{dn}>0$ holds for $n≧5$ since $g_1(n)>1$ holds for $n>(e+\sqrt{e^2+8e})/2=4.0579…$, $g_2(n)>0$ holds for $n>3.7540…$ and $2/\log\ n-1/\log(n+2)=\log((n+2)^2/n)/((\log\ n)\log(n+2))>0$ holds for $n>e$. $f(n)>0$ holds for $n≧11$ since $f(11)>0$ holds and $\frac{df(n)}{dn}>0$ holds for $n≧5$. 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 regard $n≧5$ and $r≧3$ 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 $0≦i≦r-1$ 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.
$$M=a_ip_{n+i}+2\prod_{k=0,k≠i}^{r-1}(p_{n+k}+2)$$
$$\prod_{i=0}^{r-1}(M-a_ip_{n+i})=2^rM^{r-1}\ …(1)$$
$$M\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})≡2^rM^{r-1}\ (mod\ p_{n+i})$$
$$(\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})-2^rM^{r-2})M≡0\ (mod\ p_{n+i})$$
$ $We assume that $M≢0\ (mod\ p_{n+i})$ holds for all $i$ where $0≦i≦r-1$ holds after this. Let $b_i$ be a positive integer. We consider the following equation.
$$\prod_{k=0,k≠i}^{r-1}(M-a_k p_{n+k})=b_ip_{n+i}+2^rM^{r-2}\ …(2)$$
By the equation (1),
$$(M-a_ip_{n+i})(b_ip_{n+i}+2^rM^{r-2})=2^rM^{r-1}$$
$$(M-a_ip_{n+i})b_i=a_i2^rM^{r-2}$$
$$(b_i-a_i2^rM^{r-3})M=a_ib_ip_{n+i}$$
holds. Let $c_i$ be a positive integer.
$$b_i=c_ip_{n+i}+a_i2^rM^{r-3}\ …(3)$$
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 $M≢0\ (mod\ p_{n+i})$ 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 $r≧3$ holds, $M≡0\ (mod\ p_{n+i})$ holds for at least one $i$.
$ $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 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 $n≧5$ and $r≧3$ hold and the sets of numbers between $p_n$ and $2p_n$ can be taken infinitely by increasing $n$. (Q.E.D.)