Twin Primes by an amateur mathematician 
About 20 years ago in a bookstore of Tokyo I found a book titled "Twin Primes" by Seiji Yasui, an amateur mathematician. He insisted that he proved the infinity of twin primes.
  The proof is attached to the book, but I cannot understand it. I hope the proof or disproof of the following by another person. I have confirmed their truths numerically $\leq 3000$. 

His proposition: 
Let $ n $ be a natural number. Denote as follows. 
$\pi (n) = $ the number of primes $ \leq n$. 
$G(n) = $ the number of composites $ \leq n$. 
$\pi_2(n) =$ the number of twin primes $ \leq n$. 
$G_2(n) = $ the number of twin composites $ \leq n$. 


*

*Yasui’s identity: 
$$G(n) - G_2(n) = \pi(n) - \pi_2(n) +c.$$
$$c = 0 \text{ or } -1.$$

*Yasui’s $1^\text{st}$ inequality:
$$G_2(n) > G(G(n)).$$

*Yasui’s $2^\text{nd}$ inequality:
$$\pi_2(n) > \pi(n) - \pi(G(n)).$$
Then,
$$\pi_2(n) > n /(\log(n))2.$$

This is an example:



Mathematica code here:

 G2[x]>G[G[x]]

array\[Pi]1[i_] := Select[Range[2, i], PrimeQ[#] &];
arraygn[i_] := Complement[Range[2, i], array\[Pi]1[i]];
int[i_] := Intersection[arraygn[i], arraygn[i] + 2];

ax1 = ListPlot[Table[Length[int[i]], {i, 2, 3000}], 
  AxesLabel -> {"n", "G2(n)"}, PlotLabel -> "Twin Primes", PlotStyle -> Red]
bx1 = ListPlot[Table[Length[arraygn[Length[arraygn[i]]]], {i, 2, 3000}], 
  AxesLabel -> {"n", "G(G(n))"}, PlotLabel -> "Twin Primes", 
  PlotStyle -> Blue]

fig1 = Show[ax1, bx1, AxesLabel -> {"n", "G2(n)(Red),G(G(n))(Blue)"}, 
  PlotLabel -> "Twin Primes"]

    Pi2[n] > Pi[n] - Pi[G[n]]

pi2[i_] := Length[Select[Select[Range[2, i], PrimeQ[#] &], PrimeQ[# + 2] &]];
pi1[i_] := Length[Select[Range[2, i], PrimeQ[#] &]];
gn[i_] := i - Length[Select[Range[2, i], PrimeQ[#] &]];
ax1 = ListPlot[Table[pi2[i], {i, 2, 3000}], AxesLabel -> {"n", "pi2(n)"}, 
  PlotLabel -> "Twin Primes", PlotStyle -> Red]
bx1 = ListPlot[Table[pi1[i] - pi1[gn[i]], {i, 2, 3000}], 
  AxesLabel -> {"n", "pi(n)-pi(G(n))"}, PlotLabel -> "Twin Primes", 
  PlotStyle -> Blue]
cx1 = Plot[x/(Log[x])^2, {x, 2, 3000}, AxesLabel -> {"n", "n/(log(n))^2"}, 
  PlotLabel -> "Twin Primes", PlotStyle -> Green]


fig2 = Show[ax1, bx1, cx1, 
  AxesLabel -> {"n", 
    "pi2(n)(Red), pi(n)-pi(G(n))(Blue), n/(log(n))^2(Green)"}, 
  PlotLabel -> "Twin Primes"]

 A: Incomplete answers

The first identity $G(n)−G_2(n)=\pi(n)−\pi_2(n)+c$ is true. Basically


*

*a number is either composite or prime, except $1$: $$G(n) + \pi(n)=n-1 \tag{1}$$

*a pair $(m,m+2)$ is either twin composites, twin primes or one prime one composite (let's say $k$-pairs): $$G_2(n)+\pi_2(n)+k=n-2$$


and $$G(n) - G_2(n) + \pi(n) - \pi_2(n)=k+1 \tag{2}$$
Now, let's show $$\pi(n)=\pi_2(n)+\frac{k+1-c}{2} \tag{3}$$
$$\begin{array}{|c|c|c|}
\hline
\pi(n) \text{ primes}& \pi_2(n) \text{ twin primes pairs} & k \text{ one prime one composite pairs}  \\ \hline
2 & & (2,4)\\ \hline
3 & (3,5) & (1,3) \\ \hline
5 & (3,5), (5,7) & \\ \hline
7 & (5,7) & (7,9)\\ \hline
11 & (11,13) & (9,11)\\ \hline
13 & (11,13) & (13,15)\\ \hline
... & ... & ... \\ \hline
p_{i-1} & (p_{i-1},p_i) & (p_{i-1}-2,p_{i-1})\\ \hline
p_i & (p_{i-1},p_i), (p_i,p_{i+1}) & \\ \hline
p_{i+1} & (p_i,p_{i+1})  & (p_{i+1},p_{i+1}+2)\\ \hline
... & ... & ... \\ \hline
p_j &  & (p_{j}-2,p_j),(p_j,p_{j}+2)\\ \hline
p_{j+1} &  & (p_{j+1}-2,p_{j+1}),(p_{j+1},p_{j+1}+2)\\ \hline
... & ... & ... \\ \hline
p_{\pi(n)} & ... & ...\\ \hline
\end{array}$$
or


*

*every row contains exactly $2$ pairs; either $2$ in "$\pi_2(n)$ column", $2$ in "$k$ column", or $1$ in "$\pi_2(n)$" and $1$ in "$k$". Altogether $2\pi(n)$ pairs.

*for the "$\pi_2(n)$ column", if $(p_i,p_{i+1})$ is in the "row $p_i$" then $(p_i,p_{i+1})$ is also in the "row $p_{i+1}$". Thus column $\pi_2(n)$ always contains $2\pi_2(n)$ pairs.

*for the "$k$ column" there are no repeating pairs.


As a result $$2\pi(n)=2\pi_2(n) + k + 1-c$$
where $c$ is either $0$ or $1$ to "balance parity of $k$".
Note: The original questions asks for $c$ to be either $0$ or $-1$, but this is only because I considered $p_1=2$. Original questions ignores $2$.
Now, injecting $(3)$ in $(2)$
$$G(n) - G_2(n) + \pi(n) - \pi_2(n)=2\pi(n)-2\pi_2(n)+c \Rightarrow \\ G(n)−G_2(n)=\pi(n)−\pi_2(n)+c \tag{4}$$

There is a link between 1st and 2nd inequalities. From 
$$(1) \Rightarrow G(G(n))+\pi(G(n))=G(n)-1 \Rightarrow G(G(n))=G(n)-\pi(G(n))-1$$
$$(4) \Rightarrow G_2(n)=G(n)-\pi(n)+\pi_2(n)-c$$
Subtracting one from another
$$G_2(n)-G(G(n))=\pi_2(n)+\pi(G(n))-\pi(n)+1-c$$
As a result 
$$\pi_2(n)>\pi(n)-\pi(G(n)) \Rightarrow G_2(n)>G(G(n))$$
and
$$G_2(n)>G(G(n)) \Rightarrow \pi_2(n)\geq \pi(n)-\pi(G(n))$$
