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$.

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

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

3. 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"]

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 $$G_2(n)>G(G(n)) \Rightarrow \pi_2(n)\geq \pi(n)-\pi(G(n))$$