All

「$${All}$$」

$${Translation}$$ $${of}$$ $${two}$$ $${odd}$$ $${numbers.}$$

$${P\geqq{}Q}$$
$${P\geqq{3}}$$
$${Q\geqq{3}}$$

$${m\geqq{X}}$$
$${m\geqq{Y}}$$

$${X+Y=2Z}$$

$${n\geqq{0}}$$

$${\vert(m+X)+(Y-m)\vert=X+Y}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=(m+Z)+(Z-m)}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=2n\geqq0}$$

$${0\rightarrow\infty}$$

$${AB-CD\quad{}or}$$
$${P-CD\quad{}or}$$
$${AB-Q\quad{}or}$$
$${P-Q}$$
$${All.}$$

$${\therefore}$$
$${\vert(\alpha+X)+(Y-\alpha)\vert=P-Q}$$

$${\{P-Q\}=\{2n\}\geqq{0}}$$

$${\vert(m+X)+(Y-m)\vert=X+Y}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=(m+Z)+(Z-m)}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=2}$$

$${0\rightarrow\infty}$$

$${AB-CD\quad{}or}$$
$${P-CD\quad{}or}$$
$${AB-Q\quad{}or}$$
$${P-Q}$$
$${All.}$$

$${\therefore}$$
$${\vert(\beta+X)+(Y-\beta)\vert=P-Q}$$

$${There}$$ $${are}$$ $${infinitely}$$ $${many}$$ $${twin}$$ $${primes.}$$

$${\vert(m+X)-(Y-m)\vert=2m+(X-Y)}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=(m+Z)-(Z-m)}$$
$${\quad\quad\quad\quad\quad\quad\quad\quad}$$$${　}$$$${=2n\geqq{0}}$$

$${\infty\rightarrow{}0}$$

$${AB+CD\quad{}or}$$
$${P+CD\quad{}or}$$
$${AB+Q\quad{}or}$$
$${P+Q}$$
$${All.}$$

$${\therefore}$$
$${\vert(\gamma+X)-(Y-\gamma)\vert=P+Q}$$

$${\{P+Q\}-\{P-Q\}=\{Q+Q\}-\{2Q\}=0}$$

$${\{P-Q\}=\{+2n\}}$$
$${\{Q-P\}=\{-2n\}}$$

$${\{P+Q\}=\{+2n\}+\{2Q\}}$$
$${\{Q+P\}=\{-2n\}+\{2P\}}$$

$${\{P-Q\}=\{P\}-\{n\}+\{n\}-\{Q\}}$$
$${\{P+Q\}=\{P\}-\{n\}+\{n\}+\{Q\}}$$

$${\{P-Q\}=\{P\}-\{\frac{P\pm{}Q}{2}\}+\{\frac{P\pm{}Q}{2}\}-\{Q\}=\{2n\}}$$
$${\{P+Q\}=\{P\}-\{\frac{P\pm{}Q}{2}\}+\{\frac{P\pm{}Q}{2}\}+\{Q\}=\{2n\}}$$

$${\{\frac{P+Q}{2}\}=\{n\}}$$
$${\{\frac{P-Q}{2}\}=\{n\}}$$

$${\{P\}-\{\frac{P+Q}{2}\}=\{\frac{P+Q}{2}\}-\{Q\}=\{\frac{P-Q}{2}\}}$$

$${\{P+Q\}=\{P-n\}+\{n+Q\}}$$
$${\quad\quad\quad\quad=\{n-Q\}+\{n+Q\}}$$
$${\quad\quad\quad\quad=\{2n\}\geqq{6}}$$

$${cf.}$$

$${3-3=0}$$
$${3+3=6}$$

$${\therefore}$$
$${Goldbach's}$$ $${conjecture}$$ $${is}$$ $${true.}$$

$${(Koan)}$$