“ALL”
TRANSLATION OF TWO ODD NUMBERS
P≧Q
P≧3
Q≧3
m≧X
m≧Y
X+Y=2Z
n≧0
|(m+X)+(Y-m)|=X+Y
=(m+Z)+(Z-m)
=2n≧0
0→$\infty$
AB-CD OR
P-CD OR
AB-Q OR
P-Q
ALL
∴
|(α+X)+(Y-α)|=P-Q
{P-Q}={2n}≧0
|(m+X)+(Y-m)|=X+Y
=(m+Z)+(Z-m)
=2
0→$\infty$
AB-CD OR
P-CD OR
AB-Q OR
P-Q
ALL
∴
|(β+X)+(Y-β)|=P-Q
THERE ARE INFINITELY MANY TWIN PRIMES
|(m+X)-(Y-m)|=2m+(X-Y)
=(m+Z)-(Z-m)
=2n≧0
$\infty$→0
AB+CD OR
P+CD OR
AB+Q OR
P+Q
ALL
∴
|(γ+X)-(Y-γ)|=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}
={n-Q}+{n+Q}
={2n}≧6
cf.
3-3=0
3+3=6
∴
GOLDBACH’S CONJECTURE IS TRUE
(KOAN)