collection-of-papers-by-deta
/
PDE_LAW.txt

Initons
FIGURE 1
A 分析   O 操作   P 处理   M 管理
V 感知   E 执行   C 控制   S 静态
I 增加   D 减少   U 改变   Q 查找

FIGURE 2
PDE肽展公式3.0 in DeMorgan 结合律 加法
A= V + S = U + Q + I + Q = U + Q + I = V + I =
O= E + S = I + U + I + Q = I + U + Q = E + Q =
P= E + C = I + U + I + D = I + U + D = E + D =
M= C + S = I + D + I + Q = I + D + Q = C + Q =

FIGURE 3
A= S - I = I - S
O= S - Q = Q - S
P= C - D = D - S
M= S - Q = Q - S

FIGURE 4
V + S = V + I => S = I
E + S = E + Q => S = Q
E + C = E + D => C = D
C + S = C + Q => S = Q
=> 竟然和人类的ACGTU 腺吻合!
S 已经彻底解码为 A 腺嘌呤
A 腺嘌呤在dna中属于原生静态物质

V + S = V + I => S = I
E + S = E + Q => S = Q
E + C = E + D => C = D
C + S = C + Q => S = Q

S 为 A 腺嘌呤   在dna中属于原生活性物质
Q 为 T 胸腺嘧啶 在dna中属于感应活性物质
I 为 U 尿嘧啶   在dna中属于增生活性物质
C 为 G 鸟嘌呤   在dna中属于控制活性物质
D 为 C 胞嘧啶   在dna中属于降解活性物质
=>
嘌呤 生物多样化特征 属于VPCS INTIONS肽!
嘧啶 生物应激性特征 属于IDUQ INTIONS肽!

FIGURE 5
V= U + Q
E= I + U
C= I + D
S= I + Q

I= !D
U= !Q

罗瑶光
2020年10月25日 6:00 AM D8+

FIGURE 6
我得到严谨的论证结果:
A 分析            O 操作             P 处理             M 管理
V 感知            E 执行             C 控制(G 鸟嘌呤)   S 静态(A 腺嘌呤)
I 增加(U 尿嘧啶)  D 减少(C 胞嘧啶)   U 改变             Q 查找(T 胸腺嘧啶)

可以推断:
A 分析(TA 变感腺嘌呤) O 操作(UA 增变腺嘌呤)  P 处理(UG 增变鸟嘌呤)   M 管理(GA 鸟腺嘌呤)
V 感知(T 变感嘌呤)    E 执行(U 增变嘌呤)     C 控制(G 鸟嘌呤)        S 静态(A  腺嘌呤)
I 增加(U 尿嘧啶)      D 减少(C 胞嘧啶)       U 改变(变嘧啶)          Q 感应(T  胸腺嘧啶)

U 改变(变嘧啶) Named by yaoguangluo 20201025

FIGURE 7
//20201025 18:22 AM D8+
来继续持续绝对专注论证肽增公式1.0 - 2 BITS WAY
S = I
S = Q
C = D

DNA肽码->DNA肽码INITONS->DNA肽码INITONS-CSIDQ加密变换!->掩码加密-~>补码加密  ! ->掩码      ~->补码返回
A= V + S = U + Q + I + Q = U + Q + I  -> 101101 !->010010 IDU ~-> 010011 IDQ !->101100 UQD ~-> 101101 UQI = A (OLD)(肽减)
O= E + S = I + U + I + Q = I + U + Q  -> 011011 !->100100 UID ~-> 100101 UID !->011010 IUU ~-> 011011 IUQ = O (OLD)(肽减)
P= E + C = I + U + I + D = I + U + D  -> 011000 !->100111 UIQ ~-> 101000 UUD !->010111 IIQ ~-> 011000 IUD = P
M= C + S = I + D + I + Q = I + D + Q  -> 010011 !->101100 UQD ~-> 101101 UQI !->010010 IDU ~-> 010011 IDU = M (OLD)(肽减)

V 感知->U + Q -> 1011 !-> 0100 ID ~-> 0101 II !-> 1010 IU ~-> 1011 UQ = V
E 执行->I + U -> 0110 !-> 1001 UI ~-> 1010 UU !-> 0101 II ~-> 0110 IU = E
C 控制->I + D -> 0100 !-> 1011 UQ ~-> 1100 QD !-> 0011 DQ ~-> 0100 ID = C

S 静态->I + Q -> 0111 !-> 1000 UD ~-> 1001 UI !-> 0110 IU ~-> 0111 IQ  = S
S 静态->I     -> 01   !-> 10   UD   ~-> 11 Q  !-> 00   D  ~-> 01   I  != S (OLD)(肽减)
S 静态->Q     -> 11   !-> 00   UD   ~-> 01 I  !-> 10   U  ~-> 11   Q  != S (OLD)(肽减)

假设
I 增加-> 01   !-> 10   U   ~->   11  Q  !->   00   D  ~->   01  I
D 减少-> 00   !-> 11   Q   ~-> 0100 ID  !-> 1011  UQ  ~-> 1100 QD (肽增)
U 改变-> 10   !-> 01   I   ~->   10  U  !->   01   I  ~->   10  U
Q 查找-> 11   !-> 00   D   ~->   01  I  !->   10   U  ~->   11  Q
=>

V + S = V + I => S = I => I = S
E + S = E + Q => S = Q => Q = S
E + C = E + D => C = D => D = C
C + S = C + Q => S = Q
...


FIGURE 8
=>
...........................................................................................
//SORT 20201025 19:47 AM D8+
来继续持续绝对专注论证肽增公式1.0 BY USING ENGLISH FOR - 4 BITS DIUQ WAY
PDE SWAP LAW
S =  I
S =  Q
C =  D

PDE MASK LAW
I = !D
D = !I
U = !Q
Q = !U

PDE COMPS LAW
I = ++D
U = ++I
Q = ++U
DD= ++Q

VECS PDE LAW
V= U + Q
E= I + U
C= I + D
S= I + Q

AOPM PDE LAW
A= V + S
O= E + S
P= E + C
M= C + S

SO :
I 增加 !->  D   ~->  I  !->    D  ~->  I
D 减少 !->  Q   ~-> DD  !->   II  ~->  I + U = E (肽增)
U 改变 !->  I   ~->  U  !->    I  ~->  U
Q 查找 !->  D   ~->  I  !->    U  ~->  Q

THEN WE FIND?
PDE SWAP NEW LAW
D =  E

V 感知->U + Q !-> QU ~-> QQ !-> UU ~-> UQ  = V
E 执行->I + U !-> DQ ~-> ID !-> DI ~-> DU  = D + U (肽展) = E + U(肽增)
C 控制->I + D !-> DI ~-> DU !-> IQ ~-> IDD = C + D (肽增) = C + E (肽增)

S 静态->I + Q !->  DU ~-> DQ !->  IU ~->  IQ  = S
S 静态->I     !->  D  ~-> I  !->  D  ~->  I  ~= S (OLD)(肽减)
S 静态->Q     !->  U  ~-> Q  !->  U  ~->  Q  ~= S (OLD)(肽减)

THEN
A= V + S = U + Q + I + Q = UQI !-> QUD ~-> QUI !-> UQD ~-> UQI  = A
O= E + S = I + U + I + Q = IUQ !-> DQU ~-> DQQ !-> IUU ~-> IUQ  = O
P= E + C = I + U + I + D = IUD !-> DQI ~-> DQU !-> IUQ ~-> IUDD = P + D (肽增)
M= C + S = I + D + I + Q = IDQ !-> DIU ~-> DIQ !-> IDU ~-> IDQ  = M


IT SEEMS 4BIT-PDE IS A GOOD WAY~

FIGURE 9
ABOVE ALL, WE FIND PDE LAW LIST AS BELOW:

PDE MASK LAW
I = D!
D = I!
U = Q!
Q = U!

PDE COMP'S LAW
DD = ++Q
I  = ++D
U  = ++I
Q  = ++U

PDE (肽减) LAW
C = D (肽减)
S = I (肽减)
S = Q (肽减)

PDE (肽增) LAW
D = E (肽增)
E = E + U (肽增)
P = P + D (肽增)
C = C + D (肽增)
C = C + E (肽增)

PDE (肽展) LAW
A = V + S (肽展)
A = U + Q + I (肽展)
O = E + S (肽展)
O = I + U + Q (肽展)
P = E + C (肽展)
P = I + U + D (肽展)
M = C + S (肽展)
M = I + D + Q (肽展)
V = U + Q (肽展)
E = I + U (肽展)
E = D + U (肽展)
C = I + D (肽展)
S = I + Q (肽展)

FLASH A NEW NAME 可以推断:
A 分析(LTA 变胸腺腺苷)  O 操作(UCLA 尿胞变腺苷)  P 处理(UCLG 尿胞变鸟苷)  M 管理(GA 鸟腺苷)
V 感知(LT 变胸腺嘌呤)   E 执行(UCL 尿胞变嘌呤)   C 控制(G 鸟嘌呤)         S 静态(A  腺嘌呤)
I 增加(U  尿嘧啶)       D 减少(C 胞嘧啶)         U 改变(L 变嘧啶)         Q 感应(T  胸腺嘧啶)

U 改变(L 变嘧啶)定义为 L the first char of luo and liang, Named by yaoguangluo 20201025


END
2020-10-25 21:30 PM, D8+, YAOGUANGLUO LIUYANG