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 !-> I ~-> U !-> Q ~-> DD (肽增) U 改变 !-> I ~-> U !-> I ~-> U Q 查找 !-> U ~-> Q !-> U ~-> Q THEN WE FIND? PDE SWAP NEW LAW D = DD V 感知->U + Q !-> QU ~-> QQ !-> UU ~-> UQ = V E 执行->I + U !-> DQ ~-> DDD !-> III ~-> IIU = I + E (肽增) C 控制->I + D !-> DI ~-> DU !-> IQ ~-> UDD = U + D (肽展) = U + D + D (肽增) THEN WE FIND U = E I = U THEN 1 A= V + S = U + Q + I + Q = UQIQ !-> QUDU ~-> QUDQ !-> UQIU ~-> UQIQ = A O= E + S = I + U + I + Q = IUIQ !-> DQDU ~-> DQDQ !-> IUIU ~-> IUIQ = O P= E + C = I + U + I + D = IUID !-> DQDI ~-> DQDU !-> IUIQ ~-> IUIDD = P + D (肽增) M= C + S = I + D + I + Q = IDIQ !-> DIDU ~-> DIDQ !-> IDIU ~-> IDIQ = M THEN 2 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 = DD (肽增) U = E (肽增) I = U (肽增) E = I + E (肽增) P = P + D (肽增) C = U + D + D(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 YG and liang BY, Named by yaoguangluo 20201025 END 2020-10-25 21:30 PM, D8+, YAOGUANGLUO LIUYANG