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Abstract

From a cognitive perspective in the semantic space, we proposed the revelation of the semantics of Collatz Conjecture or the 3x+1 Problem based on our proposed Existence Computation and Reasoning(EXCR) and Essence Computation and Reasoning(ESCR) mechanism following our previous revelation of the semantics axioms of Conservation of Existence Set Axiom (CEX , Consistency of Compounded Essential Set Axiom (CES) and Inheritance of Existence Semantics Axiom (IHES ).
Existence Computation and Reasoning(EXCR) and Essence Computation and
Reasoning(ESCR) based Revelation of Collatz Conjecture
基于存在计算(EXCR)与语义计算(ESCR)的关于 3x+1 问题的语义空间(SCR)解释
By Yucong Duan,
DIKW research group, Hainan University
Email: duanyucong@hotmail.com
Abstract: From a cognitive perspective in the semantic space, we proposed the
revelation of the semantics of Collatz Conjecture or the 3x+1 Problem based on our
proposed Existence Computation and Reasoning(EXCR) and Essence Computation
and Reasoning(ESCR) mechanism following our previous revelation of the semantics
axioms of Conservation of Existence Set Axiom (CEX , Consistency of
Compounded Essential Set Axiom (CES) and Inheritance of Existence Semantics
Axiom (IHES ).
Collatz Conjecture 的语义解释:
(a)Collatz Conjecture 语义可以从类型的实例的语义角度等价于:
从类型语义的实例层面任何一个自然数 N的实例 INS(N)=n它或者是一个奇数 O
的实例 INS(O)=o,或者是一个偶数 E的实例 INS(E)=e。当 n是奇数 o,则对它乘
3再加 1获得 n:=3o+1n是偶数 e则对它除以 2获得 n:=e/2如此循环,
终都能够得到 n=1
INS(N)
:=ASS({INS(O),INS(E)},{REL(+),REL(/)})
:=ASS({INS(O)*3+1, INS(E)/2})
:=ASS({INS(O)*3+1, INS(E)/2})
:=ASS({o*3+1, e/2}, 1)
:=ASS({n*3+1, n/2}, 1)
=>n->1
(b)Collatz Conjecture 语义可以从实例的整体类型的语义角度等价于:
从实例的整体类型语义的层面任何一个自然数 N的实例 TYPE(INS(N)=n 都可以在
确认自身的存在语义的基础上,由跨类型的奇数 O或偶数 E实例层面的语义
INS(N):=ASS({INS(O),INS(E)},{REL(+),REL(/)})关联,根据存在计算与推理 EXCR 的基
础假设公理就是存在的守恒公理(CEX, Conservation of Existence Set)等价推导出类
型层面的对应语义关联 TYPE(N):=ASS({TYPE(O),TYPE(E})
INS(N):=ASS({INS(O),INS(E)},{REL(+),REL(/)})
=>TYPE(N):=ASS({TYPE(O),TYPE(E)},{REL(+),REL(/)})
TYPE(N):=ASS({TYPE(O),TYPE(E})蕴含类型层面自然数类型 N与奇数类型 O与偶数
类型 E整体之间的存在语义上的等价性。
TYPE(N):=ASS({TYPE(O),TYPE(E)},{REL(+),REL(/)})
=>EXCR(N):=EXCR(TYPE(O),TYPE(E))
=>EXCR(N):=EXCR(O,E)
奇数类型 O或偶数类型 E由于可以通过类型层面语义关联 N(E):=N(O)+1 建立相互
之间的联系。
ASS(TYPE(O),TYPE(E))
:=ASS((TYPE(O),TYPE(E)),{REL(+),REL(/)})
:=ASS(((TYPE(O),TYPE(E)),1),REL(+))
=>N(E):=N(O)+1
类型层面语义关联 N(E):=N(O)+1,根据存在计算与推理 EXCR 的基础假设公理就
是存在的守恒公理(CEX, Conservation of Existence Set)等价推导出类型层面奇数类
O与偶数类型 E之间的存在语义上的等价性。
N(E):=N(O)+1
=>EXCR(TYPE(O)):=EXCR(TYPE(E))
=>EXCR(O):=EXCR(E)
EXCR(N):=EXCR(O,E)以及 EXCR(O):=EXCR(E)我们可以可以依托存在计算与推理
EXCR 的基础假设公理存在的守恒公理 CEX 确定自然数类型 N与奇数类型 O与偶
数类型 E整体之间的存在语义上的等价性。
ASS(EXCR(N):=EXCR(O,E), EXCR(O):=EXCR(E))
=>EXCR(N):=EXCR(O):=EXCR(E)
结合存在计算与推理 EXCR 的基础假设公理就是存在的守恒公理(CEX,
Conservation of Existence Set)与本质计算与推理 ESCR 的基础假设公理就是本质集
合整体完整性的组合一致性公理(CES, Consistency of Compounded Essential Set),
我们面向存在语义提出存在语义继承公理(IHES, Inheritance of Existence
Semantics):类型层面的存在语义在纯类型层面的语义处理过程中,对于具有存
在语义依赖关系或者语义等价关系的目标 A和目标 B例如类型 A的存在语义依
赖于或等价于类型 B的存在语义语义集合 EX(B) {ex(b)}目标 A继承或保有目
B的所有存在语义 EX(A) {ex(a)}
IHES
ASS({EXCR(A):=EXCR(B), EXCR(A)=>EXCR(B)})
=>EX(B)=>EX(A)
结合使用存在计算与推理 EXCR 的基础假设公理就是存在的守恒公理(CEX,
Conservation of Existence Set)本质计算与推理 ESCR 的基础假设公理就是本质集
合整体完整性的组合一致性公理(CES, Consistency of Compounded Essential Set)
存在语义继承公理(IHES, Inheritance of Existence Semantics)依据自然数类型
N的实例 INS(N)之间的自然数操作加 Z(+)、乘法 Z(*)、除法 Z(/),也即
ASS(INS(N),{Z(+), Z(*), Z(/)} ),作用在自然数类型 N的连续实例 INS(N)的整体
{INS(N)}上不改变其中任何一个类型的本质语义关系 ASS(CES(EXCS(N)))的存在语
义我们可以推出一系列的本质语义关系。
ASS(CES(EXCS(N)))
:=ASS(EXCR(CES(N)))
:=CES(ASS(EXCS(N)))
:=CES(EXCS(ASS(N)))
:=EXCS(CES(ASS(N)))
由于从存在语义层面自然数类型的实例 INS(N)=n 之间,例如 nn+1 具有有界
的连续性或有限性,也就是 ASS(n, n+1)在一个有限的自然数操作加 Z(+)、乘法
Z(*)、除法 Z(/)之后获得的边界 bound(ASS(n, n+1))会与 ASS(n-1, n)以及 ASS(n+1,
n+2)对应的边界 bound(ASS(n-1, n))以及 bound(ASS(n+1, n+2))相邻。这个相邻语义
的递归实现蕴含着每个 bound(ASS(n, n+1))的有限性。这个有限性按 CEX 公理,
在纯类型变换 ASS({n*3+1, n/2}, 1)下也将保持不变,ASS({n*3+1, n/2},
1)=>bound(ASS({n*3+1, n/2}, 1))。由存在语义继承公理(IHES, Inheritance of
Existence Semantics)基于自然数类型 N与奇数类型 O与偶数类型 E整体之间的
存在语义上的等价性可得 EX(N)=>EX(O)EX(N)=>EX(E)。将 EX(N)=>EX(O)
EX(N)=>EX(E)代入 ASS({n*3+1, n/2}, 1)=>bound(ASS({n*3+1, n/2}, 1))
,即可得到 Collatz Conjecture 的有界语义 bound(ASS({O*3+1, E/2}, 1)):
ASS({n*3+1, n/2}, 1)=>bound(ASS({n*3+1, n/2}, 1))
=>
ASS({O*3+1, E/2}, 1)=>bound(ASS({O*3+1, E/2}, 1))
综上,bound(ASS({O*3+1, E/2}, 1))的有限性也就蕴含着相关操作的处理过程的有
限性,也即得证 Collatz Conjecture
References:
(1) Yucong Duan: Towards a Periodic Table of conceptualization and formalization
on State, Style, Structure, Pattern, Framework, Architecture, Service and so on.
SNPD 2019: 133-138
(2) Yucong Duan: Existence Computation: Revelation on Entity vs. Relationship for
Relationship Defined Everything of Semantics. SNPD 2019: 139-144
(3) Yucong Duan: Applications of Relationship Defined Everything of Semantics on
Existence Computation. SNPD 2019: 184-189
(4) Yucong Duan, Xiaobing Sun, Haoyang Che, Chunjie Cao, Zhao Li, Xiaoxian Yang:
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in Studies of Computing Intelligence, Springer, 2009.
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358 607773_The_end_of_Objective
_mathematics_as_a_return_to_Subjective/stats
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itations
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Article
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