How to define membership relation from subset relation and power set operation

In my talk at 2018 Chinese Mathematical Logic Conference, I asked if \((V,\subset,P)\) is epsilon-complete, namely if the membership relation can be recovered in the reduct. Professor Joseph S. Miller approached to me during the dinner and pointed out that it is epsilon-complete. Let me explain how.

Theorem

Let \((V,\in)\) be a structure of set theory, \((V,\subset,P)\) is the structure of the inclusion relation and the power set operation, which are defined in \((V,\in)\) as usual. Then \(\in\) is definable in \((V,\subset,P)\).

Proof.

Fix a set \(x\). Define \(y\) to be the \(\subset\)-least such that

\[\forall z \big((z\subset x\wedge z\neq x)\rightarrow P(z)\subset y\big).\]

Actually, \(y=P(x)-\{x\}\), so \(\{x\}= P(x) – y\). Since set difference can be defined from subset relation and \((V,\subset,\{x\})\) can define \(\in\), we are done.

\(\Box\)

Here is another argument figured out by Jialiang He and me after we heard Professor Miller’s Claim.

Proof.
Since \(\in\) can be defined in \((V,\subset,\bigcup)\) (see the slides). Fix a set \(A\), it suffices to show that we can define \(\bigcup A\) from \(\subset\) and \(P\).

Let \(B\) be the \(\subset\)-least set such that there is \(c\), \(B=P(c)\) and \(A\subset B\). Note that
\[
\bigcap\big\{P(d)\bigm|A\subset P(d)\big\}= P\big(\bigcap\big\{d\bigm|A\subset P(d)\big\}\big).
\]
Therefore, \(B\) is well-defined. Next, we show that
\[
\bigcap\big\{d\bigm|A\subset P(d)\big\}=\bigcup A.
\]
Clearly, \(A\subset P(\bigcup A)\). This proves the direction from left to right. For the other direction, if \(x\) is in an element of \(A\), then it is in an element of \(P(d)\) given \(A\subset P(d)\), i.e. it is an element of such \(d\).

Therefore \(\bigcup A\) is the unique set whose power set is \(B\).

\(\Box\)

Modal Logic 2018

Lecture: HGX205, M 18:30-21
Section: HGW2403, F 18:30-20

Syllabus

Exercise 01

  1. Prove that \(\neg\Box(\Diamond\varphi\wedge\Diamond\neg\varphi)\) is equivalent to \(\Box\Diamond\varphi\rightarrow\Diamond\Box\varphi\). What you have assumed?
  2. Define strategy and winning strategy for modal evaluation games. Prove Key Lemma: \(M,s\vDash\varphi\) iff V has a winning strategy in \(G(M,s,\varphi)\). Prove that modal evaluation games are determined, i.e. either V or F has a winning strategy.

And all exercises for Chapter 2 (see page 23, open minds)

Exercise 02

  1. Let \(T\) with root \(r\) be the tree unraveling of some possible world model, and \(T’\) be the tree unraveling of \(T,r\). Show that \(T\) and \(T’\) are isomorphic.
  2. Prove that the union of a set of bisimulations between \(M\) and \(N\) is a bisimulation between the two models.
  3. We define the bisimulation contraction of a possible world model \(M\) to be the “quotient model”. Prove that the relation links every world \(x\) in \(M\) to the equivalent class \([x]\) is a bisimulation between the original model and its bisimulation contraction.

And exercises for Chapter 3 (see page 35, open minds): 1 (a) (b), 2.

Exercise 03

  1. Prove that modal formulas (under possible world semantics) have ‘Finite Depth Property’.

And exercises for Chapter 4 (see page 47, open minds): 1 – 3.

Exercise 04

    1. Prove the principle of Replacement by Provable Equivalents: if \(\vdash\alpha\leftrightarrow\beta\), then \(\vdash\varphi[\alpha]\leftrightarrow\varphi[\beta]\).
    2. Prove the following statements.
      • “For each formula \(\varphi\), \(\vdash\varphi\) is equivalent to \(\vDash\varphi\)” is equivalent to “for each formula \(\varphi\), \(\varphi\) being consistent is equivalent to \(\varphi\) being satisfiable”.
      • “For every set of formulas \(\Sigma\) and formula \(\varphi\), \(\Sigma\vdash\varphi\) is equivalent to \(\Sigma\vDash\varphi\)” is equivalent to “for every set of formulas \(\Sigma\), \(\Sigma\) being consistent is equivalent to \(\Sigma\) being satisfiable”.
    3. Prove that “for each formula \(\varphi\), \(\varphi\) being consistent is equivalent to \(\varphi\) being satisfiable” using the finite version of Henkin model.

And exercises for Chapter 5 (see page 60, open minds): 1 – 5.

Exercise 05

Exercises for Chapter 6 (see page 69, open minds): 1 – 3.

Exercise 06

  1. Show that “being equivalent to a modal formula” is not decidable for arbitrary first-order formulas.

Exercises for Chapter 7 (see page 88, open minds): 1 – 6. For exercise 2 (a) – (d), replace the existential modality E with the difference modality D. In the clause (b) of exercise 4, “completeness” should be “correctness”.

Exercise 07

  1. Show that there are infinitely many non-equivalent modalities under T.
  2. Show that GL + Id is inconsistent and Un proves GL.
  3. Give a complete proof of the fact: In S5, Every formula is equivalent to one of modal depth \(\leq 1\).

Exercises for Chapter 8 (see page 99, open minds): 1, 2, 4 – 6.

 Exercise 08

  1. Let \(\Sigma\) be a set of modal formulas closed under substitution. Show that \[(W,R,V),w\vDash\Sigma~\Leftrightarrow~ (W,R,V’),w\vDash\Sigma\] hold for any valuation \(V\) and \(V’\). Define a \(p\)-morphism between \((W,R),w\) and \((W’,R’),w’\) as a “functional bisimulation”, namely bisimulation regardless of valuation. Show that if there is a \(p\)-morphism between \((W,R),w\) and \((W’,R’),w’\), then for any valuation \(V\) and \(V’\), we have \[(W,R,V),w\vDash\Sigma~\Leftrightarrow~ (W’,R’,V’),w\vDash\Sigma.\]

Exercises for Chapter 9 (see page 99, open minds).

为什么能行可计算的就是图灵可计算的(递归的)


这是对知乎问题为什么能行可计算的就是图灵可计算的 的回答。

如果我没理解错的话,题主想要问的实际上就是丘奇-图灵论题(Church-Turing Thesis)为什么成立。丘奇-图灵论题简单地说就是:

一个自然数上的函数\(f:\mathbb{N}^n\to\mathbb{N}\)是能行可计算的(effectively computable),当且仅当它是图灵可计算的(Turing computable)。
Continue reading “为什么能行可计算的就是图灵可计算的(递归的)”

集合论和一阶逻辑的关系?


有朋友在知乎上问:

http://logic.fudan.edu.cn/doc/Course/2014/MathLogic/Lecture01.pdf 最后一段:

  • 集合论可以被看作是一种一阶逻辑理论
  • 一阶逻辑的语法、语义概念都可以在集合论中定义, 关于一阶逻辑的定理可以被看作是集合论的定理

建立一阶逻辑时貌似很多定义都用了集合论描述。而集合论本身又可以被看作是一种一阶逻辑理论。不是有循环定义的嫌疑吗?

我在知乎的回答如下: Continue reading “集合论和一阶逻辑的关系?”