分类：数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|MHF5306

Posted on 2023年8月12日2023年8月28日 by statistics-lab

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数理逻辑Mathematical logic概念——模型和结构、完备性和不完备性——在数学的每一个分支中都被数学家使用。此外，逻辑在数学和哲学之间，以及数学和理论计算机科学之间提供了联系。这是一门应用日益广泛、具有重大内在意义的学科。

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数学代写|数理逻辑代写Mathematical logic代考|Examples of the Proof of Theorems from the Axioms

We now return to our system of axioms consisting of the primitive formulas a) through d) and the Rules of Inference $\alpha$ ) and $\beta$ ).

We will give a series of examples for the formal proof of theorems from the axioms. We will dwell on this point at some length, since experience shows that the maintenance of the purely formal point of view is especially difficult for the beginner.
In the proof of theorems it is found advisable to embody certain frequently recurring operations in the form of derived rules. By such a rule the result of the formal transformation in question will be anticipated once and for all, and the proof of the rule consists in stating the general procedure by which in each individual case the transformation is to be carried out in accordance with the primitive rules.

RULE I. If $\mathfrak{H} \mathrm{v} \mathfrak{H}$ is a theorem, then the same is true of $\mathfrak{T}$.
The proof is obtained immediately from Axiom a). By substitution in a) one obtains :

Further, since $\mathfrak{A} \vee \mathfrak{A}$ is a theorem, the Rule of Implication furnishes the formula $\mathcal{N}$.

RULE II. If $\mathfrak{P}$ is a theorem and $\mathfrak{B}$ any other formula whatsoever, then $\mathrm{F} \mathrm{v} \&$ is also a theorem.

This rule is obtained from Axiom b) in the same manner as Rule I from a).

In a like manner, Rules III and IV correspond to Axioms c) and d), and, more generally, a corresponding rule is associated with each formula which expresses a relation of implication.
RULE III. If $\mathfrak{A} \mathrm{v} \mathfrak{B}$ is a theorem, then the same is true of $\mathfrak{B} \mathrm{v} \mathfrak{A}$.
RULE IV. If $\mathfrak{A} \rightarrow \mathfrak{B}$ is a theorem and $\mathfrak{E}$ any other formula whatsoever, then $\mathbb{\mathfrak { U }} \rightarrow \mathbb{\mathfrak { C B }}$ is also a theorem.

数学代写|数理逻辑代写Mathematical logic代考|The Consistency of the System of Axioms

The axiomatic treatment of the sentential calculus makes it possible to ask the questions and introduce the considerations which are peculiar to the axiomatic method. The most important of the questions which arise are those concerning the consistency, independence, and completeness of the system of axioms. We shall consider first the consistency of the axioms.

The question of consistency may here be posed in an appropriate form. We shall call the axioms consistent if it is impossible to prove, by means of the calculus, two sentential combinations which are mutually contradictory, i.e. which result from the pair of sentences $X, \bar{X}$, if the same substitution for $X$ is made in both.
This definition of consistency requires an explanation. It might seem as though we were giving a preferred position to one particular logical principle-the principle of contradiction. The fact is, however, that the occurrence of a formal contradiction, i.e. the provability of two formulas $\mathfrak{A}$ and $\overline{\mathfrak{A}}$, would condemn the entire calculus as meaningless; for we have observed above that if two sentences of form $\mathfrak{A}$ and $\overline{\mathfrak{A}}$ were provable, the same would be true of any other sentences whatsoever. Thus consistency of the calculus in the sense of the definition has the same meaning as the stipulation that not every arbitrary formula be provable.

In order to prove the consistency of the calculus, we proceed as follows.

We interpret the sentential symbols $X, Y, Z, \ldots$ as arithmetical variables which assume only the values 0 and 1 . Further, we interpret $X \vee Y$ as the arithmetical product, and so define $\bar{X}$ that $\overline{0}$ shall equal 1 and $\overline{1}$ shall equal 0 . On the basis of this interpretation, every sentential combination represents an arithmetical function of the elementary sentences which assumes only the values 0 and 1 . If this function is identically 0 , then for the sake of brevity we will also speak of the symbolic expression as being identically 0 .

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Examples of the Proof of Theorems from the Axioms

现在我们回到由原始公式a)到d)和推理规则$\alpha$)和$\beta$)组成的公理系统。

我们将给出一系列从公理出发的定理的形式证明的例子。我们将详细讨论这一点，因为经验表明，对于初学者来说，保持纯粹的形式观点是特别困难的。
在定理的证明中，人们发现用派生规则的形式来体现某些经常出现的运算是可取的。根据这一规则，所讨论的形式变换的结果就可以一劳永逸地预测出来，而对这一规则的证明，就在于说明在每一个别情况下，按照原始规则来进行变换的一般程序。

规则一:如果$\mathfrak{H} \mathrm{v} \mathfrak{H}$是定理，那么$\mathfrak{T}$也是定理。
这个证明是直接由公理a)得到的。通过在a)中代入，可以得到:

更进一步，因为$\mathfrak{A} \vee \mathfrak{A}$是一个定理，隐含规则提供公式$\mathcal{N}$。

规则二。如果$\mathfrak{P}$是一个定理$\mathfrak{B}$其他任何公式，那么$\mathrm{F} \mathrm{v} \&$也是一个定理。

这条规则是从公理b)得到的，与从a)得到规则1的方式相同。

同样地，规则III和规则IV对应公理c)和公理d)，并且，更一般地说，对应的规则与每一个表示隐含关系的公式相关联。
规则三。如果$\mathfrak{A} \mathrm{v} \mathfrak{B}$是定理，那么$\mathfrak{B} \mathrm{v} \mathfrak{A}$也是定理。
规则四:如果$\mathfrak{A} \rightarrow \mathfrak{B}$是一个定理，$\mathfrak{E}$是任何其他公式，那么$\mathbb{\mathfrak { U }} \rightarrow \mathbb{\mathfrak { C B }}$也是一个定理。

数学代写|数理逻辑代写Mathematical logic代考|The Consistency of the System of Axioms

对句子演算的公理化处理，使得提出公理化方法所特有的问题和引入考虑成为可能。由此产生的最重要的问题是关于公理系统的一致性、独立性和完备性的问题。我们首先考虑这些公理的一致性。

一致性的问题可以在这里以适当的形式提出。如果不能用演算来证明两个相互矛盾的句子组合，即不能证明这两个句子组合是由一对句子$X, \bar{X}$产生的，如果在这两个句子组合中对$X$作了同样的替换，那么我们就称这两个公理是一致的。
一致性的定义需要一个解释。似乎我们是在给一个特殊的逻辑原理——矛盾原理——一个优先的位置。但事实是，如果出现形式矛盾，即两个公式$\mathfrak{A}$和$\overline{\mathfrak{A}}$的可证明性，就会使整个演算变成无意义的;因为我们在上面已经看到，如果两个形式为$\mathfrak{A}$和$\overline{\mathfrak{A}}$的句子是可证明的，那么其他任何句子也同样是可证明的。因此，在定义意义上的微积分的一致性与不是每一个任意公式都是可证明的规定具有同样的意义。

为了证明微积分的相合性，我们进行如下操作。

我们将句子符号$X, Y, Z, \ldots$解释为仅假设值0和1的算术变量。进一步，我们将$X \vee Y$解释为算术乘积，因此定义$\bar{X}$, $\overline{0}$等于1,$\overline{1}$等于0。在这种解释的基础上，每个句子组合都代表了基本句子的算术函数，它只假设0和1的值。如果这个函数等于0，那么为了简洁起见，我们也将符号表达式说成等于0。

数学代写|数理逻辑代写Mathematical logic代考请认准statistics-lab™

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金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|PHIL250

Posted on 2023年8月12日2023年8月28日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|The Principle of Duality

A remark which is important for the characterization of our calculus is based on Rule a3). From that rule it follows that given an expression which is formed from elementary sentences and their negations by means of conjunction and disjunction alone, we can obtain its negation by interchanging the symbols \& and $\mathrm{v}$, and replacing each elementary sentence by its negation.

We can make the following application of this. Let an expression of the form $\mathfrak{A} \sim \mathfrak{B}$, or as we also say, a logical equation, be established as logically true. (We use German letters to designate combinations of sentences whose exact formal structure is left undetermined, and we also use them for abbreviation.) Since $\mathfrak{I} \sim \mathfrak{B}$ has the same truth value as $\overline{\mathfrak{I}} \sim \overline{\mathbb{B}}$, we obtain another true expression by forming the negation of both sides of the equation. Now if both sides of the equation are formed from the elementary sentences and their negations by means of conjunction and disjunction only, we can apply the rule just mentioned. We obtain therefrom a formula which arises from the original equation $\mathfrak{A} \sim \mathfrak{B}$ by interchanging the signs $\&$ and $v$, and replacing each elementary sentence with its negation. Since this formula is logically true, it remains so if we replace each elementary sentence by its negation. In so doing, however, we cancel out the original replacement of the elementary sentences with their negations.

Thus we obtain the following Principle of Duality: From a formula $\mathfrak{A} \sim \mathfrak{B}$ which is logically true, and both of whose sides are formed from elementary sentences and their negations by conjunction and disjunction only, there results another true equation by the interchange of \& and $\mathrm{v}$.
Thus, for example,
$$
X(Y \& Z) \sim X Y \& X Z
$$
is logically true. This formula expresses the first distributive law. From it is derived, in accordance with the principle of duality, the formula
$$
X \& Y Z \sim(X \& Y)(X \& Z),
$$
which is also true and which expresses the second distributive law.
In the same way, the true formula
$$
(X \& \bar{X}) Y \sim Y
$$
is associated, according to the principle of duality, with the formula
$$
X \bar{X} \& Y \sim Y,
$$
which is likewise true.

数学代写|数理逻辑代写Mathematical logic代考|The Disjunctive Normal Form for Logical Expressions

There is an important application of the rule for forming the negation of a formula. We have seen that every logical expression can be brought into a normal form. This normal form consists of a conjunction of disjunctions, where each disjunct of every disjunction is either a negated or an un-negated elementary sentence. The tranformation of an expression into its normal form is effected by means of Rules a1) through a4). There is, in addition, still a second normal form, which consists of a disjunction of conjunctions. Each conjunct is a negated or an un-negated elementary sentence. We call this normal form “disjunctive,” and the preceding one, “conjunctive,” to distinguish between them.
The transformation of an expression into disjunctive normal form can be effected in the following way: Negate the original expression, then bring it into conjunctive normal form, and finally form the negation thereof by means of our rule.

One can also make use of the fact that, as far as Rules a1) through a4) are concerned, conjunction and disjunction play dual roles.

Just as one can determine by inspection whether or not an expression in conjunctive normal form is logically true, so also by means of the disjunctive normal form one can determine whether or not it is logically false. This is the case if and only if each disjunct contains an elementary sentence together with its negation.

The proof of this follows at once if one reflects that the negation of a disjunctive normal form reduces, by our rule, to a conjunctive normal form, and that a formula is logically false if and only if its negation is logically true.

As an example of the application of the disjunctive normal form, let us consider the sentential combination
$\bar{X} Y \& \bar{Y} Z \& X \& \bar{Z}$.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|The Principle of Duality

根据规则a3，有一点对我们的微积分的性质很重要。根据这条规则，我们可以得出这样的结论:给定一个表达式，这个表达式是由基本句和它们的否定仅仅通过连接和析取而形成的，我们可以通过交换符号＆和$\mathrm{v}$，并用每个基本句的否定来代替它，从而得到它的否定。

我们可以对此进行如下应用。假设一个表达式$\mathfrak{A} \sim \mathfrak{B}$的形式，或者我们也说，一个逻辑方程，被建立为逻辑真。(我们用德语字母来表示没有确定确切形式结构的句子组合，我们也用它们来表示缩写。)由于$\mathfrak{I} \sim \mathfrak{B}$与$\overline{\mathfrak{I}} \sim \overline{\mathbb{B}}$具有相同的真值，我们通过形成等式两边的负数来获得另一个真表达式。现在，如果方程的两边都是由基本句和它们的否定构成的，只通过连接和析取，我们可以应用刚才提到的规则。我们通过交换$\&$和$v$的符号，用每个基本句的否定代替每个基本句，从原方程$\mathfrak{A} \sim \mathfrak{B}$得到一个公式。既然这个公式在逻辑上是正确的，那么如果我们用每个基本句子的否定来代替它，它仍然是正确的。然而，这样一来，我们就取消了原来用否定句替换基本句的做法。

由此我们得到了对偶原理:从一个公式$\mathfrak{A} \sim \mathfrak{B}$，它的两边都是由基本句和它们的否定仅通过连接和析取而形成的，通过＆和$\mathrm{v}$的交换可以得到另一个真方程。
因此，例如，
$$
X(Y \& Z) \sim X Y \& X Z
$$
在逻辑上是正确的。这个公式表达了第一分配律。根据对偶原理，由它推导出公式
$$
X \& Y Z \sim(X \& Y)(X \& Z),
$$
这也是正确的，它表达了第二分配律。
同样的，真正的公式
$$
(X \& \bar{X}) Y \sim Y
$$
根据对偶原理，与公式有关联吗
$$
X \bar{X} \& Y \sim Y,
$$
这也是对的。

数学代写|数理逻辑代写Mathematical logic代考|The Disjunctive Normal Form for Logical Expressions

形成公式否定的规则有一个重要的应用。我们已经看到，每一个逻辑表达式都可以化为标准形式。该范式由析取词的合取词构成，其中每个析取词的每个合取词要么是否定的基本句，要么是未否定的基本句。将表达式转换为其标准形式是通过规则a1)到a4)实现的。此外，还有第二种范式，它由连词的析取组成。每个连词都是一个否定或非否定的基本句。我们称这种范式为“析取的”，而前一种范式为“合取的”，以区分它们。
将一个表达式转化为析取范式可以通过以下方式实现:对原表达式进行否定，然后将其转化为合取范式，最后通过我们的规则对原表达式进行否定。

我们还可以利用这样一个事实:就规则a1)到a4)而言，合取和析取扮演双重角色。

就像我们可以通过检查确定一个合取范式的表达式在逻辑上是否为真一样，我们也可以通过析取范式来确定它在逻辑上是否为假。当且仅当每个析取词包含一个基本句及其否定时，才会出现这种情况。

如果一个人考虑到，根据我们的规则，析取范式的否定可以归结为合取范式，并且当且仅当一个公式的否定在逻辑上为真，那么这个证明就立刻产生了。

作为析取范式应用的一个例子，让我们考虑一下句子组合
$\bar{X} Y \& \bar{Y} Z \& X \& \bar{Z}$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|MATH455

Posted on 2023年8月12日2023年8月12日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|George Boolos on the Second Incompleteness Theorem

George Boolos published a wonderful article on the Second Incompleteness Theorem. Here is the first page of his
Gödel’s Second Incompleteness Theorem Explained in Words of One Syllable ${ }^1$
First of all, when I say “proved”, what I will mean is “proved with the aid of the whole of math”. Now then: two plus two is four, as you well know. And, of course, , it can be proved that two plus two is four (proved, that is, with the aid of the whole of math, as I said, though in the case of two plus two, of course we do not need the whole of math to prove that it is four). And, as may not be quite so clear, it can be proved that it can be proved that two plus two is four, as well. And it can be proved that it can be proved that it can be proved that two plus two is four. And so on. In fact, if a claim can be proved, then it can be proved that the claim can be proved. And that too can be proved.

Now: two plus two is not five. And it can be proved that two plus two is not five. And it can be proved that it can be proved that two plus two is not five, and so on.
Thus: it can be proved that two plus two is not five. Can it be proved as well that two plus two is five? It would be a real blow to math, to say the least, if it could. If it could be proved that two plus two is five, then it could be proved that five is not five, and then there would be no claim that could not be proved, and math would be a lot of bunk.

So, we now want to ask, can it be proved that it can’t be proved that two plus two is five? Here’s the shock: no, it can’t. Or, to hedge a bit: if it can be proved that it can’t be proved that two plus two is five, then it can be proved as well that two plus two is five, and math is a lot of bunk. In fact, if math is not a lot of bunk, then no claim of the form “claim $X$ can’t be proved” can be proved.

So, if math is not a lot of bunk, then, though it can’t be proved that two plus two is five, it can’t be proved that it can’t be proved that two plus two is five.

By the way, in case you’d like to know: yes, it can be proved that if it can be proved that it can’t be proved that two plus two is five, then it can be proved that two plus two is five.

数学代写|数理逻辑代写Mathematical logic代考|Summing Up, Looking Ahead

This chapter is the capstone of the book. We have stated and proved (with only a small bit of handwaving) the two incompleteness theorems. I am sure that at this point you have a strong understanding of the theorems as well as an appreciation for the delicate arguments that are used in their proofs. These theorems have had a profound impact on the philosophical understanding of the subject of mathematics, and they involve some wonderful mathematics in and of themselves.

When I was in college, I was constantly asked by my mother what I was studying, and it became sort of a game to try to distill the essence of a course into a few phrases that would get the idea across without bogging down into technical details. As I think through this course, I think the summary I would give my mom would be something like this:
We looked at the language in which mathematics is written and looked at different kinds of mathematical universes, or structures. We thought about axioms and proofs. We worked through the proof of Gödel’s Completeness Theorem, which shows that any set of axioms that is not self-contradictory is true in some mathematical universe.

Then we changed our focus and thought about the natural numbers and ordinary arithmetic. We proved that $1+1$ is 2 . We proved that we can prove that $1+1$ is 2. But the amazing thing is that even though $1+1$ is not 3 , and even though we can prove that $1+1$ is not 3 , we cannot prove that we cannot prove that $1+1$ is 3 , unless mathematics is inconsistent. That is the core of Gödel’s Second Incompleteness Theorem.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|George Boolos on the Second Incompleteness Theorem

乔治·布洛斯发表了一篇关于第二不完备定理的精彩文章。这是他的第一页
Gödel用单音节词解释的第二不完备定理${}^1$
首先，当我说“证明”时，我的意思是“通过整个数学的帮助证明”。那么，二加二等于四，你们都知道。当然，二加二等于四是可以证明的(也就是说，正如我所说的，用全部数学来证明，虽然在二加二的情况下，我们当然不需要用全部数学来证明它是四)。可能不是很清楚，我们可以证明2加2等于4。它可以被证明可以被证明可以被证明2加2等于4。等等……事实上，如果一个主张可以被证明，那么就可以证明这个主张是可以被证明的。这也是可以证明的。

2加2不是5。可以证明2加2不等于5。可以证明2加2不等于5，等等。
因此:可以证明二加二不等于五。二加二等于五也能证明吗?至少可以这么说，这对数学来说是一个沉重的打击。如果可以证明二加二等于五，那么也可以证明五不是五，那么就没有不能被证明的命题了，数学就会变成一堆废话。

我们现在想问，能否证明2加2等于5不能被证明?令人震惊的是:不，它不能。或者，稍微回避一下:如果可以证明2加2等于5不能被证明，那么也可以证明2加2等于5，而数学就是一堆废话。事实上，如果数学不是一堆废话，那么“断言X不能被证明”这种形式的断言就不可能被证明。

所以，如果数学不是很多废话，那么，虽然不能证明二加二等于五，也不能证明二加二等于五。

顺便说一下，如果你想知道，是的，它可以被证明，如果2加2等于5不能被证明，那么它可以被证明，2加2等于5。

数学代写|数理逻辑代写Mathematical logic代考|Summing Up, Looking Ahead

这一章是本书的高潮。我们已经陈述并证明了这两个不完备定理(只用了一点手势)。我确信，现在你们对这些定理已经有了深刻的理解，并且对它们的证明中使用的微妙论点也有了欣赏。这些定理对数学这一主题的哲学理解产生了深远的影响，它们本身就包含了一些奇妙的数学。

当我上大学的时候，我妈妈经常问我学的是什么，这就变成了一种游戏，试着把一门课程的精髓提炼成几个短语，这样就可以在不陷入技术细节的情况下表达出想法。在我思考这门课程的过程中，我想我会给我妈妈的总结是这样的:
我们研究了书写数学的语言，研究了不同类型的数学宇宙，或结构。我们考虑了公理和证明。我们完成了Gödel完备性定理的证明，它表明在某些数学宇宙中，任何不自相矛盾的公理集都是正确的。

然后我们把注意力转移到自然数和普通算术上。我们证明了1+1 = 2。我们证明了1+1 = 2。但令人惊奇的是，即使1+1不是3，即使我们可以证明1+1不是3，我们不能证明我们不能证明1+1是3，除非数学是不一致的。这就是Gödel第二不完备性定理的核心。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|MATH301

Posted on 2023年7月27日2023年7月27日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|Summing Up, Looking Ahead

Well, in all likelihood you are exhausted at this point. This chapter has been full of dense, technical arguments with imposing definition piled upon imposing definition. We have established our axioms, discussed recursive sets, and talked about $\Delta$-definitions. You have just finished wading through an unending stream of $\Delta$-definitions that culminated with the formula Deduction $(c, f)$ which holds if and only if $c$ is a code for a deduction of the formula with Gödel number $f$. We have succeeded in coding up our deductive theory inside of number theory.

Let me reiterate this. If you look at that formula Deduction, what it looks like is a disjunction of a lot of equations and inequalities. Everything is written in the language $\mathcal{L}_{N T}$, so everything in that formula is of the form $S S S 0<S S O+x$ (with, it must be admitted, rather more $S$ ‘s than shown here). Although we have given these formulas names which suggest that they are about formulas and terms and tautologies and deductions, the formulas are formulas of elementary number theory, so the formulas don’t know that they are about anything beyond whether this number is bigger than that number, no matter how much you want to anthropomorphize them. The interpretation of the numbers as standing for formulas via the scheme of Gödel numbering is imposed on those numbers by us.

The next chapter brings us to the statement and the proof of Gödel’s Incompleteness Theorem. To give you a taste of things to come, notice that if we define the statement
$T h m_N(f)$ is:
$(\exists c)(\operatorname{Deduction}(c, f))$,
then $T h m_N(f)$ should hold if and only if $f$ is the Gödel number of a formula that is a theorem of $N$. I am sure you noticed that $T h m_N$ is not a $\Delta$-formula, and there is no way to fix that-we cannot bound the length of a deduction of a formula. But $T h m_N$ is a $\Sigma$-formula, and Proposition 4.3.9 tells us that true $\Sigma$-sentences are provable. That will be one of the keys to Gödel’s proof.

数学代写|数理逻辑代写Mathematical logic代考|The Self-Reference Lemma

Given any formula with only one free variable, we show how to construct a sentence that asserts that the given formula applies to itself:
Lemma 5.2.1 (Gödel’s Self-Reference Lemma). Let $\psi\left(v_1\right)$ be an $\mathcal{L}{N T}$-formula with only $v_1$ free. Then there is a sentence $\phi$ such that $$ N \vdash(\phi \leftrightarrow \psi(\overline{\Gamma \phi\urcorner})) . $$ Chaff: Look at how neat this is! Do you see how $\phi$ “says” $\psi$ is true of me? And we can do this for any formula $\psi$ ! What a cool idea! Proof. We will explicitly construct the needed $\phi$. Recall that we have recursive functions Num : $\mathbb{N} \rightarrow \mathbb{N}$ and Sub: $\mathbb{N}^3 \rightarrow \mathbb{N}$ such that $\left.\operatorname{Num}(n)=r{\bar{n}}\right\urcorner$ and $\left.\left.\left.\left.\operatorname{Sub}(r \alpha\urcorner, r_x\right\urcorner, r_t\right\urcorner\right)=r_{\alpha_t^x}\right\urcorner$. Since these functions are represented by our $\Delta$-definitions of Section 4.9 , we know that
$$
\begin{aligned}
& N \vdash[N u m(\bar{a}, y) \leftrightarrow y=\overline{\operatorname{Num}(a)}], \text { and that } \
& N \vdash[\operatorname{Sub}(\bar{a}, \bar{b}, \bar{c}, z) \leftrightarrow z=\overline{\operatorname{Sub}(a, b, c)}] .
\end{aligned}
$$
Chaff: Remember, Num is the function and Num is the $\mathcal{L}{N T}$-formula that represents the function! $\mathrm{Oh}$, and just because we’re going to need it, $\left\ulcorner{v_1}\right\urcorner=4$.
Now suppose that $\psi\left(v_1\right)$ is given as in the statement of the lemma. Let $\gamma\left(v_1\right)$ be
$$
\forall y \forall z\left[\left[N u m\left(v_1, y\right) \wedge S u b\left(v_1, \overline{4}, y, z\right)\right] \rightarrow \psi(z)\right]
$$

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Summing Up, Looking Ahead

很可能你现在已经精疲力尽了。本章充满了密集的、技术性的论证，以及一个接一个的定义。我们已经建立了公理，讨论了递归集合，并讨论了$\Delta$ -定义。您刚刚涉水完成了无穷无尽的$\Delta$定义流，这些定义以公式Deduction $(c, f)$告终，该公式当且仅当$c$是对数字为Gödel的公式进行演绎的代码$f$。我们已经成功地将我们的演绎理论编码在数论中。

让我重申一下。如果你看一下这个公式的演绎，它看起来就像很多方程和不等式的分离。所有内容都是用$\mathcal{L}_{N T}$语言编写的，因此该公式中的所有内容都是$S S S 0<S S O+x$的形式(必须承认，其中包含的$S$比这里显示的要多)。虽然我们给这些公式取的名字暗示着它们是关于公式、项、重言式和演绎的，但这些公式是初等数论的公式，所以这些公式不知道它们是关于这个数是否比那个数大之外的任何东西，不管你多么想把它们拟人化。通过Gödel编号方案将数字解释为代表公式是由我们强加给这些数字的。

下一章将给我们带来Gödel不完备定理的陈述和证明。为了让您了解将要发生的事情，请注意，如果我们定义语句
$T h m_N(f)$是:
$(\exists c)(\operatorname{Deduction}(c, f))$，
那么$T h m_N(f)$当且仅当$f$是$N$定理公式的Gödel个数时成立。我相信你注意到$T h m_N$不是一个$\Delta$公式，没有办法解决这个问题——我们不能限制公式的演绎长度。但是$T h m_N$是一个$\Sigma$ -公式，命题4.3.9告诉我们真$\Sigma$ -句子是可证明的。这将是Gödel证明的关键之一。

数学代写|数理逻辑代写Mathematical logic代考|The Self-Reference Lemma

给定任何只有一个自由变量的公式，我们展示如何构造一个断言给定公式适用于自身的句子:
引理5.2.1 (Gödel的Self-Reference引理)。让$\psi\left(v_1\right)$成为一个$\mathcal{L}{N T}$ -公式，其中只有$v_1$是空闲的。然后有一个句子$\phi$这样的$$ N \vdash(\phi \leftrightarrow \psi(\overline{\Gamma \phi\urcorner})) . $$ Chaff:看看这是多么整洁!你看到$\phi$“说”$\psi$对我来说是真的吗?我们可以对任何公式做这个$\psi$ !多酷的主意啊!证明。我们将显式地构造所需的$\phi$。回想一下，我们有递归函数Num: $\mathbb{N} \rightarrow \mathbb{N}$和Sub: $\mathbb{N}^3 \rightarrow \mathbb{N}$，因此$\left.\operatorname{Num}(n)=r{\bar{n}}\right\urcorner$和$\left.\left.\left.\left.\operatorname{Sub}(r \alpha\urcorner, r_x\right\urcorner, r_t\right\urcorner\right)=r_{\alpha_t^x}\right\urcorner$。由于这些函数由第4.9节的$\Delta$ -定义表示，因此我们知道这一点
$$
\begin{aligned}
& N \vdash[N u m(\bar{a}, y) \leftrightarrow y=\overline{\operatorname{Num}(a)}], \text { and that } \
& N \vdash[\operatorname{Sub}(\bar{a}, \bar{b}, \bar{c}, z) \leftrightarrow z=\overline{\operatorname{Sub}(a, b, c)}] .
\end{aligned}
$$
记住，Num是函数，Num是表示函数的$\mathcal{L}{N T}$ -公式!$\mathrm{Oh}$，因为我们需要它，$\left\ulcorner{v_1}\right\urcorner=4$。
现在假设$\psi\left(v_1\right)$是根据引理的陈述给出的。让$\gamma\left(v_1\right)$去吧
$$
\forall y \forall z\left[\left[N u m\left(v_1, y\right) \wedge S u b\left(v_1, \overline{4}, y, z\right)\right] \rightarrow \psi(z)\right]
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|MATH230

Posted on 2023年7月27日2023年8月25日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|Recursive Sets and Computer Programs

In this section we shall investigate the relationship between recursive sets and computer programs. Our discussion will be rather informal and will rely on your intuition about computers and computation.
One of the reasons that we must be rather informal when discussing computation is that the idea of a computation is rather

vague. In the mid-1930s many mathematicians developed theoretical constructs that tried to capture the idea of a computable function. Kurt Gödel’s recursive functions, the Turing Machines of Alan Turing, and Alonzo Church’s $\lambda$-calculus are three of the best-known models of computability.

One of the reasons that mathematicians accept these formal constructs as accurately modeling the intuitive notion of computability is that all of the formal analogs of computation that have been proposed have been proved to be equivalent. Thus it is known that a function is Turing computable if and only if it general recursive if and only if it is $\lambda$-computable. It is also known that these formal notions are equivalent to the idea of a function being computable on a computer, where we will say that a function $f$ is computable on an idealized computer if there is a computer program $P$ such that if the program $P$ is run with input $n$, the program will cause the computer to output $f(n)$ and halt.

数学代写|数理逻辑代写Mathematical logic代考|Coding Is Recursive

A basic part of our coding mechanism will be the ability to code finite sequences of numbers as a single number. A number $c$ is going to be a code for a sequence of numbers $\left\langle k_1, k_2, \ldots, k_n\right\rangle$ if and only if $c=2^{k_1} 3^{k_2} \cdots p_n^{k_n}$, where $p_n$ is the $n$th prime number. We assume, for convenience, that none of the $k_i$ are equal to 0 .

We show now that $N$ is strong enough to recognize code numbers. In other words, we want to establish

Proposition 4.6.1. The collection of numbers that are codes for finite sequences is a recursive set.

Proof. It is easy to write a $\Delta$-definition for the set of code numbers:
$$
\begin{gathered}
\text { Codenumber }(c) \text { is: } \
\text { Divides }(S S 0, c) \wedge(\forall z<c)(\forall y<z) \
{[(\text { Prime }(z) \wedge \text { Dvides }(z, c) \wedge \text { Prmepair }(y, z)) \rightarrow \text { Divides }(y, c)] .}
\end{gathered}
$$
Notice that Codenumber(c) is a formula with one free variable, c. If you look at it carefully, all the formula says is that $c$ is divisible by 2 and if any prime divides $c$, so do all the earlier primes. Since the definition above is a $\Delta$-definition, Corollary 4.3 .11 tells us that the set CODENUMBER is a recursive set.

Since CoDenUmber is recursive and Codenumber is a $\Delta$-formula, we now know (for example) that
$$
N \vdash \text { Codenumber }(\overline{18})
$$
and
$$
N \vdash \neg \text { Codenumber(45). }
$$

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Recursive Sets and Computer Programs

在本节中，我们将研究递归集与计算机程序之间的关系。我们的讨论将是非正式的，将依赖于你对计算机和计算的直觉。
我们在讨论计算时必须相当不正式的原因之一是，计算的概念相当

模糊的。在20世纪30年代中期，许多数学家发展了理论结构，试图捕捉可计算函数的概念。Kurt Gödel的递归函数，Alan Turing的图灵机和Alonzo Church的$\lambda$ -微积分是三个最著名的可计算模型。

数学家之所以接受这些形式结构，因为它们准确地模拟了可计算性的直观概念，其中一个原因是，所有已经提出的计算的形式类比都被证明是等价的。因此，我们知道一个函数是图灵可计算的当且仅当它是一般递归的当且仅当它是$\lambda$ -可计算的。我们也知道，这些形式化的概念等价于一个函数在计算机上是可计算的，我们会说一个函数$f$在一个理想化的计算机上是可计算的，如果有一个计算机程序$P$，如果该程序$P$在输入$n$的情况下运行，该程序将导致计算机输出$f(n)$并停止。

数学代写|数理逻辑代写Mathematical logic代考|Coding Is Recursive

我们编码机制的一个基本部分将是将有限的数字序列编码为单个数字的能力。一个数字$c$将是一个数字序列$\left\langle k_1, k_2, \ldots, k_n\right\rangle$的代码当且仅当$c=2^{k_1} 3^{k_2} \cdots p_n^{k_n}$，其中$p_n$是$n$第一个素数。为了方便，我们假设，没有一个$k_i$等于0。

现在我们证明$N$足够强大，可以识别代码。换句话说，我们要建立

提案4.6.1作为有限序列编码的数字集合是一个递归集合。

证明。很容易为代码集编写$\Delta$ -定义:
$$
\begin{gathered}
\text { Codenumber }(c) \text { is: } \
\text { Divides }(S S 0, c) \wedge(\forall z<c)(\forall y<z) \
{[(\text { Prime }(z) \wedge \text { Dvides }(z, c) \wedge \text { Prmepair }(y, z)) \rightarrow \text { Divides }(y, c)] .}
\end{gathered}
$$
注意到codennumber (c)是一个有一个自由变量c的公式。如果你仔细看，所有的公式都说$c$能被2整除，如果任何质数能整除$c$，那么所有之前的质数也能整除。由于上面的定义是一个$\Delta$ -定义，推论4.3 .11告诉我们集合codennumber是一个递归集合。

由于codennumber是递归的，并且codennumber是一个$\Delta$ -公式，我们现在知道(例如
$$
N \vdash \text { Codenumber }(\overline{18})
$$
和
$$
N \vdash \neg \text { Codenumber(45). }
$$以上翻译结果来自有道神经网络翻译（YNMT）· 通用场景

逐句对照

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|M-781

Posted on 2023年7月27日2023年8月25日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|Extending to a Maximal Consistent Set of Sentences

As you recall, we were going to construct our model $\mathfrak{A}$ in such a way that the sentences that were true in $\mathfrak{A}$ were exactly the elements of a set of sentences $\boldsymbol{\Sigma}^{\prime}$. It is time to build $\boldsymbol{\Sigma}^{\prime}$. Since every sentence is either true or false in a given model, it will be necessary for us to make sure that for every sentence $\sigma \in \mathcal{L}^{\prime}$, either $\sigma \in \Sigma^{\prime}$ or $\neg \sigma \in \Sigma^{\prime}$. Since we can’t have both $\sigma$ and $\neg \sigma$ true in any structure, we must also make sure that we don’t put both $\sigma$ and $\neg \sigma$ into $\Sigma^{\prime}$. Thus, $\Sigma^{\prime}$ will be a maximal consistent extension of $\hat{\Sigma}$.

To build this extension, fix an enumeration of all of the $\mathcal{L}^{\prime}$ sentences
$$
\sigma_1, \sigma_2, \ldots, \sigma_n, \ldots
$$
We can do this as $\mathcal{L}^{\prime}$ is countable, being a countable union of countable sets. Now we work our way through this list, one sentence at a time, adding either $\sigma_n$ or the denial of $\sigma_n$ to our growing list of sentences, depending on which one keeps our collection consistent.
Here are the details. Let $\Sigma^0=\hat{\Sigma}$, and assume that $\Sigma^k$ is known to be a consistent set of $\mathcal{L}^{\prime}$-sentences. We will show how to build $\Sigma^{k+1} \supseteq$ $\Sigma^k$ and prove that $\Sigma^{k+1}$ is also a consistent set of $\mathcal{L}^{\prime}$-sentences. Then we let
$$
\Sigma^{\prime}=\Sigma^0 \cup \Sigma^1 \cup \Sigma^2 \cup \cdots \cup \Sigma^n \cup \cdots .
$$
You will prove in Exercise 4 that $\Sigma^{\prime}$ is a consistent set of sentences. It will be obvious from the construction of $\Sigma^{k+1}$ from $\Sigma^k$ that $\Sigma^{\prime}$ is maximal, and thus we will have completed our task of producing a maximal consistent extension of $\hat{\Sigma}$.

So all we have to do is describe how to get $\Sigma^{k+1}$ from $\Sigma^k$ and prove that $\Sigma^{k+1}$ is consistent. Given $\Sigma^k$, consider the set $\Sigma^k \cup\left{\sigma^{k+1}\right}$, where $\sigma_{k+1}$ is the $(k+1)$ st element of our fixed list of all of the $\mathcal{L}^{\prime}$ sentences. Let
$$
\Sigma^{k+1}= \begin{cases}\Sigma^k \cup\left{\sigma_{k+1}\right} & \text { if } \Sigma^k \cup\left{\sigma_{k+1}\right} \text { is consistent } \ \Sigma^k \cup\left{\neg \sigma_{k+1}\right} & \text { otherwise. }\end{cases}
$$

数学代写|数理逻辑代写Mathematical logic代考|Construction of the Model-Preliminaries

I have mentioned a few times that the model of $\Sigma$ that we are going to construct will have as its universe the collection of variable-free terms of the language $\mathcal{L}^{\prime}$. It is now time to confess that I have lied. It is easy to see why the plan of using the terms as the elements of the universe is doomed to failure. Suppose that there are two different terms $t_1$ and $t_2$ of the language and somewhere in $\Sigma^{\prime}$ is the sentence $t_1=t_2$. If the terms were the elements of the universe, then we could not model $\Sigma^{\prime}$, as the two terms $t_1$ and $t_2$ are not the same (they are typographically distinct), while $\Sigma^{\prime}$ demands that they be equal. Our solution to this problem is to take the collection of variable-free terms, define an equivalence relation on that set, and then construct a model from the equivalence classes of the variable-free terms.

So let $T$ be the set of variable-free terms of the language $\mathcal{L}^{\prime}$, and define a relation $\sim$ on $T$ by
$$
t_1 \sim t_2 \text { if and only if }\left(t_1=t_2\right) \in \Sigma^{\prime} .
$$
It is not difficult to show that $\sim$ is an equivalence relation. We will verify that $\sim$ is symmetric, leaving reflexivity and transitivity to the Exercises.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Extending to a Maximal Consistent Set of Sentences

正如您所记得的，我们将以这样一种方式构建我们的模型$\mathfrak{A}$，即$\mathfrak{A}$中为真的句子正是一组句子$\boldsymbol{\Sigma}^{\prime}$的元素。是时候建立$\boldsymbol{\Sigma}^{\prime}$了。因为在给定的模型中，每个句子要么为真要么为假，所以我们有必要确保对于每个句子$\sigma \in \mathcal{L}^{\prime}$，要么$\sigma \in \Sigma^{\prime}$要么$\neg \sigma \in \Sigma^{\prime}$。因为我们不能在任何结构中同时拥有$\sigma$和$\neg \sigma$为真，我们还必须确保不要将$\sigma$和$\neg \sigma$都放入$\Sigma^{\prime}$中。因此，$\Sigma^{\prime}$将是$\hat{\Sigma}$的最大一致扩展。

要构建这个扩展，修复所有$\mathcal{L}^{\prime}$句子的枚举
$$
\sigma_1, \sigma_2, \ldots, \sigma_n, \ldots
$$
我们可以这样做因为$\mathcal{L}^{\prime}$是可数的，是可数集合的可数并集。现在我们逐句逐句地处理这个列表，将$\sigma_n$或否认$\sigma_n$添加到我们不断增长的句子列表中，这取决于哪一个使我们的集合保持一致。
详情如下。设$\Sigma^0=\hat{\Sigma}$，并假设已知$\Sigma^k$是一组一致的$\mathcal{L}^{\prime}$ -句子。我们将展示如何构建$\Sigma^{k+1} \supseteq$$\Sigma^k$并证明$\Sigma^{k+1}$也是一组一致的$\mathcal{L}^{\prime}$ -句子。然后我们让
$$
\Sigma^{\prime}=\Sigma^0 \cup \Sigma^1 \cup \Sigma^2 \cup \cdots \cup \Sigma^n \cup \cdots .
$$
您将在练习4中证明$\Sigma^{\prime}$是一组一致的句子。从$\Sigma^k$构造$\Sigma^{k+1}$可以明显看出，$\Sigma^{\prime}$是最大的，因此我们将完成生成$\hat{\Sigma}$的最大一致扩展的任务。

所以我们所要做的就是描述如何从$\Sigma^k$得到$\Sigma^{k+1}$并证明$\Sigma^{k+1}$是一致的。给定$\Sigma^k$，考虑集合$\Sigma^k \cup\left{\sigma^{k+1}\right}$，其中$\sigma_{k+1}$是所有$\mathcal{L}^{\prime}$句子的固定列表的$(k+1)$ st元素。让
$$
\Sigma^{k+1}= \begin{cases}\Sigma^k \cup\left{\sigma_{k+1}\right} & \text { if } \Sigma^k \cup\left{\sigma_{k+1}\right} \text { is consistent } \ \Sigma^k \cup\left{\neg \sigma_{k+1}\right} & \text { otherwise. }\end{cases}
$$

数学代写|数理逻辑代写Mathematical logic代考|Construction of the Model-Preliminaries

我已经提到过几次，我们将要构建的$\Sigma$模型将把$\mathcal{L}^{\prime}$语言的无变量项集合作为它的宇宙。是时候承认我撒了谎了。很容易看出为什么使用术语作为宇宙元素的计划注定要失败。假设语言中有两个不同的术语$t_1$和$t_2$，而$\Sigma^{\prime}$中的某个地方有一个句子$t_1=t_2$。如果这两个术语是宇宙的元素，那么我们就不能对$\Sigma^{\prime}$进行建模，因为两个术语$t_1$和$t_2$不相同(它们在排版上不同)，而$\Sigma^{\prime}$要求它们相等。我们的解决方法是取一组自由变量项，在该集合上定义一个等价关系，然后由这些自由变量项的等价类构造一个模型。

因此，设$T$为$\mathcal{L}^{\prime}$语言的无变量项的集合，并在$T$ by上定义一个关系$\sim$
$$
t_1 \sim t_2 \text { if and only if }\left(t_1=t_2\right) \in \Sigma^{\prime} .
$$
不难证明$\sim$是一个等价关系。我们将验证$\sim$是对称的，将反身性和及物性留给练习。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|Propositional Consequence

Posted on 2023年7月7日2023年7月7日 by statistics-lab

如果你也在怎样代写数理逻辑 Mathematical logic 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。数理逻辑Mathematical logic对数学中形式逻辑的研究。主要子领域包括模型理论、证明理论、集合理论和递归理论。数学逻辑的研究通常涉及形式逻辑系统的数学属性，如其表达或演绎能力。

数理逻辑Mathematical logic在19世纪中期作为数学的一个子领域出现，反映了两个传统的交汇：形式化的哲学逻辑和数学。 “数理逻辑，也被称为’逻辑学’、’符号逻辑’、’逻辑代数’，最近还被简单地称为’形式逻辑’，是在上个世纪过程中借助人工符号和严格的演绎方法阐述的一套逻辑理论。”在这次出现之前，逻辑是与修辞学、计算学、通过三段论和哲学一起研究。20世纪上半叶出现了基本结果的爆发，同时伴随着对数学基础的激烈争论。

数学代写|数理逻辑代写Mathematical logic代考|Propositional Consequence

In all likelihood you are familiar with tautologies of propositional logic. They are simply formulas like $(A \rightarrow B) \leftrightarrow(\neg B \rightarrow \neg A)$. If you are comfortable with tautologies, feel free to skip over the next couple of paragraphs. If not, what follows is a very brief review of a portion of propositional logic.

We work with a restricted language $\mathcal{P}$, consisting only of a set of propositional variables $A, B, C, \ldots$ and the connectives $\mathrm{V}$ and $\neg$. Notice there are no quantifiers, no relation symbols, no function symbols, and no constants. Formulas of propositional logic are defined as being the collection of all $\phi$ such that either $\phi$ is a propositional variable, or $\phi$ is $(\neg \alpha)$, or $\phi$ is $(\alpha \vee \beta)$, with $\alpha$ and $\beta$ being formulas of propositional logic.

Each propositional variable can be assigned one of two truth values, $T$ or $F$, corresponding to truth and falsity. Given such an assignment (which is really a function $v:$ propositional variables $\rightarrow$ ${T, F}$ ), we can extend $v$ to a function $\bar{v}$ assigning a truth value to any propositional formula as follows:
$$
\bar{v}(\phi)= \begin{cases}v(\phi) & \text { if } \phi \text { is a propositional variable } \ F & \text { if } \phi \text { is }(\neg \alpha) \text { and } \bar{v}(\alpha)=T \ F & \text { if } \phi \text { is }(\alpha \vee \beta) \text { and } \bar{v}(\alpha)=\bar{v}(\beta)=F \ T & \text { otherwise. }\end{cases}
$$
Now we say that a propositional formula $\phi$ is a tautology if and only if $\bar{v}(\phi)=T$ for any truth assignment $v$.

数学代写|数理逻辑代写Mathematical logic代考|Quantifier Rules

The motivation behind our quantifier rules is very simple. Suppose, without making any particular assumptions about $x$, that you were able to prove ” $x$ is an ambitious aardvark.” Then it seems reasonable to claim that you have proved ” $\forall x) x$ is an ambitious aardvark.” Dually, if you were able to prove the Riemann Hypothesis from the assumption that ” $x$ is a bossy bullfrog,” then from the assumption ” $\exists x) x$ is a bossy bullfrog,” you should still be able to prove the Riemann Hypothesis.

Definition 2.4.6. Suppose that the variable $x$ is not free in the formula $\psi$. Then both of the following are rules of inference of type (QR):
$$
\begin{aligned}
& \langle{\psi \rightarrow \phi}, \psi \rightarrow(\forall x \phi)\rangle \
& \langle{\phi \rightarrow \psi},(\exists x \phi) \rightarrow \psi\rangle .
\end{aligned}
$$
The “not making any particular assumptions about $x$ ” comment is made formal by the requirement that $x$ not be free in $\psi$.
Chaff: Just to make sure that you are not lost in the brackets of the definition, what we are saying here is that if $x$ is not free in $\psi$ :

From the formula $\psi \rightarrow \phi$, you may deduce $\psi \rightarrow$ $(\forall x \phi)$.
From the formula $\phi \rightarrow \psi$, you may deduce $(\exists x \phi) \rightarrow$ $\psi$.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Propositional Consequence

你很可能熟悉命题逻辑的重言式。它们是简单的公式，比如$(A \rightarrow B) \leftrightarrow(\neg B \rightarrow \neg A)$。如果你对重言式感到舒服，可以随意跳过接下来的几段。如果没有，下面是对部分命题逻辑的简要回顾。

我们使用一种受限制的语言$\mathcal{P}$，它仅由一组命题变量$A, B, C, \ldots$和连接词$\mathrm{V}$和$\neg$组成。注意这里没有量词，没有关系符号，没有函数符号，也没有常量。命题逻辑的公式定义为所有$\phi$的集合，使得$\phi$为命题变量，或者$\phi$为$(\neg \alpha)$，或者$\phi$为$(\alpha \vee \beta)$，其中$\alpha$和$\beta$为命题逻辑的公式。

每个命题变量可以被赋予两个真值中的一个，$T$或$F$，对应于真和假。给定这样一个赋值(它实际上是一个函数$v:$命题变量$\rightarrow$${T, F}$)，我们可以将$v$扩展为一个函数$\bar{v}$，为任意命题公式赋真值，如下所示:
$$
\bar{v}(\phi)= \begin{cases}v(\phi) & \text { if } \phi \text { is a propositional variable } \ F & \text { if } \phi \text { is }(\neg \alpha) \text { and } \bar{v}(\alpha)=T \ F & \text { if } \phi \text { is }(\alpha \vee \beta) \text { and } \bar{v}(\alpha)=\bar{v}(\beta)=F \ T & \text { otherwise. }\end{cases}
$$
现在我们说一个命题公式$\phi$是一个重言式当且仅当$\bar{v}(\phi)=T$对于任何真值赋值$v$。

数学代写|数理逻辑代写Mathematical logic代考|Quantifier Rules

我们的量词规则背后的动机非常简单。假设，在不对$x$做任何特殊假设的情况下，您能够证明“$x$是一只雄心勃勃的食蚁兽”。那么，声称你已经证明了“$\forall x) x$是一只雄心勃勃的食蚁兽”似乎是合理的。双重地，如果你能够从“$x$是一只专横的牛蛙”的假设中证明黎曼假设，那么从“$\exists x) x$是一只专横的牛蛙”的假设中，你应该仍然能够证明黎曼假设。

2.4.6.定义假设变量$x$在公式$\psi$中不是自由的。则以下两个都是类型(QR)的推理规则:
$$
\begin{aligned}
& \langle{\psi \rightarrow \phi}, \psi \rightarrow(\forall x \phi)\rangle \
& \langle{\phi \rightarrow \psi},(\exists x \phi) \rightarrow \psi\rangle .
\end{aligned}
$$
“不对$x$做任何特殊假设”的注释通过要求$x$在$\psi$中不是免费的而变得正式。
Chaff:只是为了确保你没有迷失在定义的括号中，我们在这里说的是，如果$x$在$\psi$中不是免费的:

从公式$\psi \rightarrow \phi$，你可以推导出$\psi \rightarrow$$(\forall x \phi)$。

从公式$\phi \rightarrow \psi$，你可以推导出$(\exists x \phi) \rightarrow$$\psi$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|Logical Implication

Posted on 2023年7月7日2023年7月7日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|Logical Implication

At first glance it seems that a large portion of mathematics can be broken down into answering questions of the form: If I know this statement is true, is it necessarily the case that this other statement is true? In this section we will formalize that question.

Definition 1.9.1. Suppose that $\Delta$ and $\Gamma$ are sets of $\mathcal{L}$-formulas. We will say that $\Delta$ logically implies $\Gamma$ and write $\Delta \models \Gamma$ if for every $\mathcal{L}$-structure $\mathfrak{A}$, if $\mathfrak{A} \models \Delta$, then $\mathfrak{A} \models \Gamma$.

This definition is a little bit tricky. It says that if $\Delta$ is true in $\mathfrak{A}$, then $\Gamma$ is true in $\mathfrak{A}$. Remember, for $\Delta$ to be true in $\mathfrak{A}$, it must be the case that $\mathfrak{A} \models \Delta[s]$ for every assignment function $s$. See Exercise 4 .
If $\Gamma={\gamma}$ is a set consisting of a single formula, we will write $\Delta \vDash \gamma$ rather than the official $\Delta \models{\gamma}$.

Definition 1.9.2. An $\mathcal{L}$-formula $\phi$ is said to be valid if $\emptyset \vDash \phi$, in other words, if $\phi$ is true in every $\mathcal{L}$-structure with every assignment function $s$. In this case, we will write $\vDash \phi$.
Chaff: It doesn’t seem like it would be easy to check whether $\Delta \vDash \Gamma$. To do so directly would mean that we would have to examine every possible $\mathcal{L}$-structure and every possible assignment function $s$, of which there will be many.

I’m also sure that you’ve noticed that this double turnstyle symbol, $\models$, is getting a lot of use. Just remember that if there is a structure on the left, $\boldsymbol{A} \models \sigma$, we are discussing truth in a single structure. If there is a set of sentences on the left, $\Gamma \models \sigma$, then we are discussing logical implication.

数学代写|数理逻辑代写Mathematical logic代考|Logical Implication

At first glance it seems that a large portion of mathematics can be broken down into answering questions of the form: If $I$ know this statement is true, is it necessarily the case that this other statement is true? In this section we will formalize that question.

Definition 1.9.1. Suppose that $\Delta$ and $\Gamma$ are sets of $\mathcal{L}$-formulas. We will say that $\Delta$ logically implies $\Gamma$ and write $\Delta \vDash \Gamma$ if for every $\mathcal{L}$-structure $\mathfrak{A}$, if $\mathfrak{A} \models \Delta$, then $\mathfrak{A} \models \Gamma$.

This definition is a little bit tricky. It says that if $\Delta$ is true in $\mathfrak{A}$, then $\Gamma$ is true in $\mathfrak{A}$. Remember, for $\Delta$ to be true in $\mathfrak{A}$, it must be the case that $\mathfrak{A}=\Delta[s]$ for every assignment function $s$. See Exercise 4 .
If $\Gamma={\gamma}$ is a set consisting of a single formula, we will write $\Delta \models \gamma$ rather than the official $\Delta \models{\gamma}$.

Definition 1.9.2. An $\mathcal{L}$-formula $\phi$ is said to be valid if $\emptyset \vDash \phi$, in other words, if $\phi$ is true in every $\mathcal{L}$-structure with every assignment function $s$. In this case, we will write $\models \phi$.
Chaff: It doesn’t seem like it would be easy to check whether $\Delta \vDash \Gamma$. To do so directly would mean that we would have to examine every possible $\mathcal{L}$-structure and every possible assignment function $s$, of which there will be many.

I’m also sure that you’ve noticed that this double turnstyle symbol, $\models$, is getting a lot of use. Just remember that if there is a structure on the left, $\mathfrak{A} \models \sigma$, we are discussing truth in a single structure. If there is a set of sentences on the left, $\Gamma \models \sigma$, then we are discussing logical implication.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Logical Implication

乍一看，大部分数学似乎可以分解为回答这样的问题:如果我知道这个陈述是真的，那么另一个陈述是否也一定是真的?在本节中，我们将把这个问题形式化。

1.9.1.定义假设$\Delta$和$\Gamma$是$\mathcal{L}$ -公式的集合。我们将说$\Delta$在逻辑上暗示$\Gamma$，并为每个$\mathcal{L}$ -结构$\mathfrak{A}$写$\Delta \models \Gamma$ if，如果$\mathfrak{A} \models \Delta$，那么$\mathfrak{A} \models \Gamma$。

这个定义有点棘手。它说，如果$\Delta$在$\mathfrak{A}$中为真，那么$\Gamma$在$\mathfrak{A}$中为真。记住，要使$\Delta$在$\mathfrak{A}$中为真，对于每个赋值函数$s$，必须是$\mathfrak{A} \models \Delta[s]$。参见练习4。
如果$\Gamma={\gamma}$是由单个公式组成的集合，我们将写成$\Delta \vDash \gamma$，而不是官方的$\Delta \models{\gamma}$。

1.9.2.定义如果$\emptyset \vDash \phi$，换句话说，如果$\phi$在具有每个赋值函数$s$的每个$\mathcal{L}$ -结构中为真，则认为$\mathcal{L}$ -公式$\phi$是有效的。在本例中，我们将写入$\vDash \phi$。
Chaff:似乎不太容易检查$\Delta \vDash \Gamma$。要直接这样做，就意味着我们必须检查每一个可能的$\mathcal{L}$ -结构和每一个可能的分配函数$s$，这样的函数有很多。

我也确信你已经注意到这个双转弯风格的符号，$\models$，得到了很多的使用。只要记住，如果左边有一个结构，$\boldsymbol{A} \models \sigma$，我们讨论的是单一结构中的真理。如果左边有一组句子，$\Gamma \models \sigma$，那么我们正在讨论逻辑暗示。

数学代写|数理逻辑代写Mathematical logic代考|Logical Implication

乍一看，大部分数学似乎可以分解为回答这样的问题:如果$I$知道这个陈述是真的，那么另一个陈述是否必然是真的?在本节中，我们将把这个问题形式化。

1.9.1.定义假设$\Delta$和$\Gamma$是$\mathcal{L}$ -公式的集合。我们将说$\Delta$在逻辑上暗示$\Gamma$，并为每个$\mathcal{L}$ -结构$\mathfrak{A}$写$\Delta \vDash \Gamma$ if，如果$\mathfrak{A} \models \Delta$，那么$\mathfrak{A} \models \Gamma$。

这个定义有点棘手。它说，如果$\Delta$在$\mathfrak{A}$中为真，那么$\Gamma$在$\mathfrak{A}$中为真。记住，要使$\Delta$在$\mathfrak{A}$中为真，对于每个赋值函数$s$，必须是$\mathfrak{A}=\Delta[s]$。参见练习4。
如果$\Gamma={\gamma}$是由单个公式组成的集合，我们将写成$\Delta \models \gamma$，而不是官方的$\Delta \models{\gamma}$。

1.9.2.定义如果$\emptyset \vDash \phi$，换句话说，如果$\phi$在具有每个赋值函数$s$的每个$\mathcal{L}$ -结构中为真，则认为$\mathcal{L}$ -公式$\phi$是有效的。在本例中，我们将写入$\models \phi$。
Chaff:似乎不太容易检查$\Delta \vDash \Gamma$。要直接这样做，就意味着我们必须检查每一个可能的$\mathcal{L}$ -结构和每一个可能的分配函数$s$，这样的函数有很多。

我也确信你已经注意到这个双转弯风格的符号，$\models$，得到了很多的使用。只要记住，如果左边有一个结构，$\mathfrak{A} \models \sigma$，我们讨论的是单一结构中的真理。如果左边有一组句子，$\Gamma \models \sigma$，那么我们正在讨论逻辑暗示。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|Naïvely

Posted on 2023年7月7日2023年7月7日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|Naïvely

Let us begin by talking informally about mathematical structures and mathematical languages. There is no doubt that you have worked with mathematical models in several previous mathematics courses, although in all likelihood it was not pointed out to you at the time. For example, if you have taken a course in linear algebra, you have some experience working with $\mathbb{R}^2, \mathbb{R}^3$, and $\mathbb{R}^n$ as examples of vector spaces. In high school geometry you learned that the plane is a “model” of Euclid’s axioms for geometry. Perhaps you have taken a class in abstract algebra, where you saw several examples of groups: The integers under addition, permutation groups, and the group of invertible $n \times n$ matrices with the operation of matrix multiplication are all examples of groups-they are “models” of the group axioms. All of these are mathematical models, or structures. Different structures are used for different purposes.

Suppose we think about a particular mathematical structure, for example $\mathbb{R}^3$, the collection of ordered triples of real numbers. If we try to do plane Euclidean geometry in $\mathbb{R}^3$, we fail miserably, as (for example) the parallel postulate is false in this structure. On the other hand, if we want to do linear algebra in $\mathbb{R}^3$, all is well and good, as we can think of the points of $\mathbb{R}^3$ as vectors and let the scalars be real numbers. Then the axioms for a real vector space are all true when interpreted in $\mathbb{R}^3$. We will say that $\mathbb{R}^3$ is a model of the axioms for a vector space, whereas it is not a model for Euclid’s axioms for geometry.

As you have no doubt noticed, our discussion has introduced two separate types of things to worry about. First, there are the mathematical models, which you can think of as the mathematical worlds, or constructs. Examples of these include $\mathbb{R}^3$, the collection of polynomials of degree 17 , the set of $3 \times 2$ matrices, and the real line. We have also been talking about the axioms of geometry and vector spaces, and these are something different. Let us discuss those axioms for a moment.

数学代写|数理逻辑代写Mathematical logic代考|Languages

We will be constructing a very restricted formal language, and our goal in constructing that language will be to be able to form certain statements about certain kinds of mathematical structures. For our work, it will be necessary to be able to talk about constants, functions, and relations, and so we will need symbols to represent them.
Chaff: Let me emphasize this once more. Right now we are discussing the syntax of our language, the marks on the paper. We are not going to worry about the semantics, or meaning, of those marks until later-at least not formally. But it is silly to pretend that the intended meanings do not drive our choice of symbols and the way in which we use them. If we want to discuss left-hemisemi-demi-rings, our formal language should include the function and relation symbols that mathematicians in this lucrative and exciting field customarily use, not the symbols involved in chess, bridge, or right-hemi-semipara-fields. It is not our goal to confuse anyone more than is necessary. So you should probably go through the exercise right now of taking a guess at a reasonable language to use if our intended field of discussion was, say, the theory of the natural numbers. See Exercise 1.
Definition 1.2.1. A first-order language $\mathcal{L}$ is an infinite collection of distinct symbols, no one of which is properly contained in another, separated into the following categories:

Parentheses: (,).
Connectives: $\vee, \neg$.
Quantrfier: $\forall$.
Variables, one for each positive integer $n: v_1, v_2, \ldots, v_n, \ldots$ The set of variable symbols will be denoted Vars.
Equality symbol: =.
Constant symbols: Some set of zero or more symbols.
Function symbols: For each positive integer $n$, some set of zero or more $n$-ary function symbols.
Relation symbols: For each positive integer $n$, some set of zero or more $n$-ary relation symbols.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|Naïvely

让我们从非正式地讨论数学结构和数学语言开始。毫无疑问，你在之前的几门数学课上都用过数学模型，尽管当时很可能没有人向你指出这一点。例如，如果你上过线性代数的课程，你会有一些使用$\mathbb{R}^2， \mathbb{R}^3$和$\mathbb{R}^n$作为向量空间例子的经验。在高中几何中，你学过平面是欧几里得几何公理的“模型”。也许你上过抽象代数的课，在那里你看到了几个群的例子:加法下的整数、置换群、可逆的n × n矩阵的群以及矩阵乘法运算都是群的例子——它们是群公理的“模型”。所有这些都是数学模型或结构。不同的结构用于不同的目的。

假设我们考虑一个特定的数学结构，例如$\mathbb{R}^3$，实数的有序三元组的集合。如果我们尝试在$\mathbb{R}^3$中做平面欧几里得几何，我们会惨败，因为(例如)平行假设在这个结构中是假的。另一方面，如果我们想在$\mathbb{R}^3$中做线性代数，一切都很好，因为我们可以把$\mathbb{R}^3$中的点看作向量，让标量为实数。那么实向量空间的公理在$\mathbb{R}^3$中解释时都为真。我们会说$\mathbb{R}^3$是一个向量空间公理的模型，而它不是欧几里得几何公理的模型。

你无疑已经注意到了，我们的讨论引入了两种不同的需要担心的事情。首先是数学模型，你可以把它想象成数学世界或结构。这些例子包括$\mathbb{R}^3$， 17次多项式的集合，$3 \乘以2$矩阵的集合，以及实线。我们也讨论过几何公理和向量空间，这些是不同的东西。让我们讨论一下这些公理。

数学代写|数理逻辑代写Mathematical logic代考|Languages

我们将构建一种非常有限的形式语言，我们构建这种语言的目标是能够对特定类型的数学结构形成特定的陈述。对于我们的工作，有必要能够讨论常数、函数和关系，因此我们需要符号来表示它们。
Chaff:让我再强调一次。现在我们正在讨论我们语言的语法，也就是试卷上的标记。我们稍后才会考虑这些标记的语义或意义，至少不会正式地考虑。但是，如果假装我们对符号的选择和使用方式没有受到预期意义的影响，那就太愚蠢了。如果我们想要讨论左半半半环，我们的形式语言应该包括数学家在这个有利可图和令人兴奋的领域中习惯使用的函数和关系符号，而不是国际象棋、桥牌或右半半半半域中涉及的符号。我们的目标不是让任何人感到困惑，除非是必要的。所以你们现在应该做一个练习，假设我们要讨论的领域是，比如说，自然数理论，那么你们应该猜测一下，应该使用什么合理的语言。参见练习1。
1.2.1 “的定义。一阶语言$\mathcal{L}$是不同符号的无限集合，其中任何一个符号都不能正确地包含在另一个符号中，分为以下类别:

括号:(,)。

连接词:$\vee， $负$。

Quantrfier: \尽管美元。

变量，每个正整数$n对应一个:v_1, v_2， \ldots, v_n， \ldots$这组变量符号将表示为Vars。

等号:=。

常数符号:零个或多个符号的集合。

函数符号:对于每一个正整数$n$，一组零个或多个$n$的函数符号。

关系符号:对于每一个正整数$n$，一组零个或多个$n$的关系符号。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

数学代写|数理逻辑代写Mathematical logic代考|We Specify Ω

Posted on 2023年6月21日2023年6月21日 by statistics-lab

数学代写|数理逻辑代写Mathematical logic代考|We Specify Ω

We come back to Theorem 2. Now it is time to specify the value of the L-cardinal $\Omega$, so far left rather arbitrary by Definition 2 on page 13 .
Definition 7 (in $\mathbf{L}$ ). Recall that $1 \leq M<\omega$ is a number considered in Theorem 2.

We let $\Omega=\omega_{\mathbb{Q}}^{\mathbf{L}}$, and accordingly define $\Omega^{\ominus}=\omega_{\mathrm{M}-1}^{\mathbf{L}}, \Omega^{\oplus}=\omega_{\mathrm{M}+1}^{\mathbf{L}}$,
$$
\mathbb{H}=\left(\mathbf{H} \Omega^{\oplus}\right)^{\mathbf{L}}=\left(\mathbf{H} \omega_{\mathrm{M}+1}^{\mathbf{L}}\right)^{\mathbf{L}}=\left{x \in \mathbf{L}: \operatorname{card}(\mathrm{TC}(x))<\omega_{\mathrm{M}+1}^{\mathbf{L}} \text { in } \mathbf{L}\right}
$$
by Definition 2. Applying Definition 6 with $\Omega=\omega_{\mathbb{M}}^{\mathbf{L}}$, we accordingly fix:

$A \preccurlyeq$-increasing sequence of $\mathbb{R}$-systems $\left{\mathbb{U}{\xi}^{\Omega}\right}{\xi<\Omega^{\oplus}}$ satisfying (i), (ii), (iii), (iv) of Theorem 6 for the chosen $\mathbf{L}$-cardinal $\Omega=\omega_{\mathrm{M}}^{\mathbf{L}}$,

The basic forcing notion $\mathbb{P}^{\Omega}=\mathbf{P}\left[\mathbb{U}^{\Omega}\right]$, and the subforcings $\mathbb{P}\gamma^{\Omega}=\mathbf{P}\left[\mathbb{U}\gamma^{\cap}\right], \gamma<\mathbb{R}^{\oplus}$, and define restrictions $\mathbb{P}^{\Omega}\left|z(z \subseteq \mathcal{I}), \mathbb{P}^{\Omega}\right| \geq n, \mathbb{P}^{\Omega} \mid<n, \mathbb{P}^{\Omega} \uparrow \neq\langle n, i\rangle$ etc. as in Section 3.2.

数学代写|数理逻辑代写Mathematical logic代考|The Model

To prove Theorem 2 we make use of a certain submodel of a $\left(\mathbb{C} \times \mathbb{P}^n\right)$-generic extension of $\mathbf{L}$. First of all, if $g: \omega \rightarrow \mathscr{P}(\omega)$ is any function then we put:
$$
w[g]={\langle k, j\rangle: k<\omega \wedge j \in g(k)}
$$
Now consider a pair $\langle\zeta, G\rangle,\left(\mathbb{C} \times \mathbb{P}^{\Omega}\right)$-generic over $\mathbf{L}$. Thus $\zeta: \omega \stackrel{\text { onto }}{\longrightarrow} \Xi$ is a generic collapse function, while the set $G \subseteq \mathbb{P}^{\Omega}$ is $\mathbb{P}^{\Omega}$-generic over $\mathbf{L}[\zeta]$. The set:
$$
w[\zeta]={\langle k, j\rangle: k<\omega \wedge j \in \zeta(k)} \subseteq \mathcal{I}=\omega \times \omega
$$
obviously belongs to the model $\mathbf{L}[\zeta]=\mathbf{L}[\boldsymbol{w}[\zeta]]$, but not to $\mathbf{L}$. Therefore the restrictions $\mathbb{P}^{\Omega} \mid \boldsymbol{w}[\zeta]$, $G\lceil w[\zeta]$ in the next theorem have to be understood in the sense of Definition 5 on page 15 , ignoring Remark 3 since, definitely $w[\zeta] \notin \mathbf{L}$. Thus $\mathbb{P}^{\Omega} \mid w[\zeta]$ is a forcing notion in $\mathbf{L}[\zeta]$, not in $\mathbf{L}$.
The following theorem describes the structure of such generic models.
Theorem 8. Under the assumptions of Definition 7 , let a pair $\langle\zeta, G\rangle$ be $\left(\mathbb{C} \times \mathbb{P}^{\Omega}\right)$-generic over $\mathbf{L}$. Then:
(i) $G\left\lceil w[\zeta]\right.$ is a set $\left(\mathbb{P}^{\Omega} \mid w[\zeta]\right)$-generic over $\mathbf{L}[\zeta]$,
(ii) $\omega_\gamma^{\mathbf{L}[\zeta, G \mid w[\zeta]]}=\omega_{1+\gamma}^{\mathbf{L}}$ for all ordinals $\gamma \geq 1$, in particular, $\Omega^{\oplus}=\omega_M^{\mathbf{L}[\zeta, G \mid w[\zeta]]}$;
and it is true in the model $\mathbf{L}[\boldsymbol{\zeta}, G\lceil w[\zeta]]$ that
(iii) If $M \geq 2$ then $\Omega=\omega_{M-1}$ and $\Omega^{\oplus}=\Omega^{+}=\omega_M$, whereas if $M=1$ then $\omega<\Omega=\mathbb{R}^{\oplus}=\omega_1$;
(iv) $\mathrm{GCH}$ holds;
(v) Every constructible real belongs to $\mathbf{D}{1 \mathrm{M}}$, (vi) If $1 \leq m<\omega$ and $m \neq \mathbb{M}$ then $\mathbf{D}{1 m} \notin \mathbf{D}{2 m}$, and (vii) every real in $\mathbf{D}{1 \mathrm{M}}$ is constructible.

数理逻辑代写

数学代写|数理逻辑代写Mathematical logic代考|We Specify Ω

我们回到定理2。现在是时候指定l基数$\Omega$的值了，到目前为止，第13页的定义2相当随意。
定义7(见$\mathbf{L}$)。回想一下，$1 \leq M<\omega$是定理2中考虑的一个数字。

我们让$\Omega=\omega_{\mathbb{Q}}^{\mathbf{L}}$，并相应地定义$\Omega^{\ominus}=\omega_{\mathrm{M}-1}^{\mathbf{L}}, \Omega^{\oplus}=\omega_{\mathrm{M}+1}^{\mathbf{L}}$，
$$
\mathbb{H}=\left(\mathbf{H} \Omega^{\oplus}\right)^{\mathbf{L}}=\left(\mathbf{H} \omega_{\mathrm{M}+1}^{\mathbf{L}}\right)^{\mathbf{L}}=\left{x \in \mathbf{L}: \operatorname{card}(\mathrm{TC}(x))<\omega_{\mathrm{M}+1}^{\mathbf{L}} \text { in } \mathbf{L}\right}
$$
2.定义;通过$\Omega=\omega_{\mathbb{M}}^{\mathbf{L}}$应用定义6，我们相应地修复:

$A \preccurlyeq$对于所选的$\mathbf{L}$ -基数$\Omega=\omega_{\mathrm{M}}^{\mathbf{L}}$，满足定理6 (i)， (ii)， (iii)， (iv)的$\mathbb{R}$ -系统$\left{\mathbb{U}{\xi}^{\Omega}\right}{\xi<\Omega^{\oplus}}$的递增序列，

基本强制概念$\mathbb{P}^{\Omega}=\mathbf{P}\left[\mathbb{U}^{\Omega}\right]$和子强制$\mathbb{P}\gamma^{\Omega}=\mathbf{P}\left[\mathbb{U}\gamma^{\cap}\right], \gamma<\mathbb{R}^{\oplus}$，以及3.2节中定义的限制$\mathbb{P}^{\Omega}\left|z(z \subseteq \mathcal{I}), \mathbb{P}^{\Omega}\right| \geq n, \mathbb{P}^{\Omega} \mid<n, \mathbb{P}^{\Omega} \uparrow \neq\langle n, i\rangle$等。

数学代写|数理逻辑代写Mathematical logic代考|The Model

为了证明定理2，我们利用了$\mathbf{L}$的一个$\left(\mathbb{C} \times \mathbb{P}^n\right)$ -泛型扩展的某一子模型。首先，如果$g: \omega \rightarrow \mathscr{P}(\omega)$是任意函数，则输入:
$$
w[g]={\langle k, j\rangle: k<\omega \wedge j \in g(k)}
$$
现在考虑$\langle\zeta, G\rangle,\left(\mathbb{C} \times \mathbb{P}^{\Omega}\right)$ -generic对$\mathbf{L}$。因此$\zeta: \omega \stackrel{\text { onto }}{\longrightarrow} \Xi$是一个泛型折叠函数，而集合$G \subseteq \mathbb{P}^{\Omega}$是$\mathbb{P}^{\Omega}$ -generic over $\mathbf{L}[\zeta]$。套装:
$$
w[\zeta]={\langle k, j\rangle: k<\omega \wedge j \in \zeta(k)} \subseteq \mathcal{I}=\omega \times \omega
$$
显然属于模型$\mathbf{L}[\zeta]=\mathbf{L}[\boldsymbol{w}[\zeta]]$，但不属于$\mathbf{L}$。因此，下一个定理中的限制$\mathbb{P}^{\Omega} \mid \boldsymbol{w}[\zeta]$, $G\lceil w[\zeta]$必须在第15页定义5的意义上理解，忽略注释3，因为，肯定$w[\zeta] \notin \mathbf{L}$。因此$\mathbb{P}^{\Omega} \mid w[\zeta]$在$\mathbf{L}[\zeta]$中是一个强制概念，而在$\mathbf{L}$中不是。
下面的定理描述了这种通用模型的结构。
定理8。在定义7的假设下，设一对$\langle\zeta, G\rangle$为$\left(\mathbb{C} \times \mathbb{P}^{\Omega}\right)$ -泛型/ $\mathbf{L}$。然后:
(i) $G\left\lceil w[\zeta]\right.$是$\mathbf{L}[\zeta]$上的一个集$\left(\mathbb{P}^{\Omega} \mid w[\zeta]\right)$ -泛型;
(ii) $\omega_\gamma^{\mathbf{L}[\zeta, G \mid w[\zeta]]}=\omega_{1+\gamma}^{\mathbf{L}}$对于所有序数$\gamma \geq 1$，特别是$\Omega^{\oplus}=\omega_M^{\mathbf{L}[\zeta, G \mid w[\zeta]]}$;
在$\mathbf{L}[\boldsymbol{\zeta}, G\lceil w[\zeta]]$模型中
(三)如果是$M \geq 2$则$\Omega=\omega_{M-1}$和$\Omega^{\oplus}=\Omega^{+}=\omega_M$，如果是$M=1$则$\omega<\Omega=\mathbb{R}^{\oplus}=\omega_1$;
(iv) $\mathrm{GCH}$持有;
(v)所有可构造实物都属于$\mathbf{D}{1 \mathrm{M}}$， (vi)如果$1 \leq m<\omega$和$m \neq \mathbb{M}$，则$\mathbf{D}{1 m} \notin \mathbf{D}{2 m}$，以及(vii) $\mathbf{D}{1 \mathrm{M}}$中的所有实物都是可构造的。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写