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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

$$X(Y \& Z) \sim X Y \& X Z$$

$$X \& Y Z \sim(X \& Y)(X \& Z),$$

$$(X \& \bar{X}) Y \sim Y$$

$$X \bar{X} \& Y \sim Y,$$

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

$\bar{X} Y \& \bar{Y} Z \& X \& \bar{Z}$。

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

Gödel用单音节词解释的第二不完备定理${}^1$

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。

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

$T h m_N(f)$是:
$(\exists c)(\operatorname{Deduction}(c, f))$，

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

\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}

$$\forall y \forall z\left[\left[N u m\left(v_1, y\right) \wedge S u b\left(v_1, \overline{4}, y, z\right)\right] \rightarrow \psi(z)\right]$$

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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|Coding Is Recursive

$$\begin{gathered} \text { Codenumber }(c) \text { is: } \ \text { Divides }(S S 0, c) \wedge(\forall z<c)(\forall y<z) \ {[(\text { Prime }(z) \wedge \text { Dvides }(z, c) \wedge \text { Prmepair }(y, z)) \rightarrow \text { Divides }(y, c)] .} \end{gathered}$$

$$N \vdash \text { Codenumber }(\overline{18})$$

$$N \vdash \neg \text { Codenumber(45). }$$以上翻译结果来自有道神经网络翻译（YNMT）· 通用场景

## 有限元方法代写

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。

## 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。

## 数学代写|数理逻辑代写Mathematical logic代考|M-781

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

$$\sigma_1, \sigma_2, \ldots, \sigma_n, \ldots$$

$$\bar{v}(\phi)= \begin{cases}v(\phi) & \text { if } \phi \text { is a propositional variable } \ F & \text { if } \phi \text { is }(\neg \alpha) \text { and } \bar{v}(\alpha)=T \ F & \text { if } \phi \text { is }(\alpha \vee \beta) \text { and } \bar{v}(\alpha)=\bar{v}(\beta)=F \ T & \text { otherwise. }\end{cases}$$

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

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:

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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

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$\mathfrak{A} \models \Gamma$。

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。

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

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$\mathfrak{A} \models \Gamma$。

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。

## 有限元方法代写

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。

## 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。

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

statistics-lab™ 为您的留学生涯保驾护航 在代写数理逻辑Mathematical logic方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数理逻辑Mathematical logic代写方面经验极为丰富，各种代写数理逻辑Mathematical logic相关的作业也就用不着说。

## 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:

1. Parentheses: (,).
2. Connectives: $\vee, \neg$.
3. Quantrfier: $\forall$.
4. Variables, one for each positive integer $n: v_1, v_2, \ldots, v_n, \ldots$ The set of variable symbols will be denoted Vars.
5. Equality symbol: =.
6. Constant symbols: Some set of zero or more symbols.
7. Function symbols: For each positive integer $n$, some set of zero or more $n$-ary function symbols.
8. 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|Languages

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:

## 有限元方法代写

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。

## 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。