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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|The Collection of Axioms Is Recursive

In this section we will exhibit two $\Delta$-formulas that are designed to pick out the axioms of our deductive system.

Proposition 4.11.1. The collection of Gödel numbers of the axioms of $N$ is recursive.

Proof. The formula AxiomOfN is easy to describe. As there are only a finite number of $\mathrm{N}$-axioms, a natural number $a$ is in the set AxiomOFN if and only if it is one of a finite number of Gödel numbers. Thus
$\operatorname{AxiomOfN}(a)$ is:
$$\begin{gathered} a=\overline{\Gamma(\forall x) \neg S x=0\urcorner} \vee \ a=\overline{\Gamma(\forall x)(\forall y)[S x=S y \rightarrow x=y]\urcorner} \vee \ \vdots \ \vee a=\overline{\Gamma(\forall x)(\forall y)[(x<y) \vee(x=y) \vee(y<x)]\urcorner} . \end{gathered}$$
(To be more-than-usually picky, we need to change the $x$ ‘s and $y$ ‘s to $v_1$ ‘s and $v_2$ ‘s, but you can do that.)

Proposition 4.11.2. The collection of Gödel numbers of the logical axioms is recursive.

Proof. The formula that recognizes the logical axioms is more complicated than the formula AxiomOfN for two reasons. The first is that there are infinitely many logical axioms, so we cannot just list them all. The second reason that this group of axioms is more complicated is that the quantifier axioms depend on the notion of substitutability, so we will have to use our results from Section 4.10.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Coding Deductions

It is probably difficult to remember at this point of our journey, but our goal is to prove the Incompleteness Theorem, and to do that we need to write down an $\mathcal{L}{N T}$-sentence that is true in $\mathfrak{N}$, the standard structure, but not provable from the axioms of $N$. Our sentence, $\theta$, will “say” that $\theta$ is not provable from $N$, and in order to “say” that, we will need a formula that will identify the (Gödel numbers of the) formulas that are provable from $N$. To do that we will need to be able to code up deductions from $N$, which makes it necessary to code up sequences of formulas. Thus, our next goal will be to settle on a coding scheme for sequences of $\mathcal{L}{N T}$-formulas.

We have been pretty careful with our coding up to this point. If you check, every Gödel number that we have used has been even, with the exception of 3 , which is the garbage case in Definition 4.7.1. We will now use numbers with smallest prime factor 5 to code sequences of formulas.

Suppose that we have the sequence of formulas
$$D=\left\langle\phi_1, \phi_2, \ldots, \phi_k\right\rangle .$$
We will define the sequence code of $D$ to be the number
$$\left.r D\urcorner=5^{\left.r \phi_1\right\urcorner} 7^{\left.r \phi_2\right\urcorner} \cdots p_{k+2} \phi_k\right\urcorner .$$
So the exponent on the $(i+2)$ nd prime is the Gödel number of the $i$ th element of the sequence. You are asked in the Exercises to produce several useful $\mathcal{L}_{N T}$-formulas relating to sequence codes.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|The Collection of Axioms Is Recursive

$\operatorname{AxiomOfN}(a)$是:
$$\begin{gathered} a=\overline{\Gamma(\forall x) \neg S x=0\urcorner} \vee \ a=\overline{\Gamma(\forall x)(\forall y)[S x=S y \rightarrow x=y]\urcorner} \vee \ \vdots \ \vee a=\overline{\Gamma(\forall x)(\forall y)[(x<y) \vee(x=y) \vee(y<x)]\urcorner} . \end{gathered}$$
(为了比通常更挑剔，我们需要将$x$和$y$更改为$v_1$和$v_2$，但您可以这样做。)

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Coding Deductions

$$D=\left\langle\phi_1, \phi_2, \ldots, \phi_k\right\rangle .$$

$$\left.r D\urcorner=5^{\left.r \phi_1\right\urcorner} 7^{\left.r \phi_2\right\urcorner} \cdots p_{k+2} \phi_k\right\urcorner .$$

## 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。

Posted on Categories:Mathematical logic, 数学代写, 数理逻辑

## avatest™帮您通过考试

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|The Language, the Structure, and the Axioms of $N$

We work in the language of number theory
$$\mathcal{L}_{N T}={0, S,+, \cdot, E,<},$$
and we will continue to work in this language for the next two chapters. $\mathfrak{N}$ is the standard model of the natural numbers,
$$\mathfrak{N}=\langle\mathbb{N}, 0, S,+, \cdot, E,<\rangle,$$
where the functions and relations are the usual functions and relations that you have known since you were knee high to a grasshopper. $E$ is exponentiation, which will usually be written $x^y$ rather than $E x y$ or $x E y$.

We will now establish a set of nonlogical axioms, $N$. You will notice that the axioms are clearly sentences that are true in the standard structure, and thus if $T$ is any set of axioms such that $T \vdash \sigma$ for all $\sigma$ such that $\mathfrak{N} \vDash \sigma$, then $T \vdash N$. So, as we prove that several sorts of formulas are derivable from $N$, remember that those same formulas are also derivable from any set of axioms that has any hope of providing an axiomatization of the natural numbers.

The axiom system $N$ was introduced in Example 2.8.3 and is reproduced on the next page. These eleven axioms establish some of the basic facts about the successor function, addition, multiplication, exponentiation, and the $<$ ordering on the natural numbers.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Recursive Sets and Recursive Functions

For the sake of discussion, suppose that we let $f(x)=x^2$. It will not surprise you to find out that it is the case that $f(4)=16$, so I would like to write $n \models f(4)=16$. Unfortunately, we are not allowed to do this, since the symbol $f$, not to mention 4 and 16, are not part of the language.

What we can do, bowever, is to represent the function $f$ by a formula in $\mathcal{L}{N T}$. To be specific, suppose that $\phi(x, y)$ is $$y=E x S S O$$ Then, if we allow ourselves once again to use the abbreviation $\bar{a}$ for the $\mathcal{C}{N T \text {-term }}^{S S S \cdots S} 0$, we can assert that
$$\boldsymbol{n}=\phi(\overline{4}, \overline{16})$$
which is the same thing as
ๆю= SSSSSSSSSSSSSSSSOESSSSOSSO.
(Boy, aren’t you glad we don’t use the official language very often?) Anyway, the situation is even better than this, for $\phi(4, \overline{16})$ is derivable from $N$ rather than just true in $\mathfrak{n}$. In fact, if you look back at Lemma 2.8.4, you probably won’t have any trouble believing the following statements:

• $N \vdash \phi(\overline{4}, \overline{16})$
• $N \vdash \neg \phi(4,17)$
• $N \vdash \neg \phi(\overline{1}, \overline{714})$

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|The Language, the Structure, and the Axioms of $N$

$$\mathcal{L}_{N T}={0, S,+, \cdot, E,<},$$

$$\mathfrak{N}=\langle\mathbb{N}, 0, S,+, \cdot, E,<\rangle,$$

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Recursive Sets and Recursive Functions

$$\boldsymbol{n}=\phi(\overline{4}, \overline{16})$$

(天哪，你不高兴我们不经常使用官方语言吗?)无论如何，情况甚至比这更好，因为$\phi(4, \overline{16})$可以从$N$推导出来，而不仅仅是在$\mathfrak{n}$中成立。事实上，如果你回顾引理2.8.4，你可能会毫不费力地相信以下陈述:

$N \vdash \phi(\overline{4}, \overline{16})$

$N \vdash \neg \phi(4,17)$

$N \vdash \neg \phi(\overline{1}, \overline{714})$

## 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。

Posted on Categories:Mathematical logic, 数学代写, 数理逻辑

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Outline of the Proof

There will be a preliminary argument that will show that it is sufficient to prove that if $\Sigma$ is a consistent set of sentences, then $\Sigma$ has a model. Then we will proceed to assume that we are given such a set of sentences, and we will construct a model for $\boldsymbol{\Sigma}$.

The construction of the model will proceed in several steps, but the central idea was introduced in Example 1.6.4. The elements of the model will be variable-free terms of a language. We will construct this model so that the formulas that will be true in the model are precisely the formulas that are in a certain set of formulas, which we will call $\Sigma^{\prime}$. We will make sure that $\Sigma \subseteq \Sigma^{\prime}$, so all of the formulas of $\Sigma$ will be true in this constructed model. In other words, we will have constructed a model of $\Sigma$.

To make the construction work we will take our given set of $\mathcal{L}$ sentences $\Sigma$ and extend it to a bigger set of sentences $\Sigma^{\prime}$ in a bigger language $\mathcal{L}^{\prime}$. We do this extension in two steps. First, we will add in some new axioms, called Henkin Axioms, to get a collection $\hat{\Sigma}$. Then we will extend $\hat{\Sigma}$ to $\Sigma^{\prime}$ in such a way that:

1. $\Sigma^{\prime}$ is consistent.
2. For every $\mathcal{L}^{\prime}$-sentence $\theta$, either $\theta \in \Sigma^{\prime}$ or $(\neg \phi) \in \Sigma^{\prime}$.
Thus we will say that $\Sigma^{\prime}$ is a maximal consistent extension of $\Sigma$, where maximal means that it is impossible to add any sentences to $\Sigma^{\prime}$ without making $\Sigma^{\prime}$ inconsistent.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Preliminary Argument

So let us fix our setting for the rest of this proof. We are working in a language $\mathcal{L}$. For the purposes of this proof, we assume that the language is countable, which means that the formulas of $\mathcal{L}$ can be written in an infinite list $\alpha_1, \alpha_2, \ldots, \alpha_n, \ldots$. (An outline of the changes in the proof necessary for the case when $\mathcal{L}$ is not countable can be found in Exercise 6.)

We are given a set of formulas $\Sigma$, and we are assuming that $\Sigma \vDash \phi$. We have to prove that $\Sigma \vdash \phi$.

Note that we can assume that $\phi$ is a sentence: By Lemma 2.7.2, $\Sigma \vdash \phi$ if and only if there is a deduction from $\Sigma$ of the universal closure of $\phi$. Also, by the comments following Lemma 2.7.3, we can also assume that every element of $\Sigma$ is a sentence. So, now all(!) we have to do is prove that if $\Sigma$ is a set of sentences and $\phi$ is a sentence and if $\Sigma \vDash \phi$, then $\Sigma \vdash \phi$.

Now we claim that it suffices to prove the case where $\phi$ is the sentence $\perp$. For suppose we know that if $\Sigma \models \perp$, then $\Sigma \vdash \perp$, and suppose we are given a sentence $\phi$ such that $\Sigma \vDash \phi$. Then $\Sigma U$ $(\neg \phi) \vDash \perp$, as there are no models of $\Sigma \cup(\neg \phi)$, so $\Sigma \cup(\neg \phi) \vdash \perp$. This tells us, by Exercise 4 in Section 2.7.1, that $\Sigma \vdash \phi$, as needed.

So we have reduced what we need to do to proving that if $\Sigma \models \perp$, then $\Sigma \vdash \perp$, for $\Sigma$ a set of $\mathcal{L}$-sentences. This is equivalent to saying that if there is no model of $\Sigma$, then $\Sigma \vdash \perp$. We will work with the contrapositive: If $\Sigma \nvdash \downarrow$, then there is a model of $\Sigma$. In other words, we will prove:

If $\Sigma$ is a consistent set of sentences, then there is a model of $\Sigma$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Outline of the Proof

$\Sigma^{\prime}$ 是一致的。

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Preliminary Argument

$$L(\gamma)=\int_0^\theta\left|\gamma^{\prime}(t)\right| \mathrm{d} t=\int_0^\theta\left|\mathrm{i} r \mathrm{e}^{\mathrm{i} t}\right| \mathrm{d} t=\int_0^\theta r \mathrm{~d} t=r \theta$$

## 数学代写|复分析代写Complex analysis代考|The Argument of a Complex Number

$$\mathbb{C}=D_1 \cup D_2 \cup D_3$$
(画个图!)在每个域中，我们都有一种简单的方法来找到$\arg z$的值。

## 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。

Posted on Categories:Mathematical logic, 数学代写, 数理逻辑

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Equality Axioms

We have taken the route of assuming that the equality symbol, $=$, is a part of the language $\mathcal{L}$. There are three groups of axioms that are designed for this symbol. The first just says that any object is equal to itself:
$$x=x \text { for each variable } x .$$
For the second group of axioms, assume that $x_1, x_2, \ldots, x_n$ are variables, $y_1, y_2, \ldots, y_n$ are variables, and $f$ is an $n$-ary function symbol.
\begin{aligned} & {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\left(x_n=y_n\right)\right] \rightarrow } \ &\left(f\left(x_1, x_2, \ldots, x_n\right)=f\left(y_1, y_2, \ldots, y_n\right)\right) \end{aligned}
The assumptions for the third group of axioms is the same as for the second group, except that $R$ is assumed to be an $n$-ary relation symbol ( $R$ might be the equality symbol, which is seen as a binary relation symbol).
\begin{aligned} {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(R\left(x_1, x_2, \ldots, x_n\right) \rightarrow R\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}
Axioms (E2) and (E3) are axioms that are designed to allow substitution of equals for equals. Nothing fancier than that.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Quantifier Axioms

The quantifier axioms are designed to allow a very reasonable sort of entry in a deduction. Suppose that we know $\forall x P(x)$. Then, if $t$ is any term of the language, we should be able to state $P(t)$. To avoid problems of the sort outlined at the beginning of Section 1.8, we will demand that the term $t$ be substitutable for the variable $x$.

$(\forall x \phi) \rightarrow \phi_t^x$, if $t$ is substitutable for $x$ in $\phi$.
(Q1)
$\phi_t^x \rightarrow(\exists x \phi)$, if $t$ is substitutable for $x$ in $\phi$.
(Q2)
In many logic texts, axiom (Q1) would be called universal instantiation, while (Q2) would be known as existential generalization. We will avoid this impressive language and stick with the more mundane (Q1) and (Q2).
2.3.3 Recap
Just to gather all of the logical axioms together in one place, let me state them once again. The set $\Lambda$ of logical axioms is the collection of all formulas that fall into one of the following categories:
$x=x$ for each variable $x$.
(E1)
\begin{aligned} {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(f\left(x_1, x_2, \ldots, x_n\right)=f\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}
\begin{aligned} {\left[\left(x_1 \doteq y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(R\left(x_1, x_2, \ldots, x_n\right) \rightarrow R\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}
$(\forall x \phi) \rightarrow \phi_t^x$, if $t$ is substitutable for $x$ in $\phi$.
$\phi_t^x \rightarrow(\exists x \phi)$, if $t$ is substitutable for $x$ in $\phi$.
Notice that $\Lambda$ is decidable: We could write a computer program which, given a formula $\phi$, can decide in a finite amount of time whether or not $\phi$ is an element of $\Lambda$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Equality Axioms

$$x=x \text { for each variable } x .$$

\begin{aligned} & {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\left(x_n=y_n\right)\right] \rightarrow } \ &\left(f\left(x_1, x_2, \ldots, x_n\right)=f\left(y_1, y_2, \ldots, y_n\right)\right) \end{aligned}

\begin{aligned} {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(R\left(x_1, x_2, \ldots, x_n\right) \rightarrow R\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Quantifier Axioms

$(\forall x \phi) \rightarrow \phi_t^x$，如果 $t$ 可以代替 $x$ 在 $\phi$．
(q1)
$\phi_t^x \rightarrow(\exists x \phi)$，如果 $t$ 可以代替 $x$ 在 $\phi$．
(q2)

2.3.3概述

$x=x$ 对于每个变量 $x$．
(1)
\begin{aligned} {\left[\left(x_1=y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(f\left(x_1, x_2, \ldots, x_n\right)=f\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}

\begin{aligned} {\left[\left(x_1 \doteq y_1\right) \wedge\left(x_2=y_2\right) \wedge \cdots \wedge\right.} & \left.\left(x_n=y_n\right)\right] \rightarrow \ & \left(R\left(x_1, x_2, \ldots, x_n\right) \rightarrow R\left(y_1, y_2, \ldots, y_n\right)\right) . \end{aligned}

$(\forall x \phi) \rightarrow \phi_t^x$，如果 $t$ 可以代替 $x$ 在 $\phi$．
$\phi_t^x \rightarrow(\exists x \phi)$，如果 $t$ 可以代替 $x$ 在 $\phi$．

## 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。

Posted on Categories:Mathematical logic, 数学代写, 数理逻辑

## avatest™帮您通过考试

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Truth in a Structure

It is at last time to tie together the syntax and the semantics. We have some formal rules about what constitutes a language, and we can identify the terms, formulas, and sentences of a language. We can also identify $\mathcal{L}$-structures for a given language $\mathcal{L}$. In this section we will decide what it means to say that an $\mathcal{L}$-formula $\phi$ is true in an $\mathcal{L}$-structure $\mathfrak{A}$.

To begin the process of tying together the symbols with the structures, we will introduce assignment functions. These assignment functions will formalize what it means to interpret a term or a formula in a structure.

Definition 1.7.1. If $\mathfrak{A}$ is an $\mathcal{L}$-structure, a variable assignment function into $\mathfrak{A}$ is a function $\boldsymbol{s}$ that assigns to each variable an element of the universe $A$. So a variable assignment function into $\mathfrak{A}$ is any function with domain Vars and codomain $A$.

Variable assignment functions need not be injective or bijective. For example, if we work with $\mathcal{L}{N T}$ and the standard structure $\mathfrak{N}$, then the function $s$ defined by $s\left(v{\imath}\right)=i$ is a variable assignment function, as is the function $s^{\prime}$ defined by
$$s^{\prime}\left(v_{\imath}\right)=\text { the smallest prime number that does not divide } i .$$
We will have occasion to want to fix the value of the assignment function $s$ for certain variables.

Definition 1.7.2. If $s$ is a variable assignment function into $\mathfrak{A}$ and $x$ is a variable and $a \in A$, then $s[x \mid a]$ is the variable assignment function into $\boldsymbol{A}$ defined as follows:
$$sx \mid a= \begin{cases}s(v) & \text { if } v \neq x \ a & \text { if } v=x\end{cases}$$
We call the function $s[x \mid a]$ an $x$-modification of the assignment function $s$.

So an $x$-modification of $s$ is just like $s$, except that the variable $x$ is assigned to a particular element of the universe.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Substitutions and Substitutability

Suppose you knew that the sentence $\forall x \phi(x)$ was true in a particular structure $\mathfrak{A}$. Then, if $\boldsymbol{c}$ is a constant symbol in the language, you would certainly expect $\phi(c)$ to be true in $\boldsymbol{A}$ as well. What we have done is substitute the constant symbol $c$ for the variable $x$. This seems perfectly reasonable, although there are times when you do have to be careful.

Suppose that $\mathfrak{A}=\forall x \exists y \neg(x=y)$. This sentence is, in fact, true in any structure $\boldsymbol{A}$ such that $\boldsymbol{A}$ has at least two elements. If we then proceed to replace the variable $x$ by the variable $u$, we get the statement $\exists y \neg(u=y)$, which will still be true in $\boldsymbol{A}$, no matter what value we give to the variable $u$. If, however, we take our original formula and replace $x$ by $y$, then we find ourselves looking at $\exists y \sim(y=y)$,

which will be false in any structure. So by a poor choice of substituting variable, we have changed the truth value of our formula. The rules of substitutability that we will discuss in this section are designed to help us avoid this problem, the problem of attempting to substitute a term inside a quantifier that binds a variable involved in the term.

We begin by defining exactly what we mean when we substitute a term $\boldsymbol{t}$ for a variable $x$ in either a term $u$ or a formula $\phi$.

Definition 1.8.1. Suppose that $u$ is a term, $x$ is a variable, and $t$ is a term. We define the term $u_t^x$ (read ” $u$ with $x$ replaced by $t^{\prime \prime}$ ) as follows:

1. If $u$ is a variable not equal to $x$, then $u_t^x$ is $u$.
2. If $u$ is $x$, then $u_t^x$ is $t$.
3. If $u$ is a constant symbol, then $u_t^x$ is $u$.
4. If $u$ is $f u_1 u_2 \ldots u_n$, where $f$ is an $n$-ary function symbol and the $u_i$ are terms, then
$u_t^x$ is $f\left(u_1\right)_t^x\left(u_2\right)_t^x \ldots\left(u_n\right)_t^x$.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Truth in a Structure

1.7.1.定义如果$\mathfrak{A}$是一个$\mathcal{L}$结构，那么$\mathfrak{A}$中的变量赋值函数就是一个函数$\boldsymbol{s}$，该函数为每个变量赋值一个元素$A$。一个到$\mathfrak{A}$的变量赋值函数是任何有域Vars和上域$A$的函数。

$$s^{\prime}\left(v_{\imath}\right)=\text { the smallest prime number that does not divide } i .$$

$$sx \mid a= \begin{cases}s(v) & \text { if } v \neq x \ a & \text { if } v=x\end{cases}$$

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Substitutions and Substitutability

1.8.1.定义假设$u$是一个项，$x$是一个变量，$t$是一个项。我们将术语$u_t^x$(读作“$u$, $x$被$t^{\prime \prime}$取代”)定义如下:

$u_t^x$是$f\left(u_1\right)_t^x\left(u_2\right)_t^x \ldots\left(u_n\right)_t^x$。

## 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。

Posted on Categories:Mathematical logic, 数学代写, 数理逻辑

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Proof of Theorem 1 (Case II)

First, we make an obvious remark concerning the linear runtime of the Algorithm. For Stage 0, this follows from the fact that $a+b$ is constructed by one-time examination of all components in $a$ and $b$. Stage 1 requires one-time examination of all components in $a+b$. The number of interactions executed at Stage 2 is linear, since each of them reduces the number of chains in the corresponding graph, and each interaction is executed in constant time. Similarly, the number of operations executed at Stage 3 is linear, since each of them reduces the number of singular vertices or the number of chains. Linear runtime of solving the ILP problem at Stage 2 follows from [10], where it was shown that the time required for solving an ILP problem with a fixed number of variables and constraints is polynomial in the logarithm of the maximum absolute value of a coefficient of the problem. In our problem, this coefficient is not greater than the problem size.

The rest of this section is devoted to the proof of additive exactness of the Algorithm. If, when executing a standard operation, singular vertices are joined, the operation is said to be special; otherwise, it is nonspecial; a removal operation is special by definition.

Clearly, we may assume that $w=1$. For a graph $G$, we use the following notation: $d$ is the total size of all components in it (we call it the size of $G$ ), $f$ is the number of odd chains, $c$ is the number of cycles (excluding loops), $B$ is the number of singular vertices, $S$ is the sum of integral parts of halved segment lengths plus the number of extremal (on a chain) odd segments minus the number of cyclic segments, $D$ is the number of chains of types $1 a, 1 b, 3 a, 3 b$, and 3 , and $K_b$ is the number of components containing a $b$-singular vertex.

Lemma 1. Let $w_a$ and $w_b$ be the removal costs for singular $a$ – and $b$-vertices, $w_a \leq w_b$, and let all other operations have cost 1 . Then, the autonomous cost $A(G)$ of a graph $G$ is
$$A(G)=\left(1-w_a\right) \cdot(0.5 d+0.5 f-c)+w_a \cdot(B+S+D)+\left(w_b-w_a\right) \cdot K_b$$
Proof. Denote the right-hand side of this equality by $A^{\prime}(G)$. Let us check the equality for each component of the graph separately and then sum up the obtained equalities. Denote by $I_b$ the indicator function which is 1 if $G$ contains a $b$-singular vertex and 0 otherwise.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|o is the Rem operation

$o$ is the Rem operation. When passing from $G$ to $o(G), B$ reduces by 1 . Consider several cases.
1.1. An isolated singular vertex is removed (chain of type $2 a^{\prime}$ or $2 b^{\prime}$ ). Then $S, D$, and $c$ do not change; $d$ increases by $1 ; f$ reduces by 1 . $P$ does not increase when removing any chain. Therefore, $T^{\prime}+$ $T^{\prime \prime}-P$ reduces by at most $w_a$. If a $b$-vertex is removed, then $K_b$ reduces by 1 and $T$ reduces by at most $w_b$.
1.2. A singular vertex is removed from a cycle, or a loop is removed. Then, $S$ does not reduce.
1.3. An interior singular vertex (i.e., on both sides of it there are other singular vertices) is removed from a chain. The type of the chain does not change, changes from $3 b^$ to $3 b^{\prime}$, or changes from $2 a^$ to $2 a^{\prime}$. Then, $P$ does not increase, since no element increases its quality when making this change.
1.4. A hanging vertex which is the only singular vertex in a chain is removed from the chain. Then, $S$ and $D$ do not reduce, and $K_b$ reduces by 1 if a $b$-vertex is removed.
1.5. A hanging vertex that is not unique in a chain is removed from the chain. If, when passing from $G$ to $o(G), S$ does not change (the segment adjacent to the hanging edge is even), then the hanging extremity becomes nonhanging, and the following changes in the type of the chain are possible: $1 a$ changes into $3 a, 1 b$ changes into $3 b, 2$ changes into $1,2 a$ changes into $1 a, 2 b$ changes into $1 b$, or 1 changes into 3 . In the first three cases, $D$ does not change and $P$ does not increase. Indeed, all elements containing type $1 a$ do not increase their quality when making the change from $1 a$ to $3 a$; the same applies to the two other changes. In the last three cases, $D$ increases by $1, P$ either does not change or increases by at most $w_a$ (consider the inverse change), and all other quantities do not change. Then, $T^{\prime}+T^{\prime \prime}-P$ reduces by at most $w_a$. If $S$ increases by one (the segment adjacent to the hanging edge is odd), then the hanging extremity remains to be hanging, and the type of the chain does not change. Therefore, $D$ and $P$ do not change.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Proof of Theorem 1 (Case II)

$$A(G)=\left(1-w_a\right) \cdot(0.5 d+0.5 f-c)+w_a \cdot(B+S+D)+\left(w_b-w_a\right) \cdot K_b$$

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|o is the Rem operation

$o$ 是Rem操作。当从$G$传递到$o(G), B$时减少1。考虑几个例子。

## 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。