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

#### Doug I. Jones

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## 数学代写|数理逻辑代写Mathematical logic代考|The First Incompleteness Theorem

We are ready to state and prove the First Incompleteness Theorem, which tells us that if we are given any reasonable axiom system $A$, there is a sentence that is true in $\mathfrak{N}$ but not provable from $A$.

You may have been complaining all along about my choice for an axiom system. Perhaps you have been convinced from the beginning that $N$ is clearly too weak to prove every truth about the natural numbers. You are right. We know, for example, that $N$ does not prove the commutative law of addition. Since you are a diligent person, I imagine that you have come up with an axiom system of your own, let’s call it $A$. You might be convinced that $A$ is the “right” choice of axioms, a set of axioms that is strong enough to prove every truth about $\mathfrak{N}$. Although this shows admirable independence on your part, we will, unfortunately, be able to prove that $A$ is no better than $N$, as long as $A$ satisfies certain reasonable conditions.

Definition 5.3.1. A theory is a collection of sentences $T$ that is closed under deduction: For every sentence $\sigma$, if $T \vdash \sigma$, then $\sigma \in T$. If $A$ is a set of sentences, then the theory of $A$, written $\operatorname{Th}(A)$, is the smallest theory that includes $A: T h(A)={\sigma \mid A \vdash \sigma}$.

Definition 5.3.2. A theory $T$ in the language $\mathcal{L}_{N T}$ is said to be recursively axiomatized if there is a set of axioms $A$ such that

1. $T={\sigma \mid \sigma$ is a sentence and $A \vdash \sigma}$.
2. AxiomOfA $={\ulcorner\alpha\urcorner \mid \alpha \in A}$ is a recursive set.

## 数学代写|数理逻辑代写Mathematical logic代考|Extensions and Refinements of Incompleteness

If you look carefully at the First Incompleteness Theorem, it does not quite say that the collection of axioms $A$ is incomplete. All that is claimed is that there is a sentence $\theta$ such that $\theta$ is true-in- $\mathfrak{N}$ and $\theta$ is not provable from $A$. But, perhaps, $\neg \theta$ is provable from $A$. Our first result in this section brings the focus onto incompleteness.
Proposition 5.4.1. Suppose that $A$ is a consistent, recursive set of axioms that proves all of the axioms of $N$. If all of the axioms of $A$ are true in $\mathfrak{N}$, then there is a sentence $\theta$ such that $A \not \forall$ and $A \nvdash \neg \theta$.

Proof. As in the proof of the First Incompleteness Theorem, let $\theta$ be such that
$$N \vdash\left[\theta \leftrightarrow \neg T h m_A\left(\overline{\left.r^{\prime}\right\urcorner}\right)\right] .$$
We know that $\mathfrak{N} \vDash \theta$ and $A \forall \theta$. Suppose that $A$ proves $\neg \theta$. Then, as all of the axioms of $A$ are true in the structure $\mathfrak{N}$, we know that $\mathfrak{N} \vDash \neg \theta$, which contradicts the fact that $\mathfrak{N} \vDash \theta$. Thus $A \forall \neg \theta$, and $A$ is incomplete.

# 数理逻辑代写

## 数学代写|数理逻辑代写Mathematical logic代考|The First Incompleteness Theorem

5.3.1.定义理论是一组在演绎下封闭的句子$T$:对于每一个句子$\sigma$，如果$T \vdash \sigma$，那么$\sigma \in T$。如果$A$是一组句子，那么$A$的理论，写的$\operatorname{Th}(A)$，是包含$A: T h(A)={\sigma \mid A \vdash \sigma}$的最小的理论。

5.3.2.定义语言$\mathcal{L}_{N T}$中的一个理论$T$被称为递归公理化的，如果有一组公理$A$使得

$T={\sigma \mid \sigma$ 是一个句子和$A \vdash \sigma}$。

AxiomOfA $={\ulcorner\alpha\urcorner \mid \alpha \in A}$是一个递归集合。

## 数学代写|数理逻辑代写Mathematical logic代考|Extensions and Refinements of Incompleteness

$$N \vdash\left[\theta \leftrightarrow \neg T h m_A\left(\overline{\left.r^{\prime}\right\urcorner}\right)\right] .$$

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