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# 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|MATH230

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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:

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

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

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

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