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# 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|MATH160 Trakhtenbrot’s Theorem and the Incompleteness of Second-Order Logic

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## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Trakhtenbrot’s Theorem and the Incompleteness of Second-Order Logic

The object of this section is to prove that the set of valid second-order $S_{\infty}$-sentences is not enumerable, and to briefly discuss the methodological consequences. A useful tool in this context will be Trakhtenbrot’s Theorem, which says that the set of firstorder sentences valid in all finite structures is not enumerable.

5.1 Definition. (a) An $S$-sentence $\varphi$ is said to be fin-satisfiable if there is a finite $S$-structure that satisfies $\varphi$.
(b) An $S$-sentence $\varphi$ is said to be fin-valid if every finite $S$-structure satisfies $\varphi$.
For $S=S_{\infty}$ we set
$\Phi_{\mathrm{fs}}:=\left{\varphi \in L_0^{S_{\infty}} \mid \varphi\right.$ is fin-satisfiable $}$ and $\Phi_{\mathrm{fv}}:=\left{\varphi \in L_0^{S_{\infty}} \mid \varphi\right.$ is fin-valid $}$.
As an example, we note that over a finite domain any injective function is also surjective; therefore the sentence $\varphi:=\forall x \forall y(f x \equiv f y \rightarrow x \equiv y \rightarrow \forall x \exists y x \equiv f y)$ is finvalid; however, $\varphi$ is not valid. The sentence $\neg \varphi$ is satisfiable but not fin-satisfiable.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代考|Lemma. Φfs is R-enumerable

Proof. First we describe a procedure that decides, for every $S_{\infty}$-sentence $\varphi$ and every $n$, whether or not $\varphi$ is satisfiable over a domain with $n+1$ elements. Suppose $\varphi$ and $n$ are given. Since for every structure with $n+1$ elements there is an isomorphic structure with domain ${0, \ldots, n}$, we only need to check (by the Isomorphism Lemma) whether $\varphi$ is satisfiable over ${0, \ldots, n}$. Let $S$ be the (finite!) set of symbols occurring in $\varphi$ and $\mathfrak{A}0, \ldots, \mathfrak{A}_k$ the finitely many $S$-structures with domain ${0, \ldots, n}$ (cf. Exercise III.1.5). We can describe the $\mathfrak{A}_i$ explicitly by means of finite tables for the relations, functions, and constants. The sentence $\varphi$ is satisfiable over ${0, \ldots, n}$ if and only if $\mathfrak{A}_i \models \varphi$ for some $i \leq k$. Thus we only need to test whether $\mathfrak{A}_i \models \varphi$ for $i=0, \ldots, k$. These tests can be reduced to questions that can be answered from the respective tables as follows: If $\varphi=\neg \psi$, then the problem “’ $\mathfrak{A}_i \models \varphi$ ?” can be reduced to the question of whether $\mathfrak{A}_i \models \psi$. If $\varphi=(\psi \vee \chi)$, then similarly the problem can be reduced to the questions of whether $\mathfrak{A}_i \models \psi$ and whether $\mathfrak{A}_i \models \chi$. If $\varphi=\exists v_0 \psi$, we reduce to the questions ” ‘ $\mathfrak{A}_i=\psi[0]$ ?”,,$\ldots$, ” $\mathfrak{A}_i \models \psi[n]$ ?”. Continuing in this way we eventually arrive at questions of the form ” $\mathfrak{A}_i \models \psi\left[n_0, \ldots, n{m-1}\right]$ ?” for atomic formulas $\psi\left(v_0, \ldots, v_{m-1}\right)$ and $n_0, \ldots, n_{m-1} \leq n$. Clearly these can be answered effectively by inspecting the tables for $\mathfrak{A}_i$.

Now $\Phi_{\mathrm{fs}}$ can be enumerated as follows: For $m=0,1,2, \ldots$ generate the (finitely many) words over $\mathbb{A}0$ that are $S{\infty}$-sentences of length $\leq m$, and use the procedure just described to decide, for $n=0, \ldots, m$, whether they are satisfiable over a domain with $n+1$ elements. List the sentences where this is the case.

## 数学代写|数理逻辑入门代写Introduction To Mathematical logic代 考|Trakhtenbrot’s Theorem and the Incompleteness of SecondOrder Logic

5.1 定义。 $(\mathrm{a})$ 一个 $S$-句子 $\varphi$ 如果存在有限的，则称为 fin-satisfiable $S$-满足的结构 $\varphi$.
(b) 一个 $S$-句子 $\varphi$ 被称为 fin-valid 如果每个有限 $S$-结构满足 $\varphi$.

\left 缺少或无法识别的分隔符 $\quad$ 和 $\backslash$ left 缺少或无法识别的分隔符

$\varphi:=\forall x \forall y(f x \equiv f y \rightarrow x \equiv y \rightarrow \forall x \exists y x \equiv f y)$ 最終无效；然而， $\varphi$ 无效。这句话 $\varphi$ 是可满足的但不是 fin 可满足的。

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