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# 数学代写|数理逻辑代考Mathematical logic代写|MATH4810 Semantics of First-Order Languages

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## 数学代写|数理逻辑代考Mathematical logic代写|Semantics of First-Order Languages

Let $R$ be a binary relation symbol. The ${R}$-formula
$$\forall v_{0} R v_{0} v_{0}$$
is, at present, merely a string of symbols to which no meaning is attached. The situation changes if we specify a domain for the variable $v_{0}$ and if we interpret the binary relation symbol $R$ as a binary relation over this domain. There are, of course, many possible choices for such a domain and relation.

For example, suppose we choose $\mathbb{N}$ for the domain, take ” $\forall v_{0}$ ” to mean “for all $n \in \mathbb{N}$ ” and interpret $R$ as the divisibility relation $R^{\mathbb{N}}$ on $\mathbb{N}$. Then clearly (1) becomes the (true) statement
$$\text { for all } n \in \mathbb{N}, R^{\mathbb{N}} n n \text {, }$$
i.e., the statement
every natural number is divisible by itself.
We say that the formula $\forall v_{0} R v_{0} v_{0}$ holds in $\left(\mathbb{N}, R^{\mathbb{N}}\right)$.
But if we choose the set $\mathbb{Z}$ of integers as the domain and interpret $R$ as the “smallerthan” relation $R^{\mathbb{Z}}$ on $\mathbb{Z}$, then (1) becomes the (false) statement
$$\text { for all } a \in \mathbb{Z}, R^{\mathbb{Z}} a a \text {, }$$
i.e., the statement
for every integer $a, a<a$.
We say that the formula $\forall v_{0} R v_{0} v_{0}$ does not hold in $\left(\mathbb{Z}, R^{\mathbb{Z}}\right)$.
If we consider the formula
$$\exists v_{0}\left(R v_{1} v_{0} \wedge R v_{0} v_{2}\right)$$
in $\left(\mathbb{Z}, R^{\mathbb{Z}}\right)$, we must also interpret the free variables $v_{1}$ and $v_{2}$ as elements of $\mathbb{Z}$. If we interpret $v_{1}$ as 5 and $v_{2}$ as 8 we obtain the (true) statement
$$\text { there is an integer a such that } 5<a \text { and } a<8 \text {. }$$

## 数学代写|数理逻辑代考Mathematical logic代写|Structures and Interpretations

Let $A$ be a set and $n \geq 1$. An $n$-ary function on $A$ is a map whose domain is the set $A^{n}$ of $n$-tuples of elements from $A$, and whose values lie in $A$. By an $n$-ary relation $\Re$ on $A$ we mean a subset of $A^{n}$. Instead of writing $\left(a_{1}, \ldots, a_{n}\right) \in \Re$, we shall often write $\Re a_{1} \ldots a_{n}$, and we shall say that the relation $\Re$ holds for $a_{1}, \ldots, a_{n}$. According to this definition, the divisibility relation on $\mathbb{N}$ is the set
$${(n, m) \mid n, m \in \mathbb{N} \text { and there is } k \in \mathbb{N} \text { with } n \cdot k=m},$$
and the relation “smaller-than” on $\mathbb{Z}$ is the set
$${(a, b) \mid a, b \in \mathbb{Z} \text { and } a<b} .$$
In the examples given earlier, the structures $\left(\mathbb{N}, R^{\mathbb{N}}\right)$ and $\left(\mathbb{Z}, R^{\mathbb{Z}}\right)$ were determined by the domains $\mathbb{N}$ and $\mathbb{Z}$ and by the binary relations $R^{\mathbb{N}}$ and $R^{\mathbb{Z}}$ as interpretations of the symbol $R$. We call $\left(\mathbb{N}, R^{\mathbb{N}}\right)$ and $\left(\mathbb{Z}, R^{\mathbb{Z}}\right){R}$-structures, thereby specifying the set of interpreted symbols, in this case ${R}$.

Consider once more the symbol set $S_{\mathrm{gr}}={o, e}$ of group theory. If we take the real numbers $\mathbb{R}$ as the domain and interpret $\circ$ as the addition $+$ over $\mathbb{R}$ and $e$ as the element 0 of $\mathbb{R}$, then we obtain the $S_{\mathrm{gr}}$-structure $(\mathbb{R},+, 0)$. In general an $S$-structure $\mathfrak{A}$ is determined by specifying:
(a) a domain $A$,
(b) (1) an $n$-ary relation on $A$ for every $n$-ary relation symbol in $S$,
(2) an $n$-ary function on $A$ for every $n$-ary function symbol in $S$,
(3) an element of $A$ for every constant in $S$.

# 数理逻辑代写

## 数学代写|数理逻辑代考Mathematical logic代写|Semantics of First-Order Languages

$\forall v_{0} R v_{0} v_{0}$

for all $n \in \mathbb{N}, R^{\mathbb{N}} n n$,

for all $a \in \mathbb{Z}, R^{\mathbb{Z}} a a$,

$$\exists v_{0}\left(R v_{1} v_{0} \wedge R v_{0} v_{2}\right)$$

## 数学代写数理逻辑代考Mathematical logic代写|Structures and Interpretations

$(n, m) \mid n, m \in \mathbb{N}$ and there is $k \in \mathbb{N}$ with $n \cdot k=m$,

$$(a, b) \mid a, b \in \mathbb{Z} \text { and } a<b .$$

(a) 域来确定 $A$,
(b) (1)一个n-ary 关系 $A$ 对于每个 $n$-ary 关系符昊 $S$,
(2) 一个n-ary 函数 $A$ 对于每个 $n$-ary 函数符号 $S$ ，
(3) 的一个元嗉 $A$ 对于每一个常数 $S$.

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## MATLAB代写

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