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# 数学代写|数理逻辑代写Mathematical logic代考|MATH4810 Substitution

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

In this section we define how to substitute a term $t$ for a variable $x$ in a formula $\varphi$ at the places where $x$ occurs free, thus obtaining a formula $\psi$. We wish to define the substitution so that $\psi$ expresses the same about $t$ as $\varphi$ does about $x$. We start with an example to illustrate our objective and to show why a certain care is necessary. Let
$$\varphi:=\exists z z+z \equiv x .$$
In $\mathfrak{N}$ the formula $\varphi$ says that $x$ is even; more precisely:
$$(\mathfrak{N}, \beta) \models \varphi \quad \text { iff } \quad \beta(x) \text { is even. }$$
If we replace the variable $x$ by $y$ in $\varphi$, we obtain the formula $\exists z z+z \equiv y$, which states that $y$ is even. But if we replace the variable $x$ by $z$, we obtain the formula $\exists z z+z \equiv z$, which no longer says that $z$ is even; in fact, this formula is valid in $\mathfrak{N}$ regardless of the assignment for $z$ (because $0+0=0$ ). In this case the meaning is altered because at the place where $x$ occurred free, the variable $z$ gets bound. On the other hand, we obtain a formula which expresses the same about $z$ as $\varphi$ does about $x$ if we proceed as follows: First, we introduce a new bound variable $u$ in $\varphi$, and then in the formula $\exists u u+u \equiv x$ thus obtained we replace $x$ by $z$. It is immaterial which variable $u$ (distinct from $x$ and $z$ ) we choose. However, for certain technical purposes it is useful to make a fixed choice.

In the preceding example we replaced only one variable, but in our exact definition we specify the procedure for simultaneously replacing several variables: With a given formula $\varphi$, pairwise distinct variables $x_0, \ldots, x_r$ and arbitrary terms $t_0, \ldots, t_r$, we associate a formula $\varphi \frac{t_0 \ldots t_r}{x_0 \ldots x_r}$, which is said to be obtained from $\varphi$ by simultaneously substituting $t_0, \ldots, t_r$ for $x_0, \ldots, x_r$. The reader should note that $x_i$ has to be replaced by $t_i$ only if
$$x_i \in \operatorname{free}(\varphi) \quad \text { and } \quad x_i \neq t_i .$$
In the following inductive definition this is explicitly taken into account in the quantifier step; in the other steps it follows immediately.

It will become apparent that it is convenient to first introduce a simultaneous substitution for terms. Let $S$ be a fixed symbol set.

## 数学代写|数理逻辑代写Mathematical logic代考|Substitution Lemma.

Substitution Lemma. (a) For every term $t$,
$$\mathfrak{I}\left(t \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(t) .$$
(b) For every formula $\varphi$,
$$\mathfrak{I}=\varphi \frac{t_0 \ldots t_r}{x_0 \ldots x_r} \quad \text { iff } \quad \mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}=\varphi .$$
Proof. We proceed by induction on terms and formulas in accordance with the definitions $8.1$ and 8.2. We treat some typical cases.
$t=x$ : If $x \neq x_0, \ldots, x \neq x_r$, then, by Definition 8.1(a), $x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}=x$ and therefore,
$$\mathfrak{I}\left(x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{I}(x)=\mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(x) .$$
If $x=x_i$, then $x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}=t_i$ and hence,
\begin{aligned} & \mathfrak{I}\left(x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{I}\left(t_i\right)=\mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}\left(x_i\right)=\mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(x) . \ \varphi=& R t_1^{\prime} \ldots t_n^{\prime}: \mathfrak{I}=\left[R t_1^{\prime} \ldots t_n^{\prime}\right] \frac{t_0 \ldots t_r}{x_0 \ldots x_r} \ & \text { iff } \quad \mathfrak{I}(R) \text { holds for } \mathfrak{I}\left(t_1^{\prime} \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right), \ldots \quad \text { (by Definition 8.2(b)) } \ \text { iff } \quad \mathfrak{I}(R) \text { holds for } \mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}\left(t_1^{\prime}\right), \ldots \quad \text { (by (a)) } \ \text { iff } \quad \Im \mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(R) \text { holds for } \mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{J}\left(t_r\right)}{x_0 \ldots x_r}\left(t_1^{\prime}\right), \ldots \ & \text { iff } \quad \mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r} \models R t_1^{\prime} \ldots t_n^{\prime} . \end{aligned}

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

$$\varphi:=\exists z z+z \equiv x .$$

$$(\mathfrak{N}, \beta) \models \varphi \quad \text { iff } \quad \beta(x) \text { is even. }$$

$$x_i \in \operatorname{free}(\varphi) \quad \text { and } \quad x_i \neq t_i .$$

## 数学代写|数理逻辑代写Mathematical Logic代考|Substitution Lemma.

$$\mathfrak{I}\left(t \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{J} \frac{\mathfrak{J}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(t)$$
(b) 对于每个公式 $\varphi$,
$$\mathfrak{I}=\varphi \frac{t_0 \ldots t_r}{x_0 \ldots x_r} \quad \text { iff } \quad \mathfrak{I} \frac{\mathfrak{J}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}=\varphi$$

$$\mathfrak{J}\left(x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{I}(x)=\mathfrak{I} \frac{\mathfrak{J}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(x) .$$

$\mathfrak{I}\left(x \frac{t_0 \ldots t_r}{x_0 \ldots x_r}\right)=\mathfrak{I}\left(t_i\right)=\mathfrak{I} \frac{\mathfrak{J}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}\left(x_i\right)=\mathfrak{I} \frac{\mathfrak{I}\left(t_0\right) \ldots \mathfrak{I}\left(t_r\right)}{x_0 \ldots x_r}(x) \cdot \varphi=\quad R t_1^{\prime} \ldots t_n^{\prime}: \mathfrak{I}=\left[R t_1^{\prime} \ldots t_n^{\prime}\right] \frac{t_0 \ldots t_r}{x_0 \ldots x_r}$ iff $\quad \mathfrak{I}(R)$ holds for $\mathfrak{I}\left(t_1^{\prime} \frac{t_0}{x_0}\right.$

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