数学代写|数理逻辑代写Mathematical logic代考|Two Technical Lemmas

Doug I. Jones

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数学代写|数理逻辑代写Mathematical logic代考|Two Technical Lemmas

In this section we present two rather technical lemmas that we need to complete the proof of Theorem 2.5.1. The proofs that are involved are not pretty, and if you are the trusting sort, you may want to scan through this section rather quickly. On the other hand, if you come to grips with these results, you will gain a better appreciation for the details of substitutability and assignment functions.

To motivate the first lemma, consider this example: Suppose that we are working in the language of number theory and that the structure under consideration comprises the natural numbers. Let the term $u$ be $x \cdot v$ and the term $t$ be $y+z$. Then $u_t^x$ is $(y+z) \cdot v$. Now we have to fix a couple of assignment functions. Let the assignment function $s$ look like this:
\begin{tabular}{c|c}
Vars & $s$ \
\hline$x$ & 12 \
$y$ & 3 \
$z$ & 7 \
$v$ & 4 \
$\vdots$ & $\vdots$
\end{tabular}
So $s(x)=12, s(y)=3$, and so on.
Now, suppose that $s^{\prime}$ is an assignment function that is just like $s$,

except that $s^{\prime}$ sends $x$ to the value $\bar{s}(t)$, which is $\bar{s}(y+z)=3+7=10$ :
\begin{tabular}{c|c|c}
Vars & $s$ & $s^{\prime}$ \
\hline$x$ & 12 & 10 \
$y$ & 3 & 3 \
$z$ & 7 & 7 \
$v$ & 4 & 4 \
$\vdots$ & $\vdots$ & $\vdots$
\end{tabular}
Now, if you compare $\bar{s}\left(u_t^x\right)$ and $\overline{s^{\prime}}(u)$, you find that
$$\begin{gathered} \bar{s}\left(u_t^x\right)=\bar{s}((y+z) \cdot v)=(3+7) \cdot 4=10 \cdot 4=40 \ \overline{s^{\prime}}(u)=\overline{s^{\prime}}(x \cdot v)=10 \cdot 4=40 . \end{gathered}$$

数学代写|数理逻辑代写Mathematical logic代考|Properties of Our Deductive System

Having gone through all the trouble of setting out our deductive system, we will now prove a few things both in and about that system. First, we will show that we can prove, in our deductive system, that equality is an equivalence relation.
Theorem 2.7.1.

1. $\vdash x=x$.
2. $\vdash x=y \rightarrow y=x$.
3. $\vdash(x=y \wedge y=z) \rightarrow x=z$.
Proof. We show that we can find deductions establishing that $=$ is reflexive, symmetric, and transitive in turn.
4. This is a logical axiom of type (E1).
5. Here is the needed deduction. Notice that the notations off to the right are listed only as an aid to the reader.
\begin{aligned} & {[x=y \wedge x=x] \rightarrow[x=x \rightarrow y=x]} \ & x=x \ & x=y \rightarrow y=x . \end{aligned}
1. Again, we present a deduction:
\begin{aligned} & {[x=x \wedge y=z] \rightarrow[x=y \rightarrow x=z]} \ & x=x \ & (x=y \wedge y=z) \rightarrow x=z . \end{aligned}
Chaff: Notice that we have done a bit more than prove that equality is an equivalence relation. (Heck, you’ve known that since fourth grade.) Rather, we’ve shown that our deductive system, with the axioms and rules of inference that have been outlined in this chapter, is powerful enough to prove that equality is an equivalence relation. There will be a fair bit of “our deductive system is strong enough to do such-and-such” in the pages to come.

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