1
$\begingroup$

In a substep of a proof, I have a sentence $\varphi$ such that $\varphi\notin T$ and $(\neg\varphi)\notin T$. ($T$ is a consistent theory)

Information about $T$: $T$ is a consistent theory of a first order language $\mathcal{L}_\mathcal{A}$. Every model of $T$ is infinite. Assume $T$ is $\omega$-categorical.

I need to show that $T\cup\{\varphi \} $ and $T\cup\{\neg\varphi\}$ are both consistent.

How would I go about doing that?


I have thought of the following: Suppose $T\cup\{\varphi \}\vdash\psi $ for some formula $\psi$.

Case 1: $\varphi\neq\psi$.

Then we have $T\vdash\psi$. Since $T$ is consistent, so $T\not\vdash(\neg\psi)$. Thus $T\cup\{\varphi\}\not\vdash(\neg\psi)$.

Case 2: $\varphi=\psi$.

Suppose to the contrary, $T\cup\{\varphi\}\vdash(\neg\varphi)$. Then $T\vdash(\neg\varphi).$ But by consistency of $T$, $T\not\vdash\varphi$.

This would mean $T\cup\{\varphi\}\not\vdash(\neg\varphi)$, a contradiction.


Is what I am doing correct? Also, is there any shorter method?

Sincere thanks for help!

$\endgroup$
1
$\begingroup$

Suppose that $T\cup \{\phi\}$ is inconsistent. Then $T\cup\{\phi\}\vdash\{\neg\phi\}$. Therefore, $T\vdash\{\neg\phi\}\Rightarrow \neg \phi\in T $ which is a contradiction.

$\endgroup$
2
$\begingroup$

Suppose $T \cup {\varphi}$ is inconsistent. Then $T, \varphi \vdash \bot$, so [assuming $T$ has reductio] $T \vdash \neg\varphi$. So [assuming $T$ is closed under deducibility] $\neg\varphi \in T$. But by hypothesis, $\neg\varphi \notin T$. So $T \cup {\varphi}$ is consistent.

  1. That depends on the bracketed assumptions: but you haven't said exactly what you are allowed to assume about $T$ and that matters crucially! So you need to spell that out.

  2. The proof for the other case will work similarly, if you make the right assumptions about $T$.

$\endgroup$
  • $\begingroup$ what is reductio? $\endgroup$ – yoyostein Apr 24 '13 at 14:58
  • $\begingroup$ $T$ is assumed to be consistent (as stated in question), if that helps? $\endgroup$ – yoyostein Apr 24 '13 at 14:59
  • 1
    $\begingroup$ Reductio ad absurdum, the rule that if premisses $\Gamma$ together with $\varphi$ entail a contradiction, then $\Gamma$ alone entails $\neg\varphi$ $\endgroup$ – Peter Smith Apr 24 '13 at 15:00
  • 1
    $\begingroup$ Classically, an inconsistent theory entails everything, so [assuming deductive closure] an inconsistent theory includes every [relevant] sentence. Contraposing, if $\varphi \notin T$, $T$ has to be consistent. $\endgroup$ – Peter Smith Apr 24 '13 at 15:02
  • $\begingroup$ oh yes, I missed out several information about $T$, pls see edit. Thanks! $\endgroup$ – yoyostein Apr 24 '13 at 15:05
2
$\begingroup$

If $T$ is $\omega$-categorical, $T$ is complete (Vaught's test). So such a $\varphi$ does not exist.

$\endgroup$
  • $\begingroup$ After a little search, I assumed that this is related to this. Then you want to prove : $T$ not complete implies $T$ not $\omega$-categorical. So remove the $\omega$-categoricalness from the hypothesis in here. $\endgroup$ – Pece Apr 24 '13 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.