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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!

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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 $T$.

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  • $\begingroup$ 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
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    $\begingroup$ Reductio ad absurdum, the rule that if premisses $\Gamma$ together with $\varphi$ entail a contradiction, then $\Gamma$ alone entails $\neg\varphi$ $\endgroup$ Apr 24 '13 at 15:00
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    $\begingroup$ 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. $\endgroup$ 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
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If $T$ is $\omega$-categorical, $T$ is complete (Vaught's test). So such a $\varphi$ does not exist.

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  • $\begingroup$ 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
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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.

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