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$\Sigma$ is a set of sentences, the set $\mathcal{L}$ consists of all axioms of the forms:

A1) $\phi \rightarrow (\psi \rightarrow \phi)$

A2) $(\phi \rightarrow (\psi \rightarrow \theta)) \rightarrow ((\phi \rightarrow \psi) \rightarrow (\phi \rightarrow \theta))$

A3) $((\lnot \phi \rightarrow \psi) \rightarrow (( \lnot \phi \rightarrow \lnot \psi) \rightarrow \phi))$

*I can only use $\rightarrow$ with modus ponens.

I need to prove this and I would like some hints. I can get by Deduction theorem, $\Sigma \vdash \phi \rightarrow \theta$ and $\Sigma \vdash \lnot\phi \rightarrow \theta$. I think you need to use A3 with $\phi$ as $\theta$, but I'm not quite sure what else.

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  • $\begingroup$ What's the question? $\endgroup$ Commented Sep 21, 2018 at 3:52
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    $\begingroup$ See your post : you need the same preliminary work, and also contraposition : B6) $\vdash (\phi \to \psi) \to (\lnot \psi \to \lnot \phi)$. $\endgroup$ Commented Sep 21, 2018 at 8:29
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    $\begingroup$ Thus, byDT we have $\Sigma \vdash (\phi \to \theta)$ and $\Sigma \vdash (\lnot \phi \to \theta)$. $\endgroup$ Commented Sep 21, 2018 at 8:30
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    $\begingroup$ By contraposition : $\Sigma \vdash (\lnot \theta \to \lnot \phi)$ and $\Sigma \vdash (\lnot \theta \to \lnot \lnot \phi)$. $\endgroup$ Commented Sep 21, 2018 at 8:31
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    $\begingroup$ Finally, use a suitable instance of A3). $\endgroup$ Commented Sep 21, 2018 at 8:32

1 Answer 1

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For deduction theorem $\Sigma \vdash \phi \rightarrow \theta$ and $\Sigma \vdash \neg \phi \rightarrow \theta$, then $\Sigma \vdash \phi \wedge \neg \phi \rightarrow \theta$, but $\Sigma \not\vdash \phi \wedge \neg \phi$.

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  • $\begingroup$ Oh shoot. I forgot to add I can only use $\rightarrow$ and modus ponens. $\endgroup$
    – Timelined
    Commented Sep 21, 2018 at 4:03
  • $\begingroup$ Don't forget that $\phi \rightarrow \theta$ is equivalent to $\neg \phi \vee \theta$. $\endgroup$
    – Tom Ryddle
    Commented Sep 21, 2018 at 4:09
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    $\begingroup$ Does $\Sigma \vdash \phi \wedge \neg \phi \rightarrow \theta$ mean $\Sigma \vdash$ both $\phi$ and $\neg \phi \rightarrow \theta$. If so how did you get $\Sigma \vdash \phi$ $\endgroup$
    – Timelined
    Commented Sep 21, 2018 at 4:22

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