# If $\Sigma \Cup${$\phi$}$\vdash \theta$ and $\Sigma \Cup${$\lnot\phi$}$\vdash \theta$ then $\Sigma \vdash \theta.$

$$\Sigma$$ 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.

• What's the question? – Lord Shark the Unknown Sep 21 '18 at 3:52
• See your post : you need the same preliminary work, and also contraposition : B6) $\vdash (\phi \to \psi) \to (\lnot \psi \to \lnot \phi)$. – Mauro ALLEGRANZA Sep 21 '18 at 8:29
• Thus, byDT we have $\Sigma \vdash (\phi \to \theta)$ and $\Sigma \vdash (\lnot \phi \to \theta)$. – Mauro ALLEGRANZA Sep 21 '18 at 8:30
• By contraposition : $\Sigma \vdash (\lnot \theta \to \lnot \phi)$ and $\Sigma \vdash (\lnot \theta \to \lnot \lnot \phi)$. – Mauro ALLEGRANZA Sep 21 '18 at 8:31
• Finally, use a suitable instance of A3). – Mauro ALLEGRANZA Sep 21 '18 at 8:32

## 1 Answer

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 $$\Sigma \not\vdash \phi \wedge \neg \phi$$.

• Oh shoot. I forgot to add I can only use $\rightarrow$ and modus ponens. – Timelined Sep 21 '18 at 4:03
• Don't forget that $\phi \rightarrow \theta$ is equivalent to $\neg \phi \vee \theta$. – Tom Ryddle Sep 21 '18 at 4:09
• 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 $\Sigma \vdash \phi$ – Timelined Sep 21 '18 at 4:22