$\Sigma$ is a set of sentences, the set $ 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 → with modus ponens to make any deductions.

I'm having a tough time dealing with $\Sigma \vdash \lnot(\phi \rightarrow \psi)$. I can't use $\land$ or $\lor$ so I can't use DeMorgan's law. I assume I have to use axiom A3 with $\phi$ as $(\phi \rightarrow \psi)$ or something. Any hints?

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    $\begingroup$ We have to produce a lot of preliminary work ... : use A1) and A2) to prove B1) $\vdash \phi \to \phi$. $\endgroup$ – Mauro ALLEGRANZA Sep 21 '18 at 7:56
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    $\begingroup$ With B1) prove the Deduction Theorem. $\endgroup$ – Mauro ALLEGRANZA Sep 21 '18 at 7:57
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    $\begingroup$ With the DT prove the "auxiliary" rule : B2) $\phi \to \psi, \psi \to \chi \vdash \phi \to \chi$. $\endgroup$ – Mauro ALLEGRANZA Sep 21 '18 at 7:57
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    $\begingroup$ Prove Double Negation laws : B3) $\lnot \lnot \phi \vdash \phi$ and B4) $\phi \vdash \lnot \lnot \phi$. $\endgroup$ – Mauro ALLEGRANZA Sep 21 '18 at 7:59
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    $\begingroup$ Now we can prove a useful "variant" of A3) : B5) $\vdash (\phi \to \psi) \to ((\phi \to \lnot \psi) \to \lnot \phi)$. $\endgroup$ – Mauro ALLEGRANZA Sep 21 '18 at 8:01

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