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I have reduced a more complicated problem into a simpler one and now I wish to show that $\{\exists x_i (A \rightarrow B), \neg B \} \vdash \neg A$

This is equivalent to $\{ \neg \forall x_i \neg(A\rightarrow B), \neg B \} \vdash \neg A$

I'm struggling to somehow change the $\neg (A \rightarrow B)$.

We're in this system: enter image description here

and can use deduction theorem

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    only connectives are $\neg$ and $\rightarrow$ – 2017-01-11
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    Are you sure that you can ? How the semantics corresponding to your proof system manage open formuale ? – 2017-01-12
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    @user381198 Do $A$ and/or $B$ contain $x_i$ as a free variable? – 2017-01-13

1 Answers 1

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This is quite a long proof, so I'll only sketch it. If you search online, you should be able to fill in the details.

OK, so first let's point out that you can prove as a Theorem that:

If $\Gamma,\varphi \vdash \psi$ and $\Gamma,\varphi \vdash \neg \psi$, then $\Gamma \vdash \neg \varphi$ (so this is like a proof by Contradiction)

OK, it should be pretty clear that:

$\{A, \neg B, A \rightarrow B\} \vdash B$

and also that:

$\{A, \neg B, A \rightarrow B\} \vdash \neg B$

Hence, by the Theorem:

$\{A, \neg B \} \vdash \neg (A \rightarrow B)$

Since there is no free variable $x_i$ in $\neg (A \rightarrow B)$, that means that we extend that derivation by $\forall x_i \neg (A \rightarrow B)$ using principle iv). In other words:

$\{A, \neg B \} \vdash \forall x_i \neg (A \rightarrow B)$

OK, now let's use the Deduction Theorem, which says that:

If $\Gamma,\varphi \vdash \psi$, then $\Gamma \vdash \varphi \rightarrow \psi$

and apply it to what we just obtained:

$\{\neg B \} \vdash A \rightarrow \forall x_i \neg (A \rightarrow B)$

OK, let's use one final Lemma, which says that:

$\vdash (\varphi \rightarrow \psi) \rightarrow (\neg \psi \rightarrow \neg \varphi)$

As a specific instance of this Lemma we get:

$\vdash (A \rightarrow \forall x_i \neg (A \rightarrow B)) \rightarrow (\neg \forall x_i \neg (A \rightarrow B) \rightarrow \neg A)$

Adding this to the derivation of $\{\neg B \} \vdash A \rightarrow \forall x_i \neg (A \rightarrow B)$, and applying MP (Modus Ponens, principle iii)), we thus get that:

$\{\neg B \} \vdash \neg \forall x_i \neg (A \rightarrow B) \rightarrow \neg A$

But that immediately means that:

$\{ \neg \forall x_i \neg (A \rightarrow B) , \neg B \} \vdash \neg A$

as required.

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    As stated by the OP, the question does not say that there are no free vars in $A$ and $B$... – 2017-01-13
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    @MauroALLEGRANZA Hmm, good point. I'll ask the OP. – 2017-01-13