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I want to prove this statement (EFQ) using the natural deduction rules presented in the book Mathematical Logic (Chiswell & Hodges, 2007). It does not explicitly state EFQ as one of the fundamental rules of inference, so I am trying to derive it somehow.

The book provides the usual rules for $\wedge, \vee, \to$ introduction and elimination, which seem to agree with every other text that I have seen, but each text seems to differ on the rest of the rules. C&H uses the following:

($\neg$E): $P, \neg P \vdash \bot$

($\neg$I): If $P \vdash \bot$, then $\vdash \neg P$

(RAA): If $\neg P \vdash \bot$, then $\vdash P$

Is the following proof sufficient?

  1. $\bot$ (Initial premise)
  2. Assume $\neg P$
  3. Therefore, P (RAA using 1 and 2, discharges assumption on line 2)

I'm a little sketchy as to whether the use of RAA in line 3 is proper use, because the assumption is introduced after the $\bot$. Do you need to derive $\bot$ from $\neg P$ in order to use RAA? If this proof is not legal, then how would I prove it?

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    $\begingroup$ It works; if you want to avoid the "imperfection", reiterate $\bot$ after 2. It seems a "trick" but it is a sound move. Recall that $P \to (Q \to P)$ is a tautology. Thus, we can build the derivation : 1) $P$ --- assumed; 2) $Q$ --- assumed; 3) $\vdash P \to (Q \to P)$; 4) $Q \to P$ --- from 1) and 4) by $\to$-elim; 5) $P$ --- from 2) and 4) bt $\to$-elim. $\endgroup$ – Mauro ALLEGRANZA Jan 25 '17 at 8:02
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I looked at the $RAA$ rule defined by the book:

It says that if you have a derivation $D$ of $\bot$, i.e. if you have:

$D$

$-$

$\bot$

then you can have a derivation of any statement $\phi$ as follows:

$D$

$-$

$\bot$

$-$ ($RAA$)

$\phi$

And it says:

"Its assumptions are those of $D$, except possibly $\neg \phi$"

(by which they mean that if $\neg \phi$ is part of the undischarged assumptions of $D$, then you can remove $\neg \phi$)

OK, since you have $\bot$ as a premise, your 'derivation' of $\bot$ is just:

$\bot$

So, using the $RAA$ rule, this means we can now have the following derivation:

$\bot$

$-$ ($RAA$)

$P$

And that's it! So, no assumption of $\neg P$ necessary: you can immediately infer $P$ from $\bot$ as a special case of the $RAA$ rule.

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  • $\begingroup$ Thanks for the answer, maybe I'm being pedantic but after having another look at the sequent rule for RAA, it seems that technically you need to introduce an assumption in order to match the sequent. The overall outcome is identical, because you introduce the assumption and discharge it immediately. $\endgroup$ – esotechnica Jan 26 '17 at 9:37
  • $\begingroup$ @esotechnica I am indeed not 100% sure, but my interpretation of the 'possibly $\neg \phi$' phrase was to indicate that you don't need $\neg \phi$. But yes, I could be wrong about that, in which case you do indeed add it as an assumption, and then discharge it with the RAA rule, as you say, in which case your original proof was exactly right. $\endgroup$ – Bram28 Jan 26 '17 at 12:50

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