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Question: Is is possible to derive $\ \vdash \sim\forall xP(x) \leftrightarrow\exists x \sim P(x) \ $ (or any other version of the quantifier negation rule) axiomatically?

Context: I tutor college students in various subjects including symbolic logic. I recently acquired a new student who is using a text that I am unfamiliar with. This text is accompanied by a program called Logic 2010, so by way of preparation, I downloaded the software and began playing with it.

One of the exercises is the above derivation. However, I keep hitting an impasse when attempting it: I can't manage to derive the relationship unless I assume $\sim\exists x\sim P(x)\vdash \forall xP(x)$. Pretty much any logical rule or identity is fair game other than internal substitutions of logical equivalences.

These identities have always been presented to me as intuitively obvious, so now, I am beginning to think that it is not possible to derive them - that they must be axiomatic.

Are the quantifier negation rules axiomatic or derivable?

Geoffrey
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  • @MauroALLEGRANZA So if they are each individually introduced without reference to one another, then there is no way to derive their relationship using the basic axioms of PL (i.e. SL + Universal Instantiation, Existential Instantiation, etc.)? – Geoffrey Sep 05 '17 at 06:03
  • Sorry... ny mistake. There are two main possibilities: (i) Introduce one as primitive and the other as defined; e.g. $\exists$ as $\lnot \forall \lnot$. In this case, the usual equivalences are "built-in" in the system. – Mauro ALLEGRANZA Sep 05 '17 at 06:10
  • (ii) Introduce them both (independently) with their specific rules (see Natural Deduction, in which case the equivalences can be proved (in classical logic) or at least part of them can be (in intuitionistic logic). – Mauro ALLEGRANZA Sep 05 '17 at 06:11
  • @MauroALLEGRANZA If you know of a resource that provides that classical derivation (or if you can explain it yourself), I would be more than happy to accept that answer. – Geoffrey Sep 05 '17 at 06:21
  • @Geoffrey Do you have a link to the axioms the course uses? – DanielV Sep 05 '17 at 06:58
  • @DanielV Unfortunately, you have to download the software from the link in the question to get access to the (frankly) confusing text. The course is using the Kalish-Montague system. It's pretty much your standard "natural deduction" system with double negation, modus tollens, modus ponens, UI, EI, EG, and all of the standard logical equivalences. It doesn't have "axioms" per se (it's not $K_{\mathscr L}$ for instance), but it does have strict rules for structuring derivations (e.g. everything is linear so as far as I can tell it is impossible to construct dilemmas and its UG is weird). – Geoffrey Sep 05 '17 at 07:14

1 Answers1

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In Natural Deduction we can prove the "usual" relationships between quantifiers starting from the basic rules.

E.g.

1) $\forall x Px$ --- premise

2) $\exists x \lnot Px$ --- assumed [a]

3) $\lnot Py$ --- assumed [b] from 4) for $\exists$-elim

4) $Py$ --- from 1) by $\forall$-elim (Universal Instantiation)

5) $\bot$ --- contradiction: from 3) and 4)

6) $\bot$ --- from 2), 3) and 4) by $\exists$-elim, discharging [b]

7) $\lnot \exists x \lnot Px$ --- from 2) and 6) by $\to$-intro, dicharging [a]

8) $\forall x Px \vdash \lnot \exists x \lnot Px$ --- from 1) and 7).

1) $\lnot \exists x \lnot Px$ --- premise

2) $\lnot Py$ --- assumed [a]

3) $\exists x \lnot Px$ --- from 2) by $\exists$-intro

4) $\bot$ --- contradicition: from 1) and 3)

5) $\lnot \lnot Py$ --- from 2) and 4) by $\to$-intro, discharging [a]

6) $Py$ --- from 5) by Double Negation (it holds only classically)

7) $\forall x Px$ --- from 6) by $\forall$-intro

8) $\lnot \exists x \lnot Px \vdash \forall x Px$ --- from 1) and 7).

As you can see, some of the "usual" equivalences (see also; $¬∀x¬ Px \to ∃xPx$) hold only in classical logic, because their proof needs Double Negation; see Intuitionistic Logic.

For more details, see e.g.:

For a different approach you can see:

Mauro ALLEGRANZA
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  • Thanks! I'll browse these sources. I understand this portion of the proof. However, since I'm looking to prove the logical equivalence, the converse is also necessary (i.e. $\sim\exists x\sim P(x) \rightarrow \forall xP(x)$). That's the part that really holds me up since there is no way to apply UI or EI with those negations in front of both quantifiers. – Geoffrey Sep 05 '17 at 06:47
  • Perfect, this is crystal clear. Thanks so much! – Geoffrey Sep 05 '17 at 07:19