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For formal proofs, would the negation from the universal need to be removed first? Would this be done by attempting to prove that ¬∀x:X.P → Q is also not true using negation induction?

that is to arrive to two opposite conclusions using the above proof? Right to left is easier to breach but left to right is leaving me puzzled.

Zen
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  • Form Left to Right you have to "move inside" the negation sign : $\lnot \forall x$ is equivalent to : $\exists x \lnot$. – Mauro ALLEGRANZA Jan 22 '16 at 12:27
  • Yes, that much makes sense, thanks. The problem now is going past the next negation which is ¬(P → Q).

    The end result would be ¬(¬Q → ¬P) Which means that the implication must be reversed, which is easily done using the contrapositive law, but if the outer negation is in the way, I can't do that, yet...

     – Zen Jan 22 '16 at 12:39
  • This is simple propositional logic : $P \to Q$ is equivalent to $\lnot Q \to \lnot P$; thus their negations also are. Details depend on the proof system you are using : truth-table, natural deduction, etc. – Mauro ALLEGRANZA Jan 22 '16 at 13:04
  • Thanks for the reply. It's using formal proofing, and from what I could gather, we can never ignore a negation or work inside the brackets when using formal proofs, we must always include the outer operator. So unless I can reference some sort of law, I am unable to move past this using the contrapositive law you quoted. At least not according to this particular criteria.

    Edit: To clarify, what you mentioned is correct, however I would need to somehow include the negation in that, to mantain formal proofing. I have no idea how I would go on about doing this but that's what being asked

     – Zen Jan 22 '16 at 13:16

2 Answers2

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This is based purely on what I managed to figure out, by backing up and not swapping the negative universal to an existential. ¬∀x:X.P → Q ⊣⊢ ∃x:X.¬(¬Q → ¬P)

1. ¬∀x:X.P → Q  (hyp)
  2. ¬Q → ¬P (assume)
  3. ¬∀x:X.P → Q (copy 1)

  4. ¬Q → ¬P (assume)
  5. P → Q (contrapositive, 4)
  6. ∀x:X.P → Q (∀ intro, 5)
7. ¬(¬Q → ¬P) (¬ intro, 2-3, 4-6)
8. ¬(¬Q → ¬P)[x/x] (A[x/x] = A)
9. ∃x:X.¬(¬Q → ¬P) (∃ intro, 8)

I'm not sure if this is the correct solution, but based on using the simplest rules possible, this is what I ended up with.

Primary Concern is whether or not line 3 and line 6 are truly opposites or if that doesn't count as a solid basis for introducing a negation as I did in line 7.

I would appreciate input on this answer.

Zen
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  • The issue is with step 6 ($\forall$-intro) : you cannot perform it with assumption 2 IF you are not sure that $x$ is not free in it (see proviso for $\forall$-intro). The result is "classical"; thus, it must need double negation... – Mauro ALLEGRANZA Jan 22 '16 at 14:44
  • Thanks for the reply Mauro, I don't quite understand what you mean with "x is not free in it" however, could you elaborate? Is there an example answer you might provide to work around this issue? – Zen Jan 22 '16 at 14:56
  • see this post for details on the restriction regarding $\forall$-introduction (called also UG). – Mauro ALLEGRANZA Jan 22 '16 at 18:31
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Here is a proof with Natural Deduction; for the rules, see e.g.;

1) $¬∀x(Px → Qx)$ --- premise

2) $¬∃x¬ (¬Qx → ¬Px)$ --- assumed [a]

3) $¬(¬Qx → ¬Px)$ --- assumed [b]

4) $∃x¬ (¬Qx → ¬Px)$ --- from 3) by $∃$-intro

5) $\bot$ --- from 2) and 4) by $\lnot$-elim

6) $¬Qx → ¬Px$ --- from 3) and 5) by Double Negation-elim, discharging [b]

7) $Px → Qx$ --- from 6) by Contraposition [it is easily proved with a sub-derivation]

8) $∀x(Px → Qx)$ --- from 7) by $∀$-intro : no open assumptions with $x$ free

9) $\bot$ --- from 1) and 8)

10) $∃x¬ (¬Qx → ¬Px)$ --- from 2) and 9) by Double Negation, discharging [a].

Thus, from 1) and 10) we have :

$¬∀x(Px → Qx) \vdash ∃x¬ (¬Qx → ¬Px)$.

Mauro ALLEGRANZA
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  • Thanks for the reply, I've been attempting to figure out what you're trying to demonstrate, but regretfully could not since ⊥ is not a sign I've ever come across. No idea what it means, and also first time I see two assumptions being taken one after the other. If you have the patience to explain briefly (I tried searching for it, found Orthogonal Complement) but that was used as a superscript, so not sure either way, and I doubt we need to use those for our solution, considering we never covered them.

    Is my introduction considered formally incorrect in line 6, or more like incomplete?

     – Zen Jan 23 '16 at 15:07
  • @Zen - $\bot$ is the symbol for the contradiction. – Mauro ALLEGRANZA Jan 23 '16 at 20:38