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In every case of $\neg \neg P$ that I've come across, the statement $\neg P$ has been disproven. Never has such a proof merely been proof for the inexistence of a proof for $\neg P$.

Take the proposition $\neg P = 3/2 \in \Bbb Z$. $ P = 3/2 \not \in \Bbb Z$. I am now going to disprove $\neg P$.

$3/2 = 1.5$.

I have now just proven that $\neg P$ is false, or in classical terms, $\neg \neg P$. It is thus simple logic that $P$ is true, because I didn't just prove the lack of a proof for $\neg P$*, I proved directly that $\neg P$ is false.

*I think I've read that ZFC has yet to be proven consistent, so if I'm doing this in ZFC, then I haven't even proven that there is no proof for $\neg P$; yet, my proof for the falsity of $\neg P$ is still undeniable.

If I'm not mistaken, intuitionists agree that if I prove $\neg P$ false, then I automatically prove $P$ true. They just disagree on what a mathematical disproving of any/specific negative statement(s) are/is doing. They seem to think that disproving a negative statement isn't proving its falsity, just its lack of a proof. I just don't understand why they think that. As far as I understand, it must be something that only applies to specific negative statements.

user110391
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    "ZFC has yet to be proven consistent" - what proof system would be able to do that? No sufficiently useful axiom system can prove its own consistency. – aschepler Aug 11 '22 at 21:45
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    It's not obvious to me that you have a proof that $\lnot\lnot($"$3$ is odd"$)$ – aschepler Aug 11 '22 at 21:49
  • @aschepler "what proof system would be able to do that?" I don't know. Are you saying no such axiom system could be created? Even so, the point still stands. We don't know if ZFC is consistent, so disproving any statement $P$ does not rule out the possibility that proof of $P$ exists. What was the point of this comment? – user110391 Aug 11 '22 at 22:12
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    $3/2=1.5$ doesn't prove that $3/2$ is not an integer. What axiom system are you using? – Thomas Andrews Aug 11 '22 at 22:12
  • @aschepler I have edited it in response to Mark Saving's answer. I don't see how my previous formulation wasn't clear though. – user110391 Aug 11 '22 at 22:12
  • @ThomasAndrews How in God's name is $1.5$ an integer? – user110391 Aug 11 '22 at 22:14
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    What axiom system are you using? Most axiom systems dont even use decimal notation. "How in good name is $1.5$ an integer?" You have to prove it is not, not me, and you have to gove a formal system in which you prove it, with axioms. "Look at it! It's not an integer!" is not a proof. You'd need an axiom system that supports decimal notation (which is highly problematic, since the decimal notation of rational numbers is often infinite.) – Thomas Andrews Aug 11 '22 at 22:17
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    You'd also need to peove that $3/2=1.5,$ which is mere assertion without axioms. – Thomas Andrews Aug 11 '22 at 22:19
  • I hope you can't disprove $\lnot P = 3/2 \notin \mathbb{Z}$, since we ordinarily call that true. – aschepler Aug 11 '22 at 22:55
  • @aschepler Thanks for pointing out the typo, I have fixed it now – user110391 Aug 12 '22 at 00:57
  • There is no proposition $P$ where $\lnot P = (3/2 \in \mathbb{Z})$. – aschepler Aug 12 '22 at 01:49

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Intuitionists agree that if you prove $\neg Q$, you are proving that $Q$ is false. For an elaboration on this idea, see my answer here. This is true in the special case of $Q = \neg P$.

If I'm not mistaken, intuitionists agree that if I prove $\neg P$ false, then I automatically prove $P$ true.

You are mistaken. This is exactly the thing intuitionists do not accept about classical logic.

As for your purported proof of $\neg \neg (3$ is odd$)$, you would need to seriously flesh out this proof (starting with a definition of “odd”) before I can say anything about it.

Mark Saving
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  • Intuitionists agree that if you prove ¬Q, you are proving that Q is false. For an elaboration on this idea, see my answer here. This is true in the special case of Q=¬P." How is that different from saying that intuitionists agree that if I disprove $\neg P$, then I automatically prove $P$. Do intuitionists require a positive reformulation of the negative statement, so that the positive statement can then be disproven, thus providing a proof for the negation of the positive statement? Is that what you mean by $Q = \neg P$? – user110391 Aug 11 '22 at 22:05
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    Indeed, since odd can be expressed as a negative condition, $\lnot\lnot(\text{3 is odd})=\lnot\lnot\lnot(\text{3 is even}),$ and my vague memory is that intuitionists accept that $\lnot\lnot\lnot p$ implies $\lnot p.$ But my memory might be mistaken. – Thomas Andrews Aug 11 '22 at 22:22
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    Your vague memory is correct. From $P$ and $\neg \neg \neg P$ we deduce $\bot$, since $P$ implies $\neg \neg P$. – Patrick Stevens Aug 11 '22 at 22:24
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    @user110391 It is intuitionistically correct to say that if we prove $\neg Q$, this is the same as saying $Q$ is false. Applying this to $Q = \neg P$, we are saying that if we prove $\neg \neg P$, we are proving that $\neg P$ is false. But showing $\neg P$ is false is not sufficient to conclude that $P$ is true. Disproving $\neg P$ is only sufficient to show $\neg \neg P$; since intuitionists don’t accept the implication $\neg \neg P \to P$ as universally valid, they don’t consider a proof of $\neg \neg P$ sufficient to prove $P$. – Mark Saving Aug 11 '22 at 22:28
  • @MarkSaving But why don't they see it as universally valid? I thought that was because they disagreed on the semantics. That they think $\neg \neg P$ is not disproving $\neg P$, but simply a proof that there is no proof for $\neg P$. That semantical interpretation makes their rejection of LEM completely sensible, but the semantical interpretation itself is, as far as I can see, nonsensical. And as far as what you're saying, that ISN'T their semantical interpretation. You're saying that they do see $\neg \neg P$ as disproving $\neg P$. But if so, LEM is obviously true. – user110391 Aug 11 '22 at 23:04
  • @user110391 In general, intuitionists believe that if we know $A \lor B$, then we must either know $A$ or know $B$. On the other hand, knowing $\neg \neg (A \lor B)$ simply means that we know $\neg (A \lor B)$ isn’t true. This is not enough to establish whether $A$ is true, or $B$ is true. A similar thing happens for $\exists x (P(x))$. To know $\exists x (P(x))$ is to know of a specific $c$ such that $P(c)$. By contrast, to know $\neg \neg \exists x (P(x))$ is simply to know that $\neg \exists x (P(x))$ can’t be true. But this doesn’t give any insight for which specific $c$ $P(c)$ holds. – Mark Saving Aug 11 '22 at 23:27
  • @MarkSaving Your last comment makes complete sense, but I don't see how it rebuts my stance. It just seems to be moving the goalpost; it's pointing at things that remain unproven after proving the negation of a negation, yet those things weren't the goal to begin with. Proving $\neg \neg \exists x(P(x))$ does not give us $c$, but it still proves $\exists x(P(x))$. To say that to know of $\exists x(P(x))$ is to know of a $c$ is wrong. It could be your proof simply shows there MUST be such a $c$, without ever finding an exact one, right? – user110391 Aug 12 '22 at 00:21
  • @MarkSaving Knowing $A \lor B$ is not necessarily enough to know whether $A$ is true, nor is it necessarily enough to know whether $B$ is true. The statement doesn't try to assert either of those in isolation; it asserts that at least one of them is true. If my premise is that $m \in \Bbb N$, then I can say $m = 2n \lor 2n +1$. I don't know if either is true, but I do know that at least one is true. So, saying that $\neg\neg(A \lor B)$ doesn't tell us if either $A$ or $B$ is true is inconsequential, because $(A \lor B)$ doesn't tell us that either. – user110391 Aug 12 '22 at 00:26
  • @MarkSaving However, perhaps I am to interpret your arguments here as a showcase of the Intuitionist way of thought? As in, within Intuitionism, it IS required to find a $c$ such that $P(c)$, in order to prove $\exists x(P(x))$. An analogously, in Intuitionism, it is required to show that either $A$ or $B$ is true, in order to prove $A \lor B$. Perhaps this is the fundamental difference between constructivist logic and classical logic? I feel like I have a vague memory of that; in constructivism, you have to construct the (counter-)example that proves your proposition. [1/2] – user110391 Aug 12 '22 at 00:29
  • @MarkSaving Whereas in classical logic, you don't need to construct a (counter-)example at all. You can (dis)prove something indirectly. You can show that a construct is possible, and that is enough to prove its existence. Something like that, yeah? – user110391 Aug 12 '22 at 00:31
  • @user110391 My criterion relating to knowing $A \lor B$ applies to sentences $A$ and $B$ derived without assumptions. “$m$ is even or $m$ is odd” is not a sentence since it has a free variable $m$. The actual sentence proved is: for all $m$, either $m$ is even or $m$ is odd. The criterion for a sentence $\forall m P(m)$ is: if we know $\forall m P(m)$, then for all particular values of $m$, we must know that $P(m)$. So if we know $\forall m (m$ is even or $m$ is odd$)$, then for all particular values of $m$, we either know that $m$ is even, or we know that $m$ is odd. – Mark Saving Aug 12 '22 at 01:48
  • @user110391 "A proof that there is no proof" is a very different thing from a proof that a proposition is false. Normally we take "prove $P$ is false" to mean exactly "prove $\lnot P$". I believe that's also the case in intuitionist logic. – aschepler Aug 12 '22 at 13:01
  • @MarkSaving Okay, so I formulated it incorrectly, but the correctly formulated version of what I said could be proven without ever supplying a specific $m$, yes? An as for my thoughts on proving $\exists x(P(x))$; isn't it possible to do this in classical logic without every showing a $c$? – user110391 Aug 12 '22 at 13:40
  • @aschepler "I believe that's also the case in intuitionist logic." See the answer from Alex Kruckman to my answer here: https://math.stackexchange.com/questions/4420407/how-is-reductio-ad-absurdum-unintuitive This is where I got the idea from, and it is the only interpretation from which it makes sense to reject LEM in my head. The interpretation itself just doesn't make much sense IMO though, which is what I was trying to resolve with this question. – user110391 Aug 12 '22 at 14:06
  • "Helpful to interpret" has to be used in the metalogic, not within the system. It is true that if the system is consistent and a proof of $\lnot P$ exists, then no proof of $P$ exists. It is not the case that if a proof of $\lnot P$ exists, then a proof of "no proof of $P$ exists" exists. We can conclude that the proof does not exist by assuming consistency, but we can't always prove the proof does not exist (since the system can't prove its own consistency). – aschepler Aug 12 '22 at 14:46