19

Here is a paradox I have some difficulty resolving:

As far as I understand, by one of Gödel's incompleteness theorems, in a first order logic theory with Peano arithmetic, one can find some non-trivial universal closed sentences (starting with a "for all" quantifier, all variables being bound) that can be proven to be unprovable.

Consider such an unprovable universal statement of the form "For all x, P(x)". We proved that there can be no counter example of this statement, exactly because finding such a counter example would disprove the statement hence contradicting Gödel's theorem which said that this statement can not be proven nor disproven. Therefore the given statement must be true.

As one can observe, my previous paragraph is a valid sequence of arguments that explain why my considered universal statement must be true. This previous paragraph is, by the very definition of proof, a proof of the given statement. My conclusion is that either Gödel was wrong, or mathematics are inconsistent :)

What is wrong with my reasoning ? Can you explain why the second paragraph would not be a valid proof? Does it have something to do with metalanguage? Even if metalanguage is mixed with regular language, cannot all the metalanguage used here be encoded in a first order logic with Peano arithmetic, and seen as not part of a stronger theory ?

jam
  • 435
  • 1
    I'm not certain that your paragraph can be phrased within the language in question. You're looking at your logic theory from the outside-in, and when doing that you can prove things about it that it couldn't prove about itself. At least, I suspect that that's what's happening here. This is not a field where I'm very proficient. – Arthur Aug 04 '19 at 19:37
  • Yes I thought the same, but I don't see why I couldn't reiterate the same proof in such a theory that would be powerful enough to encode the metalanguage of my argument. – jam Aug 04 '19 at 19:45
  • 4
    Please see How could a statement be true without proof?. Such conflation of truth with provability is common, but would vanish once you define precisely what truth and provability are. Also, for a simplistic analogy, $∀a,b ( a·b = b·a )$ is true in an abelian group $(G,·)$ but not provable by the theory of groups (since there are groups that are not abelian), analogous to how Con(PA) is true in $\mathbb{N}$ but not provable by PA (since there are models of PA that satisfy ¬Con(PA)). – user21820 Aug 05 '19 at 04:31
  • @Arthur: Your suspicion is more or less correct. And if you're interested in a self-contained proof of the core of the incompleteness theorems, you can look at this post. One still needs Godel's β-lemma if one wants to apply it to arithmetic theories. In general, any formal system $T$ that has a proof verifier and can reason about programs but does not prove that $0=1$ can prove "There is no proof of length $n$ over $T$ that $0=1$" for each fixed $n$ of the form "$1+1+\cdots+1$", but cannot prove "There is no proof over $T$ that $0=1$"! – user21820 Aug 05 '19 at 04:54

8 Answers8

18

The key step in your argument is when you say that if there is a counterexample to $\forall x.P(x)$, then the theory $T$ we speak about can prove that it is in fact a counterexample.

The particular form of $P(x)$ that comes out of Gödel's construction has the property that whenever we have a particular numeral $\bar n$, then the formula $P(\bar n)$ will always be true, and $T$ proves it to be true.

However, that does not constitute a proof of $\forall x.P(x)$. To think it does invokes a hidden assumption that everything our $\forall x$ ranges over can be expressed as a numeral. For sure this is true about our intended interpretation of $T$ as the "actual natural numbers", but in order for $T$ to prove something it needs not only to be true in the intended interpretation, but also in every other interpretation that satisfies the axioms of $T$.

Here the reasoning runs into trouble, because we know that $T$ must have some models where some of the elements don't correspond to numerals. (This is so independently of the incompleteness theorem; the "usual compactness argument" shows that such a model must exist for any theory that can speak about the naturals). Therefore, the known fact that $P$ is true and provable about all numerals doesn't imply that $\forall x.P(x)$ is true in all models, and therefore we can't conclude that it ought to be provable. And so the entire argument falls apart.

Gödel's notion of an "$\omega$-consistent" theory requires that there is no formula $\varphi(x)$ such that $T$ both proves $\varphi(\bar n)$ about every numeral, and proves $\exists x.\neg\varphi(x)$ which is the same as $\neg \forall x.\varphi(x)$. One way to show that a $T$ is $\omega$-consistent is to show that it has some model where all elements are numerals; interpreted in such a model the Gödel sentence will be true. The original incompleteness proof explicitly assumed that $T$ is $\omega$-consistent and used that to argue that $T$ cannot prove $\neg\forall x.P(x)$. Later, Rosser found a way to avoid this assumption and instead just assume that $T$ is consistent.

hmakholm left over Monica
  • 286,031
9

The thing you're missing is that "true" and "false" are always relative to a model, but in arithmetic they are relative to the standard model, unless stated otherwise.

Now, if there is a counterexample to a $\Pi_1$ statement, then it is a witness to its negation, a $\Sigma_1$ statement. But here's the thing, $\sf PA$ is $\Sigma_1$-complete: every $\Sigma_1$ statement which is true in $\Bbb N$ is in fact provable from $\sf PA$.

Now, if $\forall x\varphi(x)$ is a $\Pi_1$ statement which is neither provable nor disprovable from $\sf PA$, that means that it is necessarily the case that $\exists x\lnot\varphi(x)$ is false (in $\Bbb N$), otherwise, it would be provable. Therefore $\forall x\varphi(x)$ is true (in $\Bbb N$).

Asaf Karagila
  • 393,674
  • "Now, if $\forall x\varphi(x)$ is a $\Pi_1$ statement which is neither provable nor disprovable from $\sf PA$, that means that it is necessarily the case that $\exists x\lnot\varphi(x)$ is false (in $\Bbb N$), otherwise, it would be provable. Therefore $\forall x\varphi(x)$ is true (in $\Bbb N$)."

    How is that what you just said not a proof by itself of $\forall x\varphi(x)$ ? Thats my whole question here. Its a valid sequence of human language arguments that end up concluding the truth of the sentence.

     – jam Aug 04 '19 at 20:11
  • 1
    It's not a proof of $\forall x\varphi(x)$. It is a proof that $\Bbb N\models\forall x\varphi(x)$. Since if there was $n$ such that $\Bbb N\models\lnot\varphi[n]$, then $\exists x\lnot\varphi(x)$ is a true $\Sigma_1$ sentence which is therefore provable from $\sf PA$, so $\forall x\varphi(x)$ is disprovable. But we know that it is not disprovable. – Asaf Karagila Aug 04 '19 at 20:33
  • I think you are putting the finger on the problem. Do you think you could include that in your answer and explain it in more thorough details so that someone unfamiliar with these matters could understand? I believe understanding this paradox is essential to understand incompleteness. – jam Aug 04 '19 at 20:40
  • 1
    Sure, but can you be more explicit regarding to what part? I'm pretty sure all of the things I've left out have been thoroughly covered in the past by more competent writers on this topic, so at the very least, if you focus the problem I can write a bit and find a link. – Asaf Karagila Aug 04 '19 at 20:42
  • I was referring to your second comment. I would like to understand clearly why as you said it is not a proof of $\forall x\varphi(x)$ . Imagine you are not familiar with this field. Its not at all obvious why what I said would be wrong. Now maybe there is probably a philosophical problem going on here. Perhaps the standard ways of modeling mathematics rules and axioms, and modeling truth artificially made my argument invalid, and godels incompleteness theorem true. Perhaps if we modeled mathematics fundamental concepts in another way, my argument would hold. Who knows? – jam Aug 04 '19 at 20:48
  • Because $\Sigma_1$-completeness, which is the key tool for this claim, is not about all models of PA, just the standard model, which is pretty unique and different from the rest of them. – Asaf Karagila Aug 04 '19 at 20:49
  • I think my problem is failure to clearly distinguish between truth and provability. And confusion about what the universal quantifier is meant to represent. I dont conceive how a universal statement could be both true and unprovable. – jam Aug 04 '19 at 21:09
  • 1
    It is *true in the standard model, so it is true about numerals*. But there are non-standard models, where there are elements which are not of the form $S\dots S0$. – Asaf Karagila Aug 04 '19 at 21:10
  • It's more that I philosophically do not want to distinguish truth and provability regarding universal statements somehow. In my model, A universal statement that is true, should by itself be a proof of itself. – jam Aug 04 '19 at 21:11
  • 3
    That's just saying "Philosophically, I want to be wrong". I can't help you with that... :-) – Asaf Karagila Aug 04 '19 at 21:12
  • I think you dismiss my point promptly. All models are constructed by humans and completely arbitrary. I wonder if someone could someday design a model of predicate logic with universal statement truth=provability as a basic assumption. Perhaps that would prevent odd theorems like godel ones from happening. But it might cause other problems to occur, Im not sure. – jam Aug 04 '19 at 21:16
  • 2
    I am not dismissing your point. You are trying to understand the incompleteness theorem. I pointed out your mistake. What you want and the way mathematics is are two independent forces of the world. – Asaf Karagila Aug 04 '19 at 21:17
  • I think I just don't know what truth means in math. I dont see a difference between proven statement and true statement. Thats the problem. – jam Aug 04 '19 at 21:34
  • Truth is relative to a structure. Normally you would quality that $\psi$ is true *in $M$. In the case of arithmetic (and a few other select cases), we have standard and very canonical models, so we measure truth relative to these models. As I wrote in my answer. In the case of PA, this is $\Bbb N$. So "true" is really "true in $\Bbb N$". But provable mean "true in all* models", not just in $\Bbb N$". – Asaf Karagila Aug 04 '19 at 21:35
  • 1
    I find this answer a bit dissatisfying (and possibly OP does too) for the following reason: I think the question can be stated only in terms of provability, without reference to models or truth. Moreover provability seems simpler to define. So, it feels like an apparent contradiction involving elementary concepts is resolved only by introducing more complex machinery. If we replace "true" by "provable" in the question, is it possible to give an answer only in terms of provability (probably of the form "once written out carefully, the conclusion is provability in a stronger theory")? – stewbasic Aug 05 '19 at 06:27
  • 2
    @stewbasic: I don't quite see how you can phrase the whole question in terms of provability. The whole question seems to me as a long way of asking "Why does a non-disprovable $\Pi_1$ sentence have to be true?". Truth gets in there, by hook or by crook. – Asaf Karagila Aug 05 '19 at 06:38
5

one can find some non-trivial universal closed sentences (starting with a "for all" quantifier, all variables being bound) that can be proven to be unprovable.

Specifically, given a logic $L_1$, you can construct (algorithmically) a statement $P(x)$ such that there is a proof of $P(0)$ and there is a proof of $P(1)$ and there is a proof of $P(2)$, etc, but there is no proof of $\forall k~P(k)$ (and all of that assumes consistency).

We proved that there can be no counter example of this statement

We have not formally proved this. There is no formal proof in $L_1$ that $\lnot \exists k ~ \lnot P(k)$. To show $\lnot \exists k ~ \lnot P(k)$, we actually need to use the assumption that $P(k)$ represents the statement "$k \text{ is not a proof of } G$" (for a specially chosen $G$). But that assumption isn't actually provable in $L_1$, it is just something we know to be true because it was designed that way.

My conclusion is that either Gödel was wrong, or mathematics are inconsistent

Godel wasn't wrong, nor is the logic the theorem deals with. The logic is just incomplete. And more importantly, it is incompletable.

Even if metalanguage is mixed with regular language, cannot all the metalanguage used here be encoded in a first order logic with Peano arithmetic, and seen as not part of a stronger theory ?

Yes, and there is huge branch of logic that deals with this approach. It is called provability logic: https://plato.stanford.edu/entries/logic-provability/ . But, even after you encode all the outer logic and use it as your new logic, then you've just created a new logic $L_2$, from which a new $P$ can be algorithmically constructed. No matter how much "meta" you keep adding, a new $P$ can be created.

I suggest reading through https://plato.stanford.edu/entries/goedel-incompleteness/ , it is an absolutely fantastic reference for Godel's first incompleteness theorem (but an unfortunately deficient reference for the second, that requires a bit more framework than the article presents).

DanielV
  • 23,556
  • Thanks for this detailed analysis, I should have use more caution indeed and not say that we have proven that there are no counter examples. Just that the fact that there are no counter examples is true. – jam Aug 04 '19 at 22:45
2

Some statements couldn't be refuted by presenting a counterexample, because that counterexample would itself be a universal claim. A counterexample to "all men are mortal" would be "nothing can kill Bob". In mathematics, it would be more like, "for any $x\in S$, there is some $y\in T$ such that..."

Let's discuss a concrete example. Write your favourite positive integer as a sum of powers of two, where the exponents are written in the same format etc. so no number $>2$ appears, e.g. $37$ becomes $2^{2^2+1}+2^2+1$. Now replace every $2$ with a $3$ and subtract one, viz. $3^{3^3+1}+3^3$. Now repeat moving $3$s to $4$s, viz. $4^{4^4+1}+3\times 4^3+3\times 4^2+3\times 4+3$. The numbers grow very fast at first, but it can be shown in ZF, a version of set theory, that eventually you'll get to $0$. On the other hand, it can also be shown that the Peano axioms, the weakest system Gödel considered in his incompleteness theorems, can't prove this result.

J.G.
  • 115,835
  • Thats why I assumed that the structure of the statement I am considering in my post is "for all x such that P(x)", there must be such unprovable statement, because given any unprovable statement, both itself and its negation are unprovable, if its closed it starts with for all/there exists, so either itself or its negation starts with a for all. – jam Aug 04 '19 at 19:41
  • @J.M. You just have the wrong kind of statement in mind if you're looking to understand example of incompleteness. Fermat's last theorem, had it been false, would have been decidable, since a counterexample could then be presented. This was why mathematicians knew, with your reasoning, that if it were undecidable they'd never even be able to prove that. But this only works because the claim takes the right form for falsity to imply decidability. – J.G. Aug 04 '19 at 19:44
  • I'm not sure why you are saying I am wrong. I am just making an assumption. Do you think my assumption is false? why do you think so? I believe there are unprovable statement of the form "for all x, P(x)". Unfortunately I don't really understand your point, or how it disproves what I said. Its not detailed enough for me. I'm not that familiar with foundations. – jam Aug 04 '19 at 19:52
  • This doesn't seem to at all address the question being asked. And even if it did, someone having trouble with GIT isn't going to even remotely understand Goodstein's theorem. – DanielV Aug 04 '19 at 22:26
  • @DanielIV It addressed the question when it was posted, but it's been edited since then in a way that makes the addressing less obvious. It's still applicable, because $P(x)$ can be of the form $\exists yQ(x,,y)$. The original version of this answer only included the first paragraph, but I thought to add an explicit example of this $\forall x\exists yQ(x,,y)$ structure. The point I'm trying to make is comprehensible from the first paragraph alone, but the second paragraph links to all the material needed to understand something specific. – J.G. Aug 05 '19 at 05:13
2

What one can prove, in PA (and in fact in even weaker arithmetical theories than that) via a careful version of the argument you give is "If PA is consistent, then there is a statement $G$ such that $G$ holds and PA cannot prove $G$." If PA could prove its own consistency, then it would prove both $G$ and that PA cannot prove $G.$ Thus the latter is a false statement and PA is unsound. Thus if PA is sound, then it cannot prove its own consistency. This is essentially a proof of (a weak form of) the second incompleteness theorem from the first.

spaceisdarkgreen
  • 58,508
  • I think this is not about this particular example, but rather about the underlying idea of "Independent $\Pi_1$ is true". – Asaf Karagila Aug 04 '19 at 20:58
  • 1
    @AsafKaragila But my point is that we can formalize all that in PA. $Con(PA)\to G\land \lnot Prov(\ulcorner G\urcorner)$ is a theorem of PA, and the reasoning within PA that $G$ follows from $Con(PA)$ is exactly that a witness to $\lnot G$ would allow a PA proof of $Prov(\ulcorner G\urcorner)$ which contradicts $\lnot Prov(\ulcorner G\urcorner)$ (which holds, and produces a contradiction under the assumption of $Con(PA)$). So OP is right that the argument works in a PA metatheory, they just forgot about the consistency part (and are confused about the metatheory part). – spaceisdarkgreen Aug 04 '19 at 21:45
0

I think people here understand the flaw in my argument so I will try to make it sound simple. To try to sum it up, I think the problem with my second paragraph, was that although it is indeed a proof of the universal statement, it is a proof assuming the statement itself, rather than just the principles of the implicit theory. It is perfectly circular in its form, it does not mean that the theory proves the statement. Just that the statement proves itself by virtue of it being true.

As one can notice "proof" is a relative notion. Unlike what I failed to realize early enough, one should always make the mental effort to ask oneself: proven with respect to what?

jam
  • 435
  • It's easy to throw a pebble into a well, but it takes sophistication to remove it. – Asaf Karagila Aug 04 '19 at 22:51
  • 2
    This is very wrong. Neither GIT nor Godel's sentence is a circular claim that assumes itself to be true. If you are actually interested in the subject then you are going to have to do the hard work of learning the proof of GIT. There is no royal road to geometry. – DanielV Aug 04 '19 at 23:32
  • 1
    I think there is some misunderstanding. In this answer I am referring to my own claim "Consider such an unprovable universal statement [...] Therefore the given statement must be true.". What I meant by this whole paragraph actually, is that "For all x, P(x)" is a proof of "For all x, P(x)". I just initially said that it was "a proof" period. I failed to notice that it was under the assumption of "For all x, P(x)" itself – jam Aug 04 '19 at 23:40
  • It's really a dumb mistake and I'm considering deleting my question. But its funny to watch all these experts arguing on the various possible interpretations of my post. – jam Aug 04 '19 at 23:47
  • 1
    We are not arguing. We just offer different explanations... And you can't delete your question once it has two answers or more. – Asaf Karagila Aug 05 '19 at 06:13
-2

Your argument breaks down when you reason, that there is no counter example. There may be very well counter examples, but all we can reason is, that we are unable to construct one of them.

aqwewteerhz
  • 1
-2

Your mistake is already in the first sentence:

As far as I understand, by one of Gödel's incompleteness theorems, in a first order logic theory with Peano arithmetic, one can find some non-trivial universal closed sentences (starting with a "for all" quantifier, all variables being bound) that can be proven to be unprovable.

Here you state "one can find" when Gödel's theorem instead is "one can prove the existence". If you could "find" it in some mathematically robust manner, it would be proven.

  • 2
    The incompleteness theorem gives a completely concrete way to construct the Gödel sentence for a theory. It is perfectly feasible to write it down explicitly -- it would probably fill several pages, but it wouldn't take hundreds of pages. – hmakholm left over Monica Aug 05 '19 at 10:30
  • 1
    (You would need a few optimizations relative to pure textbook presentations of the theorem, which generally assume, e.g., that you can just write down hundred-digit numbers in unary notation when you need to. But that doesn't affect the core of the reasoning). – hmakholm left over Monica Aug 05 '19 at 10:48