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I know no formal system containing arithmetic can prove itself consistent. But is it possible to add some ungrounded axiom in the sense described here https://en.wikipedia.org/wiki/Liar_paradox#Saul_Kripke such that the new system can somehow prove the part of itself which concerns all grounded statements, complete? In other words, can we axiomatise some concept whose only bearing on arithmetic is to facilitate the completion of the part of our new theorem that concerns itself with evaluable facts about the numbers.

By this I mean for example, we might start with Peano axioms and "incompletely" define some "arbitrarily large integer" $a$ axiomatically by the axiom $a\geq n\forall n\in\mathbb{N}$ with some suitable definition of what $\geq$ means, or some other suitable axiom.

Is there such an axiom that would render our theory "Peano+" complete in respect of all meaningful statements about any number we might actually encounter, and allows us to ring-fence in some way the part of the theory that is incomplete or possibly inconsistent, and say for certain that the part of the theory that is possibly incomplete or inconsistent is necessarily only concerned with ungrounded statements, i.e. statements about $a$?

In the example of the axiom above, we might say the theorem is "complete for all statements not mentioning $a$." Which would surely be a desirable outcome since by definition, any $n=a$ would be the greatest integer, and the axioms of Peano guarantee there is no greatest integer, so no integer we will ever meet will ever be $a$. Godel's incompleteness theorem would therefore be rendered irrelevant because our new theory would only be incomplete insofar as it would fail to make statements which are not concerned with evaluable facts about the numbers. Every evaluable fact about the numbers, it would be able to prove.

My gut feel is that such an axiom is possible, but that the grounded statements might exclude certain things we can currently say about transcendental numbers and possibly irrational numbers too.

it's a hire car baby
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Leaving aside a number of issues (which also applied to your previous question on illogicality) around foundational matters, there is arguably a sense in which the answer to your question is yes: if we pass to an appropriate theory which can talk about the set of all natural numbers - for example, $ZFC$! - then such a theory can indeed prove that $PA$ is consistent (although of course "appropriate" here is subsuming a lot . . .). Note that Godel proved a general theorem about the value of passing to a higher-type theory - see his speed-up theorem - which, while not directly relevant to what you're talking about, seems like it would be interesting to you.

That said, I think your philosophical musings are getting in the way of asking a well-posed question. I strongly recommend that you read a book on first-order logic; I think that will both clarify many issues (especially around Godel's theorem), and also help you put your questions into a precise form that lets them be answered well. In particular, re: this question, note that the passage to ZFC looks nothing like your "arbitrarily large integer $a$", and I have absolutely no idea what you mean by "incompletely define"; moreover, note that it is impossible for any recursive theory to be complete for even the set of sentences asserting that Diophantine equations have/don't have solutions, by the MRDP theorem.

Noah Schweber
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  • Indeed, though I doubt mere consistency is what the asker is hoping to have, judging from the quote Every evaluable fact about the numbers, it would be able to prove. which suggests that he wants some kind of completeness at least for arithmetical sentence. In the case of ZFC, it does prove the arithmetical sentence Con(PA), but still cannot show that itself can prove or disprove every arithmetical sentence unless ZFC is Σ1-unsound, which is what my linked post implies. And if you have time, I would greatly appreciate you checking that post. =) – user21820 Aug 21 '16 at 04:09
  • Thanks. I also have very little idea what I meant by "incompletely define"! Also thanks for the reference re the MRDP theorem, which is very interesting but moves the question for me, to what the implications are for MRDP theorem, of the possibility that some diophantine equations might have infinitely many solutions. – it's a hire car baby Aug 22 '16 at 08:30
  • Ok I see, you would think a countably infinite set of them could be listed, so it's paradoxical that they're a subset of the integers but the means to find them can be uncountable in its steps. – it's a hire car baby Aug 22 '16 at 08:42
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I know no formal system containing arithmetic can prove itself complete. $ \def\nn{\mathbb{N}} \def\t{\text} \def\th{\t{Th}} \def\con{\t{Con}} \def\prov{\square} \def\pa{\t{PA}} $

Wrong. Any inconsistent formal system that can state completeness of itself proves itself complete. [Edit: In my original post I made a stupid careless mistake and said that $\th(\nn)$ proves its own completeness. That cannot be expressed in any meaningful sense because there is no decidable proof validity for $\th(\nn)$, so it does not. What is true is that $\th(\nn)$ is complete but useless for that very reason.]

... "incompletely" define ...

Meaningless. It is akin to asking "Can we incompletely define a flying pig?". Yes we can: A flying pig is some pig that somehow flies.

Godel's incompleteness theorem would therefore be rendered irrelevant

Wrong. Godel's incompleteness theorem is not rendered irrelevant on ill-defined things.

Every evaluable fact about the numbers, it would be able to prove.

Meaningless. Your post has nowhere defined the term "evaluable" and hence it is meaningless. This applies to most of the other terms used in your post as well.

My gut feel ...

is wrong yet again (and I'm talking about your underlying idea, which is terribly distorted in your post due to your lack of understanding of logic). There is no formal system with decidable proof validity that interprets PA and can prove (the translation of) every sentence in the language of arithmetic that is satisfied by $\nn$. See this post for a precise statement and proof. (In the simplest layman terms, absolutely no practical formal system can prove every fact about even just the natural numbers; it is simply impossible to 'push the incompleteness away' from statements that are purely about arithmetic!)

Furthermore, it is absolutely unjustifiable to reject the meaningfulness of unprovable arithmetic sentences if you accept arithmetic in the first place. For any such formal system $S$, the unprovable sentence constructed in the linked post is a $Π_1$-sentence of the form $\forall x\ ( P(x) )$ where $S$ proves the translation of $P(c(n))$ for every $n \in \mathbb{N}$, where $c(n)$ is the term coding for $n$. Note that each such $P(c(n))$ is a sentence with only bounded quantifiers, and hence can be deterministically checked by brute-force, but the universal sentence cannot be proven!

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user21820
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  • Thank-you for your generous answer. Ironically definition of the 'flying pig' is a deep philosophical question debated by some of the greatest logicians. Whether 'evaluable' has meaning, I'm going to trust Saul Kripke the distinguished professor of philosophy at new York university in thinking it is a meaningful statement, as mentioned in the reference provided. Re incompletely define, where there's a will there's a way. But I think your tl;dr is 'no'. – it's a hire car baby Aug 18 '16 at 18:28
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    @RobertFrost: Yes, the short story is that it is absolutely impossible since my linked post proves the incompleteness theorem for every possible formal system that can be used, not just the ones that we know of now! This includes higher-order logic, non-classical logic, and everything else. Kripke's notion is well-defined, but unfortunately yours isn't. I'm familiar enough with his notion of "groundedness" and in fact I proposed a similar one (but precisely defined unlike on Wikipedia) to your earlier question at http://math.stackexchange.com/a/1888389/21820! Didn't you read it? – user21820 Aug 19 '16 at 01:28
  • @RobertFrost: The point is that that notion does nothing at all to solve the incompleteness, and provably cannot as my linked post shows. And I'm curious why you've not accepted my answers to your earlier question about the modified liar paradox... – user21820 Aug 19 '16 at 01:30
  • @RobertFrost: Note that I said "provably" not "probably" in case you're wondering whether there is a typographical error. There isn't. Also, I'll note that the only way out of incompleteness is to reject the real-world existence of the conceptual collection of natural numbers, but that also implies you must reject the existence of the conceptual collection of strings (with the usual string manipulation operations). Unfortunately, doing so will mean that you reject all known mathematics altogether, because all mathematics depends on proofs which can be strings of arbitrary length. – user21820 Aug 19 '16 at 01:36
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    @RobertFrost: Lastly, concerning "flying pig" I'll have to remind you once more that mathematical logic is quite separate from philosophy. Logic is precise, made possible via programs (as I previously told you), while philosophy in general is as clear as mud and hence obviously you're going to get rather meaningless debates that go in circles due to inability to even state definitions, not to say agree upon them. – user21820 Aug 19 '16 at 01:38
  • I knew you meant provably. But note I'm not proposing a way out of incompleteness, nor suggesting we reject the "concept" of strings of infinite length. I'm asking whether we can construct a theory such that the part of it that is grounded has all the properties we want, so all the things that are unprovable don't matter. I accept that is an imprecise statement but it is meaningful nevertheless. Fair enough re flying pigs. It seemed like your answer was an ad hominem attack. Note I think all definable strings of infinite length are countably infinite aren't they? – it's a hire car baby Aug 19 '16 at 08:08
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    @RobertFrost: I didn't say "strings of infinite length". The two words "infinite" and "arbitrary" are completely different. As I already said, my linked post shows that you can't have what you want ("all unprovable sentences don't matter"), which you will never be able to understand without learning logic. None of my answers are ad hominem but are based on facts. If you think any sentence is an attack on your person, point it out and I will reconsider. You must realize that when you ask a question that is very ill-formed or mis-informed, the right answer is to point out the misconceptions. – user21820 Aug 20 '16 at 00:42
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    @RobertFrost: But I still cannot understand is why you want to waste time (which I can assure you is a correct judgement) instead of just going to learn programming and logic, both of which will clarify a lot of things. If you refuse to learn programming, understanding logic will be like trying to chop firewood with bare hands; most logic textbooks end up building their own axe (a programming language) in the form of the notion of computable functions, but if you don't know programming it will be painful. – user21820 Aug 20 '16 at 01:03
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    @RobertFrost: The sentences I marked as "wrong" are indeed wrong. You cannot make a mathematically false statement and expect me not to say they are wrong! As for meaningfulness, I already said so many times that until you learn (programming and) logic you are not going to understand your serious lack of understanding. Not once did I say you have to express what you say in a formal language, but it must be clear how to convert what you say into one otherwise it remains meaningless. You don't understand this because you don't (yet) know the true nature of logic. – user21820 Aug 20 '16 at 12:24
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    @user21820 Your statement "$Th(\mathbb{N})$ proves itself complete" is false: even though it is complete, $Th(\mathbb{N})$ cannot talk about itself, since it's not arithmetically definable. The rest of your answer is correct though. – Noah Schweber Aug 20 '16 at 20:02
  • @NoahSchweber: I'm not sure what you mean by that. When I say "complete" I mean "syntactically complete", so when I say $S$ proves itself complete I meant that it proves (the translation of) $\forall n\ ( \square φ_n \lor \square \neg φ_n )$ where $φ$ is an enumeration of all sentences over S. The theory of natural numbers contains PA and hence can capture such an enumeration, and so it trivially proves such a sentence. Did you have some other notion of completeness in mind? – user21820 Aug 21 '16 at 03:38
  • @user21820 $Th(\mathbb{N})$ is complete and consistent, but it can't prove that it is complete and consistent because the language of arithmetic can't express those facts: "$Th(\mathbb{N})$" is not arithmetically definable! Think about what "$\Box$" really means in the sentence you wrote above (and this is why one should really write "$\Box_T$" instead of just "$\Box$"). You need to be able to arithmetically define a theory in order to write a sentence expressing provability in that theory, and $Th(\mathbb{N})$ is not, again, arithmetically definable. – Noah Schweber Aug 21 '16 at 06:35
  • @NoahSchweber: Oh okay sorry sorry. A stupid mistake. I myself said in another of my own posts that one should write the subscript... I've fixed my mistake in my answer so I'm going to remove my comments to you, okay? – user21820 Aug 21 '16 at 06:45
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    @RobertFrost Regarding user21820's remark that some of your sentence are "meaningless", it could be rephrased to "ambiguous" if you'd like to. For example, when you said that certain things are "evaluable" you gave no criterion of being evaluable hence the word is meaningless in that sense. To give a meaning to it is not to formulate it in a formal system, but to make the word unambiguous ("has a concrete representation in our physical world" might cut it, but "grounded" wouldn't). – BigbearZzz Aug 21 '16 at 23:04
  • @BigbearZzz: Fully agreed, though there is a slight catch with things like "has a real-world meaning". If one takes for granted that natural numbers exist in reality in some form, then all arithmetical sentences would be evaluable (and this is the sense I assume Robert is taking, since he wants to add to PA, and there is no reason for PA unless it is to axiomatize the natural numbers). In that case the incompleteness theorems hold in the strong sense I've proven. That said, it may be that the real world does not have a structure isomorphic to the natural numbers, and so PA is unphysical... – user21820 Aug 22 '16 at 04:57
  • @BigbearZzz: By the way, have you heard of Dan Willard's theories that are like PA but have subtraction and integer division instead of addition and multiplication, and hence are weaker than PA but still strong enough to state and prove its own consistency? We of course still need a meta-system that has the natural numbers to be able to observe that its statement of its consistency actually corresponds to its consistency, so the philosophical problem does not go away. But still it is interesting. – user21820 Aug 22 '16 at 05:00
  • @BigbearZzz If you think of a projection as a reduction in the information contained within something, then all mathematical concepts are projections of the tangible world. So if we have a tree with 500 apples on it, the number 500 is a certain projection of the apples. The number 1 can be looked upon as a projection of a single apple. I would for now define an ungrounded statement as a statement for which no projection from the real world exists. – it's a hire car baby Aug 22 '16 at 08:16
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    @RobertFrost: As I said already, if you accept the meaningfulness of even a single quantifier over the natural numbers, you're doomed to incompleteness. But if you don't accept that, then you need some very good explanation of why PA seems to predict accurately theorems that are used in the real world such as Fermat's little theorem underlying RSA. – user21820 Aug 22 '16 at 08:34
  • @user21820 Thank you for sharing! I've heard of the idea but never knew that it belongs to Dan Willard. The situation is indeed very interesting. – BigbearZzz Aug 22 '16 at 18:01