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I am reading my logic professor's notes, and he writes

What then is the evidence that there are no undiscovered contradictions from axioms of arithmetic? Exactly why is $0 \neq 1$ not provable in an axiomatic arithemic based on a specific list of properties expressed in a language?

This is shortly after he listed the Peano axioms. $1$ is just shorthand for $s(0)$, and we have as an axiom that $(\forall x)\neg(s(x)=0)$. So in particular, $\neg (s(0)=0)$, i.e., $1 \neq 0$. So what could my professor mean when he says $0 \neq 1$ is not provable in an axiomatic arithmetic? It is possible he has misspoken.

kccu
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    Perhaps you can ask your professor. – Michael Apr 07 '17 at 19:10
  • While I do not know what axioms you have, you might try $x \in {0,1}$ and use mod-2 arithmetic to see if your axioms are consistent in that case. So in that case $s(0)=1$ and $s(1)=0$, which seems fine as long as you remove that assumption that $s(x) \neq 0$. – Michael Apr 07 '17 at 19:14
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    The point is that there might be a long and sophisticated proof of $0 = 1$, which would be a contradiction. The question is how do we know that our axioms are consistent? That is not as obvious as it may seem. – Mark Bennet Apr 07 '17 at 19:15
  • @MarkBennet Thank you, that does seem to be the point he was trying to make. – kccu Apr 07 '17 at 19:16

2 Answers2

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Your professor misspoke. The doubt is whether "$0=1$" could somehow be proved from the axioms of arithmetic - note that this is equivalent to those axioms being inconsistent. Of course, the statement "$0\not=1$" is very easily provable from those axioms.

Noah Schweber
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    This looks good (I give plus 1, don't know why others downvote). My interpretation is a bit different though, one could ask "why do we need to include the axiom $s(x) \neq 0$, can't we just prove $s(x) \neq 0$ from the other axioms"? And I think a simple proof that "it is not provable from the other axioms" is if you define your set of numbers to be something trivial, like the set 0 alone, or the mod-2 binary field, which likely satisfies all other axioms. Overall, the professor is there and likely holds office hours, it makes sense to ask him/her directly! – Michael Apr 07 '17 at 19:24
  • @Michael That would be a very reasonable question, but I don't think it's the one the professor was mentioning: "What then is the evidence that there are no undiscovered contradictions from axioms of arithmetic?" is asking about the issue of being able to prove too many things from already given axioms, not about needing more axioms to prove new things (although of course the issue of sufficiency is a fundamental point in mathematical logic!). – Noah Schweber Apr 07 '17 at 19:27
  • I was focusing on the second sentence of the professor "Exactly why is $0\neq 1$ not provable in an axiomatic arithmetic," which suggested to me the idea of removing this as an axiom and trying to prove it as a lemma. You are convincing that, in view of the first sentence, the "$0\neq 1$" was likely a typo that should have been "$0=1$." This is kind of an interesting "max-likelihood error correction" problem on language. – Michael Apr 07 '17 at 20:09
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    @Michael Yeah, another possibility that leapt to mind was that maybe the professor meant to say "Exactly why is $0\not=1$ not disprovable in ...", which is the sort of thing I used to say before I figured out why three negatives in a row are a bad idea. – Noah Schweber Apr 07 '17 at 20:16
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$1 \not = 0$ is not the same statement as $0 \not = 1$, but that is easily fixed, assuming some standard rules of logic:

  1. $\forall x \neg s(x) = 0$ Peano Axiom 1

  2. $\quad 0 = s(0)$ (Assumption)

  3. $\quad 0 = 0$ ($=$ Intro)

  4. $\quad s(0) = 0$ ($=$ Elim 2,3)

  5. $\quad s(0) \not = 0$ ($\forall$ Elim 1) (i.e. '$1 \not = 0$')

  6. $\quad \bot$ ($\bot$ Intro 4,5)

  7. $0 \not = s(0)$ ($\neg$ Intro 2-6) (i.e. '$0 \not = 1$')

OK, so it is definitely false that '$0 \not = 1$' is not provable from the Peano Axioms.

But probably your professor meant the following: how do we know our axioms are consistent? How do we know we are not able to infer a contradiction from them (which would be the case if, e.g. '$1=0$' would be provable)?

That's a good question, and the standard answer is that we can come up with a model for the Peano axioms ... which is of course just the domain of natural numbers, together with the successor, addition , and multiplication functions as we know them. And since there is a model, that means it is impossible to derive a contradiction assuming your logical inference rules are sound (as they are for any standard proof system). So, it would be impossible to prove 1=0 or anything else like that that would lead to a contradiction.

Bram28
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    Symmetry of $\neq$ was not part of my confusion. – kccu Apr 07 '17 at 19:19
  • @kccu Understood. I updated my answer to address what I think your professor really meant – Bram28 Apr 07 '17 at 19:24
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    Why the downvote? – Bram28 Apr 07 '17 at 19:24
  • I did not downvote, but I suspect it is because your original answer did not address my question. However, is it really true that because there is a model it is impossible to derive a contradiction? I thought it was not possible to prove the consistency of arithmetic. – kccu Apr 07 '17 at 19:26
  • @kccu But we can prove the consistency of proof systems, i.e. of logic itself. So, if there is a model for a set of statements, you really cannot derive a contradiction from it (using a sound proof system). – Bram28 Apr 07 '17 at 19:27
  • @kccu Well, how did we know that model exists? Bram28's claim that the naturals form a model of PA relies on the existence of the set of natural numbers; this is a statement that needs some axioms to prove. What is true is that any (non-silly) theory which can prove the existence of the set of natural numbers proves that PA is consistent. But you're right, you can't get something from nothing - we have to prove that this strange thing called "the natural numbers" exists (and also is a model of PA, but that's usually trivial from existence). (continued) – Noah Schweber Apr 07 '17 at 19:30
  • What Bram28's argument shows it that we can conclude that PA is consistent from "basic" assumptions - it's just that one of those basic assumptions is that the set of natural numbers exists. This is certainly a basic assumption I make, and I would argue that it's a necessary one for doing mathematics, but it is implicit there. (And I would argue that the statement "PA is consistent" is not so obviously true, so it's not silly to say "PA is consistent since the set of natural numbers exists.") – Noah Schweber Apr 07 '17 at 19:31
  • @NoahSchweber Thank you for your comments! It makes me wonder: is there maybe a way to show that 1=0 cannot be derived using an argument that does not use formal semantics, and speaks to the nature of those derivations directly? ... but I am guessing any such argument would probably rely on induction? ... and thus somehow on the nature of the natural numbers after all? Do you have insights about that? – Bram28 Apr 07 '17 at 19:42
  • Wow, it is clear that there is still a lot here I don't understand. Thank you both for your comments! – kccu Apr 07 '17 at 19:45
  • @Bram28 Yes, most definitely! (Although of course that deduction itself has to take place in some background theory, so you'll never "catch your tail.") This is a very important subject in proof theory, and is essentially the beginning of ordinal analysis. The first result in this direction was due to Gentzen, who showed that the consistency of PA is provable in the weak theory PRA (a very very small fragment of PA) together with a certain transfinite induction axiom - roughly speaking, "induction along $\epsilon_0$." – Noah Schweber Apr 07 '17 at 20:09
  • The idea behind ordinal analysis is this: given a theory $T$, we analyze the structure of deductions from $T$. We first "thin" the set of deductions to "appropriately nice" ones, and show that anything provable from $T$ has a "nice" proof. It is then usually very easy to show that there are no "nice" proofs of $0=1$, so the whole meat of the argument is in showing that every proof can be made nice; this is the sort of thing cut elimination does. This step itself winds up being a transfinite induction on the "ranks" of proofs. (continued) – Noah Schweber Apr 07 '17 at 20:13
  • The supremum $\theta$ of the ranks of proofs from $T$ is then the proof-theoretic ordinal of $T$; and (if you've done everything right) the theory PRA+"induction along $\theta$" (appropriately formalized) will prove the consistency of $T$. Now by Goedel, we'll never have PRA+"induction along $\theta$" actually contained in $T$; but this is still interesting, since usually PRA+"induction along $\theta$" won't contain $T$, either. Incidentally, it's worth noting just how terribly complicated the proof-theoretic ordinals of strong theories can be! – Noah Schweber Apr 07 '17 at 20:15
  • @NoahSchweber Oh, wow, that sounds terribly complicated indeed! ... And in the end, there is still this "can't catch your tail" feeling left, isn't there? ... I guess I'll stick to my original argument: natural numbers form a model of PA ... that's a reasonable enough answer to the OP's question, right? – Bram28 Apr 07 '17 at 20:19
  • @Bram28 Yes, indeed - and you can find at various places on this site (e.g. here and here) discussion about that vicious circle (I personally think it's ultimately inescapable, there are many who disagree with me). I do think your answer was spot-on for the OP btw (and have upvoted), I just wanted to clarify based on their – Noah Schweber Apr 07 '17 at 20:25
  • later question about what is provable where. – Noah Schweber Apr 07 '17 at 20:25
  • @NoahSchweber Cool, thanks! I really appreciate you taking the time to talk about this, and thanks for the references ... and I have seen and learned from other posts from you as well, so thanks for all your contributions to the community! :) – Bram28 Apr 07 '17 at 20:29