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While preparing some lecture notes for next semester and going back to basics (set theory and proof strategies) I came along the following simple question which is about proving theorems in general but exemplified here by the proof by contradiction:

How do you see whether a proof is not only the necessary but also sufficient condition to arive at a "q.e.d"?

As an example let's take the simple example of irrationality of $\sqrt{2}$. Here you suppose that the result will be rational and arrive at a contradiction, so you say it must be irrational because there are only those two possibilities. q.e.d - case closed.

When you look at the Zermelo-Russell paradox you have a similar situation: Two cases which are mutually exclusive: Either $R \in R$ or $R \not \in R$. You then suppose e.g. that $R \in R$ and arrive at a contradiction... but you don't stop there! Otherwise it would not be a paradox! You also test the other case and, again, arrive at a contradiction!

It would not be enough to stop after the first part and arrive at the result that because you arrived at a contradiction it must be the other case, so $R \not \in R$ q.e.d.

So my question is how do you ensure that you can trust your proof and haven't run into another paradox? One possibility where something like this could happen are obviously situations that are self-referencing. But are these the only possibilites? And some proofs are quite long and involved so that even this could slip through, can it not?

Full disclosure: I asked this question on mathoverflow but it got closed (although upvoted!) so I guess it was not sophisticated enough for that forum - yet I am still looking for a satisfying answer...

Andrés E. Caicedo
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vonjd
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    For proof by contradiction there are always only two choices: a statement and its negation. If you assume something and arrive at a contradiction, then you have proved its negation. If you are working in a consistent theory, then that is all you have to do. – Tim Seguine Jan 24 '14 at 17:54
  • @TimSeguine: But there still remains the possibility that when you assume the negation and again arrive at a contradiction that your axiom system is faulty (see Zermelo-Russell paradox), so the question remains: How to ensure that you haven't run into a paradox proving a theorem? – vonjd Jan 24 '14 at 17:58
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    The working assumption is that ZFC is consistent. Questioning that is relatively pointless due to Gödel's second incompleteness theorem. – Tim Seguine Jan 24 '14 at 18:05
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    And your question got closed because this is not research level mathematics. – Tim Seguine Jan 24 '14 at 18:08
  • @Tim: $\sf ZFC$ is consistent. The question is whether or not inaccessible cardinals are consistent. Well, to be fair, inaccessible cardinals are consistent. The question is whether or not measurable cardinals are consistent. Well, to be fair, measurable cardinals are consistent. The question is whether or not strong cardinals are consistent. Well, to be fair, strong cardinals are consistent. The question is whether or not Woodin cardinals are consistent. Well, to be fair, Woodin cardinals are consistent. The question is whether or not supercompact are consistent, and we're not sure yet. – Asaf Karagila Jan 24 '14 at 18:55
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    @AsafKaragila Has ZFC been proven to be consistent? – Dan Christensen Jan 24 '14 at 19:53
  • @Dan: Yes. $\sf ZFC$ is consistent. If we work with $\sf ZFC+\exists\kappa\text{ inaccessible}$. – Asaf Karagila Jan 24 '14 at 20:23
  • @AsafKaragila Somehow, that's not very comforting. Whatever happened to the notion that axioms should be self-evident? – Dan Christensen Jan 24 '14 at 21:06
  • @Dan: Is $\sf PA$ not self evident? Can it prove its own consistency? No, we use $\sf ZFC$ to prove that. Large cardinals seem very natural and there are plenty of reasons to accept them. Lots of people work with "universes" which are essentially equivalent to asserting the existence of inaccessible cardinals. – Asaf Karagila Jan 24 '14 at 21:15
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    Sounds like fun... I guess. Has anyone ever returned from one of these universes? – Dan Christensen Jan 24 '14 at 21:20
  • @Dan: I don't think that people in category theory appreciate you mocking one of the fundamental ideas for the foundations of category theory. – Asaf Karagila Jan 24 '14 at 21:41
  • Sorry. Just having a little fun on a Friday afternoon. Anyway, how does one go about proving the consistency of this beast? (In a sentence or two.) – Dan Christensen Jan 24 '14 at 21:47
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    Besides which, naive set theory seemed pretty self-evident, and look where that got it. Self-evidence ain't what it used to be. – Malice Vidrine Jan 24 '14 at 22:25
  • @MaliceVidrine "Self-evident" has on occasion led us astray, it is true. Not self-evident, on the other hand should raise alarm bells. – Dan Christensen Jan 25 '14 at 03:36
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    @Dan: Personally, I strongly disagree; counterintuitiveness is a pretty poor litmus test for whether something is true or false, also. I trust trial and error and careful deliberation more than "self-evident" first principles. – Malice Vidrine Jan 25 '14 at 04:01
  • @Dan: It was not self evident that bacteria exists and causes illness; it was not self evident that vacuum is a real thing, rather than some "ether"; it was not self evident that light can be bent by gravity. It was self evident that the plague was caused by birds, Jews, and people with acne, and astrological occurrence. Yes, I agree that the things which are not self-evident are "alarming", much more than those which are. Rationalization have no place in mathematics, it has place in numerology. – Asaf Karagila Jan 25 '14 at 12:01
  • @Dan: "Self-evident" means "known to be true by understanding its meaning". "Not self-evident", therefore, means "known to be false by understanding its meaning... *or* its meaning is not understood". –  Jan 25 '14 at 12:40
  • If you want to play around with strange, counter-intuitive axioms just to see where they take you, there is nothing to stop you, I guess. As I see it, however, the idea of an axiomatic system is provide a starting point for your arguments that most, if not all people can agree on. – Dan Christensen Jan 25 '14 at 15:21
  • @AsafKaragila anything we have to assume in order to prove the consistency of ZFC just leads us in circles. The consistency ends up depending upon the consistency of a theory we can't prove the consistency of. Or is the independence of large cardinal axioms from ZFC a loophole? – Tim Seguine Jan 25 '14 at 15:34
  • @TimSeguine I agree -- it's an infinite regress. Realistically, the best we can hope for is that there are no inconsistencies in our rules and axioms. If you want absolute certainty, try theology. – Dan Christensen Jan 25 '14 at 15:48
  • @Dan: Why is $\sf PA$ or $\sf PA_2$ consistent? Self-evidence is not an argument. If you would ever sit down to study set theory, and large cardinals, you will find out that the large cardinal axioms are very self-evident and make a lot of sense as additional assumptions. But you choose to criticize those instead. – Asaf Karagila Jan 25 '14 at 16:51
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    @DanChristensen I don't quite get where you are headed with your comments. You seem to be simultaneously agreeing with and disagreeing with everyone at the same time. – Tim Seguine Jan 25 '14 at 19:05
  • @AsafKaragila I never said PA or PA2 was consistent. To date, however, no inconsistencies have been found. Likewise for ZFC. On large cardinals, when I see stuff at Wikipedia like "A large cardinal axiom is an axiom stating that there exists a cardinal (or perhaps many of them) with some specified large cardinal property. There is no generally agreed precise definition of what a large cardinal property is, though essentially everyone agrees that those in the List of large cardinal properties are large cardinal properties," I can only shake my head. I don't handle that kind ambiguity very well. – Dan Christensen Jan 25 '14 at 20:48
  • @Dan: What is a number? What is a sieve? If you expect everything in mathematics to be perfectly well-defined without any ambiguous notions, then you're probably going to be thoroughly disappointed. It doesn't work like that. – Asaf Karagila Jan 25 '14 at 20:51
  • @TimSeguine Many would beg to differ. – Dan Christensen Jan 25 '14 at 20:52
  • @AsafKaragila I think I now have a pretty good handle on what numbers are. There is nothing ambiguous about them. That said, ambiguity is initially necessary with any new idea. IMHO however, by the time you get to publishing proofs on a topic, there should be no ambiguity. Is that too high a standard? – Dan Christensen Jan 25 '14 at 21:04
  • @DanChristensen The type of cardinal that is needed to prove ZFC consistency(IIRC) is one that is a limit of repeated powerset operations. That sounds pretty reasonable to me. And I also don't see how it is ambiguous. – Tim Seguine Jan 25 '14 at 21:05
  • @Dan: You have a pretty good handle? Oh that's great. But what if we both try to explain to someone what are numbers, and we disagree on that handle? Or did someone appoint you high judge of mathematical ambiguity lately? – Asaf Karagila Jan 25 '14 at 21:06
  • @TimSeguine I guess, I'm just not clever enough to see it. – Dan Christensen Jan 25 '14 at 21:09
  • @Tim: Not quite. The consistency of $\sf ZFC$ is generally much much weaker than asserting that there is a well-founded model, and that is weaker (by much) than asserting that $V_\alpha$ is a model of $\sf ZFC$, for some $\alpha$. Finally, the notion of a cardinal closed under exponentiation (i.e. $\mu,\lambda<\kappa\implies\mu^\lambda<\kappa$) is not sufficient anyway, $\sf ZFC$ proves the existence of these cardinals. It is necessary though, if we want $V_\kappa$ to be a model of $\sf ZFC$. – Asaf Karagila Jan 25 '14 at 21:09
  • @AsafKaragila Readers will have to judge for themselves. – Dan Christensen Jan 25 '14 at 21:12
  • @Dan: I never trusted people without technical experience and understanding to be judges of technical issues. I'm glad for you that you are okay with that. – Asaf Karagila Jan 25 '14 at 21:13
  • @AsafKaragila I thought that ZFC can't prove the existence of limit cardinals above aleph-null? Fine if I was mistaken. – Tim Seguine Jan 25 '14 at 21:15
  • @Tim: No, it certainly can. It cannot prove the existence of regular limit cardinals, though. – Asaf Karagila Jan 25 '14 at 21:23
  • @AsafKaragila just like me to forget adjectives. – Tim Seguine Jan 25 '14 at 21:32

2 Answers2

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Your proof is irrelevant -- either the theory you're working in is consistent or it is not. If it is consistent, then you can't derive any antinomies (unless you are actually making invalid arguments).

Whether it is even possible to "know" a theory is consistent is a deep epistemological problem; e.g. see the regress problem. Mathematicians usually settle for proofs relative consistency -- e.g. ZFC can prove that Peano's axioms for the natural numbers are consistent.

That's not quite true: for many things, mathematicians don't even bother to go that far -- e.g. we just accept that our collective experience working with the natural numbers hasn't turned up any antinomies. And even if it did some day, we'd make appropriate adjustments to correct the problem, then continue along.

  • The OP's proof is not irrelevant. As you point out, the system is robust enough to handle any known contradictions. Along this line, I have shown that it can handle even what might be called a "partial resolution" of Russell's Paradox. – Dan Christensen Jan 26 '14 at 15:40
  • When you say "mathematicians don't even bother to...", I believe you are unfairly disparaging them. Consistent with GIT, proofs of consistency are probably a nothing more than a wild-goose chase, the quarry forever eluding us in an infinite regress: Prove system A is consistent using system B, and you must prove system B is consistent in yet another system, and so on. Realistically, the best we can do is to establish a robust set of axioms, taking them as far as we can, and tweaking them as required when intractable problems arise. – Dan Christensen Jan 26 '14 at 15:46
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    @Dan: It wasn't meant to be disparaging on mathematicians -- it was trying to express the fact that these foundational issues aren't as important to actual practice as they're sometimes made out to be. I suppose I was unclear that "mathematicians" was referring to the 'average' practice, rather than the totality: certainly some subset of mathematicians do spend some subset of their time delving into foundational and epistemological issues (and that's a good thing). –  Jan 26 '14 at 16:14
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When you look at the Zermelo-Russell paradox you have a similar situation: Two cases which are mutually exclusive: Either $R \in R$ or $R \not \in R$. You then suppose e.g. that $R \in R$ and arrive at a contradiction... but you don't stop there! Otherwise it would not be a paradox! You also test the other case and, again, arrive at a contradiction!

Suppose that, in your resolution of Russell's Paradox, after you obtained $R\notin R$, you did not go on to obtain $R\in R$. You would still be able to obtain a valid result on generalizing as follows:

  1. Suppose $\forall x:[x\in R\iff x\notin x]$

  2. By specifying $x=R$ in (1), we have $R\in R\iff R\notin R$

  3. Suppose $R\in R$

  4. From (2), we obtain $R\notin R$, and the contradiction $R\in R\land R\notin R$

  5. By contradiction, we must have $R\notin R$

  6. Stopping here, we could nevertheless generalize to obtain

$$\forall r: [\forall x:[x\in r\iff x\notin x] \implies r\notin r]$$

Intuitively, you can see that this will always be true since, vacuously, you will never be able to prove that $\forall x:[x\in R\iff x\notin x]$ for any R. The system works!

Of course, you could also prove

$$\forall r: [\forall x:[x\in r\iff x\notin x] \implies r\in r]$$

But that's OK, too, for the same reason.

So my question is how do you ensure that you can trust your proof and haven't run into another paradox?

You can't be absolutely certain you won't come across some internal inconsistency in whatever system of deduction you are using, but if you stick to the time-tested systems of logic and set theory, it is highly unlikely. As I demonstrate above, even a botched (or partial) resolution of RP is easily handled in ordinary predicate logic.

Dan Christensen
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    Here is a formal version this proof written with the aid of my proof checking software at: http://www.dcproof.com/PartialRussell.htm BTW, your students may this software useful. Free download at my website http://www.dcproof.com – Dan Christensen Jan 24 '14 at 19:42
  • Thank you, looks promising! – vonjd Jan 24 '14 at 19:59