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I will start off by saying that I am an elementary student of mathematics and do not possess the deep and rigorous knowledge of most members of this site. Nonetheless, whilst learning how to do a proof by contradiction, I had a fascinating thought that I would like clarification on.

I realise that, by Gödel's incompleteness theorems, mathematical axioms must be either consistent or complete but not both. If we ever encounter an inconsistency in our (correct) mathematical reasoning, then that is a sign that our mathematical axioms are complete but inconsistent.

This may be a really stupid question, but I would rather ask it anyway:

If one is attempting a proof by contradiction (say, at the most advanced levels of mathematics) and they encounter a contradiction, how would we know that it is the contradiction we were searching for and NOT an inconsistency of our axioms? I realise that, given the success of our axioms thus far, this is extremely unlikely, but if the unlikely did happen in such a situation, how would we know?

Again, I apologise if this is a stupid question, but I would greatly appreciate it if someone would entertain my thought.

The Pointer
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3 Answers3

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how would we know that it is the contradiction we were searching for and NOT an inconsistency of our axioms?

We don't. But either way, the conclusion holds (since, if the axioms themselves are inconsistent, then all statements are true).

By that, I mean that

  • if the axioms are consistent, then $\neg P\implies \bot$ is proof that $\neg \neg P$ is true, and from that, most will conclude that $P$ is true.
  • If the axioms are inconsistent, then, because $\bot\implies P$ is true and $\bot$ is true, $P$ must also be true.

So, in both cases, the conclusion, $P$, is true (in ZFC).

5xum
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    Thanks for the response. You mean everything we thought to be true/false instantly becomes unknown, since the inconsistency means we could have proved it either way (true or false) by "mistake"? – The Pointer Feb 09 '17 at 11:48
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    @ThePointer Formally, yes. However, many of the proofs may be re-used in some other set of axioms, so it might not be a disaster. – 5xum Feb 09 '17 at 11:49
  • Absolutely fascinating. I heard this before, but it provides little solace. Perhaps I should change my major to physics ... Just kidding. – The Pointer Feb 09 '17 at 11:51
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    @ThePointer My advice is just to not think about these things too much. After all, math works in the real world, so whatever formalization we choose must be something darn close to what we have right now. – 5xum Feb 09 '17 at 11:53
  • Indeed, this has been my thoughts too. – The Pointer Feb 09 '17 at 11:54
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    @ThePointer See this answer on the topic "What if ZFC is inconsistent?": http://mathoverflow.net/a/41030/103918 – Wolfram Feb 09 '17 at 12:06
  • If the axioms of ZFC are inconsistent, then they aren't true; by definition $\bot$ is not a true statement. – Carl Mummert Feb 09 '17 at 12:41
  • @CarlMummert Agreed. But my point is that even if ZFC is not consistent, it's possible that it can be replaced with something in which most existing mathematical proofs can still be "salvaged". – 5xum Feb 09 '17 at 12:50
  • Sure, but that does not mean that if ZFC is inconsistent then everything provable in ZFC must still be true, which is what this answer appears to claim in the second bullet. Even if we directly took $\bot$ as an axiom, it is never a true statement in any model. – Carl Mummert Feb 09 '17 at 12:54
  • @CarlMummert No, the second bullet stays within ZFC. I.e., in ZFC, $P$ is true if ZFC is inconsistent (because all statements are true in an inconsistent system). I will add this into my answer. – 5xum Feb 09 '17 at 12:56
  • It is not true that "all statements are true in an inconsistent system". On one hand, statements are not "true in a system"; they can only be true or false in a model. An inconsistent system has no models. More importantly, when we don't mention a model, we implicitly refer to the standard model, and so saying that a statement "is true" is just a property of the statement itself, independent of any formal systems. – Carl Mummert Feb 09 '17 at 12:58
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    By true in a system, I mean "provable". I thought that much was clear. – 5xum Feb 09 '17 at 12:59
  • @ThePointer: You wrote that finding a proof of a contradiction in ZFC implies that "everything we thought to be true/false instantly becomes unknown". This is incorrect. A statement only has a truth-value if it is meaningful to begin with. Mathematically, we define "meaning" to be with respect to a model, but to do so we already have to be working in a formal system (usually ZFC). One may reject the meaningfulness of such "meaning" on philosophical grounds, but then one has to provide an alternative. See http://math.stackexchange.com/a/1888389 and http://math.stackexchange.com/a/1808558. – user21820 Feb 09 '17 at 14:09
  • @5xum: And I disagree with your claim that mathematics works in the real-world. The mathematics that pertain to the real world seems to lie completely within predicative mathematics. I have yet to see a single example of an empirically verifiable experiment that corresponds to the truth of a mathematical statement that cannot be stated naturally and proven in ACA, which is extremely far below ZFC. In other words we do not have evidence that ZFC works. We merely have evidence that the minuscule fragment ACA works. If you know counter-examples to my claim, please tell me! – user21820 Feb 09 '17 at 14:19
  • @user21820 This isn't really the best place for a pretty deep debate like this. – 5xum Feb 09 '17 at 14:21
  • @5xum: Not trying to debate, but just stating that I haven't seen evidence. That's why I would be extremely interested to see any in either direction. And if you would like to have any discussion about such stuff, I'll be very glad to hear your views in chat. =) – user21820 Feb 09 '17 at 14:24
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    @5xum: the distinction between "true" and "provable" is one of the key issues in logic. Unfortunate terminology such as "true in a system" only confuses the issue. Clear and precise terminology is vital for mathematical communication, particularly in areas like this that are so prone to misunderstandings. There is an established, standard meaning of "true", which is not "provable". – Carl Mummert Feb 09 '17 at 16:07
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The short answer is that there is no simple way to distinguish an inconsistency that comes from a proof by contradiction from an inconsistency that comes from working with inconsistent axioms.

In some cases, if the axioms were the source of the inconsistency, an examination of the proof in question might show it. But if the proof is sufficiently subtle, this analysis may not be straightforward.

In many cases the proof we are working with uses very simple axioms. In these cases, it may be clear that the axioms are consistent. This is the case, for example, with the axioms for a field or a vector space.

In other cases, such as ZFC, there is a general belief in the consistency of the system, because we have been working with it for so long. Many people have tried to construct a contradiction in ZFC, and none has been found. This is not a proof that ZFC is consistent, of course (although there are some arguments that do aim to prove the consistency).

Even if ZFC was found to be inconsistent, it would not affect the vast majority of mathematical results, which do not really rely on special features of ZFC. We could reprove these results using many sets of axioms. If an inconsistency in ZFC was found, it would be interesting for logicians, but unlikely to affect most mathematicians.

Carl Mummert
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    "If an inconsistency in ZFC was found, it would be interesting for logicians, but unlikely to affect most mathematicians" which is why, dear OP, you shouldn't be too worried about inconsistencies in ZFC. – 5xum Feb 09 '17 at 13:01
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When you're doing proof by contradiction, you explicitly assume some statement $A$ and after some logical process obtain a contradiction, that is, for some statement $B$ you prove both $B$ and $\neg B$. This only proves that you assumption was false, that is, it proves $\neg A$. The contradiction in ZFC would be entirely different matter - you doesn't assume anything but the axioms of ZFC and obtain the proof of both $B$ and $\neg B$. In some sense, when you prove by contradiction, you prove that ZFC, if you add $A$ to it, is inconsistent and therefore ZFC implies $\neg A$. However, inconsistency of $ZFC+A$ doesn't say anything about inconsistency of ZFC itself.

Wolfram
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  • Thanks for the response. Excuse my ignorance, but it sounds like your answer disagrees with that of 5xum? Can you please elaborate on what you disagree with, if anything? – The Pointer Feb 09 '17 at 12:05
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    No, I agree with the answer, that any statement we have proved by contradiction is true in ZFC, no matter is ZFC consistent or not (because any statement is true in inconsistent system). – Wolfram Feb 09 '17 at 12:11
  • I understand. Thank you for the answer and clarification. – The Pointer Feb 09 '17 at 12:15
  • @Wolfram: "true in ZFC" is not usual terminology. Statements can only be true or false in a model; if ZFC is inconsistent it has no model. Normally we are concerned with whether statements are true in the standard model. Importantly for this question, it is not the case that "every statement is true in an inconsistent system". – Carl Mummert Feb 09 '17 at 12:43
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    @Wolfram: In the first part of your answer ("When you're doing ... it proves $\neg A$.") you should make clear if, besides the explicit assumption $A$, also axioms of ZFC are implicitly assumed. Because then, if you prove $B$ and $\neg B$, you know only that $A$ and axioms of ZFC cannot be simultaneously true in any model. I believe that the situation you had in mind was the first one, when no implicit assumptions were made. However, OP asks about a proof "say, at the most advanced levels of mathematics", where axioms of ZFC are almost certainly assumed. – Peter Elias Feb 09 '17 at 15:13