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After reading and being inspired by, Can every proof by contradiction also be shown without contradiction? and after some thought, I still don't have an answer to this.

Does every theorem with a true proof have a proof by contradiction?

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TripleA
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    Let $P$ be a proof of the theorem. Assume the theorem is false. However we can exhibit $P$, which contradicts the assumption. – Stahl Jan 03 '17 at 04:11
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    I think @Stahl is correct – MPW Jan 03 '17 at 04:22
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    @Masacroso note that I did not post it as an answer, but rather as a comment ;) I'm neither a logician nor particularly interested in the subtleties of logic, and I assume someone can and will give a better explanation than I can. However, I have a hunch that one can formalize what I've said. – Stahl Jan 03 '17 at 04:26
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    I think @Stahl is correct. Many so-called proofs by contradiction amount to assuming the result is false, proceeding to prove the result, then saying that the proven result then contradicts the hypothesis that it is false. It's a classic overkill used by many students when they could just prove the result directly. – MPW Jan 03 '17 at 04:28
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    Ah, sorry, I missread the question as the opposite direction. Anyway it is not clear to me if this is possible or not, I mean that we can construct a proof by contradiction derived from any other proof, but then the status of this proof to be "by contradiction" is not very clear (because the statement is already proved before to "prove it but this derived proof"). – Masacroso Jan 03 '17 at 04:57
  • @Masacroso Yeah, I'm waiting for someone who knows haha, it's quite frustrating :) – TripleA Jan 03 '17 at 04:59
  • Stahl is absolutely correct. – Noah Schweber Jan 03 '17 at 05:16
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    @MPW: if you have assumed the negation of the desired result, you have one more statement to base your derivation of the desired result on, generating the desired contradiction. – Ross Millikan Jan 03 '17 at 06:12
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    @Masacroso: Yes you can certainly say that a proof by construction constructed as you described is unsatisfying, but formally speaking it is still a proof. See my answer for various avenues one can explore when attempting to avoid this trivial answer. – user21820 Jan 03 '17 at 07:00
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    @Stahl: See my answer. In particular note that we always work in a meta-system and can easily, given any proof in a natural deduction system, convert it to a proof of the same theorem that has a proof by contradiction at its outermost shell. The same goes with many other deductive systems for classical logic, but to be really precise one has to define what a "proof by contradiction" means. Of course the key aspect that permits the conversion is the ability to use (or convert) a proof as a subproof. – user21820 Jan 03 '17 at 07:11
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    @RossMillikan : Quite so; but if the theorem is proved without actually using that additional assumption, and then the assumption is invoked in order to produce a contradiction, then that additional statement is in fact unneeded as the theorem has already been proved without using it. – MPW Jan 03 '17 at 08:04
  • @MPW Look at the new answer – TripleA Jan 03 '17 at 19:53
  • @Masacroso ^^^^ – TripleA Jan 03 '17 at 19:54
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    Possible duplicate of Can every proof by contradiction also be shown without contradiction?. Go and read http://math.stackexchange.com/questions/243770/can-every-proof-by-contradiction-also-be-shown-without-contradiction again. If you accept proof by contradiction, then anything you can prove is provable by contradiction. – Rob Arthan Jan 04 '17 at 23:25
  • Does this answer your question? Direct proof and contradicttion – user95921 Dec 26 '23 at 06:51

3 Answers3

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In classical logic, the answer is yes. Take any theorem $T$ and any proof $P$ for $T$. Now write the following proof:

If $\neg T$:

  [Write $P$ here.]

  Thus $T$.

  Thus a contradiction.

Therefore $\neg \neg T$, by negation introduction.

Thus $T$, by double negation elimination.

One may object that this proof is essentially the same as $P$, and is just wrapped up. That is true, but it is a perfectly legitimate proof of $T$, even if it is longer than $P$, and it is indeed of the form of a proof by contradiction. A natural question that arises is whether the shortest proof of $T$ is a proof by contradiction. That is a much harder question to answer in general, but there are some easy examples, at least for any reasonable natural deduction system.

For instance, the shortest proof of "$A \to A$" for any given statement $A$ is definitely not a proof by contradiction but rather just:

If $A$:

  $A$.

Therefore $A \to A$, by implication introduction.

On the other hand, the shortest proof of "$\neg ( A \land \neg A )$" for any given statement $A$ is definitely a proof by contradiction:

If $A \land \neg A$:

  $A$, by conjunction elimination.

  $\neg A$, by conjunction elimination.

  Thus a contradiction.

Therefore $\neg( A \land \neg A )$.

The first part of this post shows that the shortest proof by contradiction is at most a few lines longer than the shortest proof, but nothing much else interesting can be said about the shortest proof unless...

Well what if we do not allow the use of double negation elimination? If you have only the other usual rules (the first-order logic rules here but excluding ¬¬elim and including ex falso), then the resulting logic is intuitionistic logic, which is strictly weaker than classical logic, and cannot even prove the law of excluded middle, namely "$A \lor \neg A$" for any statement $A$. So if you instead ask the more interesting question of whether every true theorem can be proven in intuitionistic logic, then the answer is no.

Note that intuitionistic logic plus the rule "$\neg A \to \bot \vdash A$" gives back classical logic, and one could say that this rule embodies the 'true principle' of proof by contradiction, in which case one can say that some true theorems require the use of a proof by contradiction somewhere.

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user21820
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  • The only downvote so far comes from a troll who came here in Apr 2020 because he(?) just wanted something to downvote. In fact, trolls tend to downvote posts about logic, perhaps because they can't think straight. – user21820 Aug 08 '20 at 17:34
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If you can prove a statement $A$ directly, you can prove it by contradiction. Just assume $\lnot A$, perform your proof of $A$, note the contradiction, and derive $A$.

Ross Millikan
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I find the common thought here unsettling, even if seemingly widely held. Perhaps others share my view, perhaps not (and note, it is just an intuitive view, one which--as explained to me in the comments--isn't in line with the notion of proof!)

The proof in these purported proofs by contradiction must necessarily rely on the content of the direct proof. Yes, we have ourselves a contradiction, but in a proof by contradiction what convinces us (i.e., what we count as the proof) is not merely that we have a contradiction, but that we have derived a contradiction in a particular manner--from particular assumption(s)--which leads us to conclude something about these assumptions....that they must not be true.

Here, all we do is prove a statement directly from the premises of the argument, and make a note that our conclusion contradicts the conclusion's_negation--leading us to believe that the negation of the conclusion's_negation (i.e., our conclusion) must be true. We then say to ourselves "aha! Indeed this is the case...since we have already (directly) proven our conclusion to be true!"

It is not simply that our conclusion need not be derived via contradiction (and that it can be done directly and then "wrapped up" in a so called proof by contradiction), but rather that our conclusion has not been derived in this manner.

A contradiction arising in a proof does not necessarily warrant that the method of proof being employed is what we call "proof by contradiction." It is the derivation of this contradiction which warrants the name "proof by contradiction." It is how the contradiction arises. And here, our contradiction doesn't really arise so much as the situation is one in which we are pointing out that a directly derived conclusion is contradictory to its negation.

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    Your intuition makes sense, but raises the challenge of how to precisely define what a "proof by contradiction" actually is then. – Eric Wofsey Jan 03 '17 at 23:10
  • @MorganRodgers I'm arguing that we have not rephrased the original proof to be a proof by contradiction, but that we have merely inserted a remark about the directly derived conclusion into the proof--namely, that it is contradictory with its negation. The problem, as Eric Wofsey says, is that we would need to precisely define "proof by contradiction." – Parry Jan 03 '17 at 23:14
  • I'm reading it the same way. Within a new proof, start with an assumption of the negation and repeat the argument found in the first proof (without making reference to this first proof). The problem I have is that within those arguments the conclusion gets derived independently of any contradiction. Where does the proof by contradiction come into play then? The answer put forth is that after the conclusion gets derived, we note that it contradicts our assumption that we started with, leaving us to infer the negation of the assumption...which is the conclusion. – Parry Jan 04 '17 at 02:43
  • My point is that since we have already established the conclusion in the proof, then at this point we are doing nothing more then adding superfluous remarks about the relation between it and it's negation (that they are contradictory). – Parry Jan 04 '17 at 02:43
  • One might reply that a proof (which we describe as following a particular method of proof) with unnecessary information still constitutes a proof done via that particular method. I wouldn't argue that, but I would argue that the unnecessary information can't be the grounds from which we determine how to describe which particular method of proof we are using, for if we established the conclusion without using that information in the proof--how can it constitute the method of proof? – Parry Jan 04 '17 at 02:57
  • The sentiment in your last 2 comments is precisely why in my answer I suggested that we could instead ask for what form the shortest proof takes rather than just whether there is a proof of a certain form or not. As I described, it is in general a hard question to answer, but possibly addresses your unsettledness because wrapping proofs up in another form will make it longer. Also, the last paragraph in my answer states a possible precise definition of a proof by contradiction, which can easily be proven equivalent to double negation elimination over intuitionistic logic. – user21820 Jan 04 '17 at 04:18
  • Another thing is that perhaps you do not have a precise understanding of "proof". You may think that a proof is some writing that convinces another person of the truth of some statement, but that would be very wrong. In modern mathematics, a proper proof is a writing that provides enough information so that any mathematician sufficiently trained in the relevant fields can easily translate it into a formal proof in some fixed formal system, usually ZFC set theory plus on-the-fly definitorial expansion (abbreviation power). A formal proof is a string accepted by the formal system. [continued] – user21820 Jan 04 '17 at 04:29
  • [continued] In this precise sense of "proof" there is no notion of 'method of proof' or 'superflous' or 'already proven by a subproof' or 'remarks'. Look up natural deduction and Hilbert-style systems to get an idea of what a formal system is actually like. The reason we need such kind of notion of proof is because it is in some sense necessary for the completeness theorem of classical first-order logic to hold. Of course, after defining "proof" this way we can then define "short proof" as a proof with minimal length, or "core proof" as one that is minimal in some sense to be defined by you. – user21820 Jan 04 '17 at 04:38
  • One thing I would appreciate you clarifying before I respond more in depth is what you meant in your answer when saying that your first proof "indeed [is] of the form of proof by contradiction"? Do you consider it (the form of the proof, in this case, by contradiction) synonymous with "method of proof" which you are saying is not a notion pertaining to modern math's sense of "proof"? If so, then in what notion can you talk of this form and if not, then in what sense do you mean by the statement? Thanks. – Parry Jan 04 '17 at 04:52
  • What I mean to ask is: In what sense does the notion of "form of a proof" pertain to modern mathematics' definition/view of "proof?" When you talk of the form of the proof that you detailed in your answer as being of the form of a proof by contradiction, did you mean that the method of proof employed here is proof-by-contradiction? If not, then could you clarify what you meant. If so, then taking into account your comments here saying that the notion of "method of proof" is not taken into consideration in math's definition of "proof", then what use were you making of the statement? Thanks – Parry Jan 04 '17 at 05:09
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    I meant that it is of a specific syntactic form; it begins with a subproof of a contradiction under an assumption of the form $\neg A$, and then deduces $\neg \neg A$ and then applies DNE to get $A$. So when I said "form" in my answer I do not refer to the notion of "method" that most non-logicians have in mind. Is what I say clearer now? – user21820 Jan 04 '17 at 05:15
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    The specific details of the syntactic form will depend on the specific formal system chosen, so you can consider the proof sketch in my answer akin to pseudo-code. If you use the rules in my linked post, then "form of a proof by contradiction" in my answer corresponds to a formal proof whose last 3 lines are of the form: $$¬A→⊥. \quad [→intro] \ ¬¬A. \quad \quad \quad [¬intro] \ A. \quad \quad \quad \quad [¬¬elim]$$. By the way, your answer is related to http://mathoverflow.net/questions/3776/when-are-two-proofs-of-the-same-theorem-really-different-proofs. – user21820 Jan 04 '17 at 05:38
  • Let us continue this discussion in chat. – Parry Jan 04 '17 at 05:38