Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?

As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians. Is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.

Simon Fraser
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    Let us assume every proof by contradiction can also be shown without contradiction... – GeoffDS Nov 24 '12 at 18:21
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    Since this topic came up in some of the answers and comments, Andrej Bauer's blog post [Proof of negation and proof by contradiction](http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/) seems relevant. –  Nov 24 '12 at 18:21
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    ... and Gowers's blog post: [When is proof by contradiction necessary?](http://gowers.wordpress.com/2010/03/28/). –  Nov 24 '12 at 18:29
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    I should know this, but I am not sure. You are basically asking whether intuitionistic logic is equivalent to classical logic? I am pretty sure it is not. – Panayiotis Karabassis Nov 25 '12 at 00:36
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    It would be mathematically ironic if a question about the weakness of proof by contradiction would be closed as *not constructive*. [Insert a rimshot sound here] – Asaf Karagila Nov 25 '12 at 16:01
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    Meta-question: Can a proof that "every proof by contradiction can also be shown without contradiction" be shown without proof by contradiction. – Eric Nov 25 '12 at 22:26
  • If your goal is to prove some statement or idea, as long as the method of proof is valid, I think anyones feelings on that method are irrelevent. – Ethan Nov 26 '12 at 08:51
  • But that's the point, not every proof method is valid/invalid to everyone. I think it always pays to be explicit about your assumptions. – Panayiotis Karabassis Nov 27 '12 at 16:56
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    This is exactly the point of departure for **constructive** maths (esp. of the **intuitionistic** flavor). That simply "proving" that some object *must* exist, because it is a contradiction, not to (usually by applying LEM in non-obvious, non-verifiable way) is not adequate since it does not provide any other informaton (if it is valid of course in the 1st place) – Nikos M. May 27 '14 at 23:10

15 Answers15


To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$.

There are two key logical principles that express different parts of what we call "proof by contradiction":

  1. The principle of explosion: for any statement $A$, we can take "$\bot$ implies $A$" as an axiom. This is also called ex falso quodlibet.

  2. The law of the excluded middle: for any statement $A$, we can take "$A$ or $\lnot A$" as an axiom.

In proof theory, there are three well known systems:

  • Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.

  • Intuitionistic logic includes minimal logic and the principle of explosion

  • Classical logic includes intuitionistic logic and the law of the excluded middle

It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.

The scheme "If $A$ implies a contradiction, then $\lnot A$ must hold" is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says "if $A \to \bot$ then $A \to \bot$". But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.

Carl Mummert
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    +1 Great explanation. Can you emphasize the affirmative answer?: ___So there are statements that are provable by contradiction that are not provable directly.___ – ypercubeᵀᴹ Nov 29 '12 at 20:10
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    What is "identically false proposition"? – SasQ Jul 11 '13 at 17:08
  • @SasQ: a proposition that is never true. The exact choice doesn't matter, but it depends on the language. In a language with equality and negation, you can use $(\exists x)[x = x \land x \not = x]$. Or you can just directly add a new atomic formula, $\bot$, with the rule that $\bot$ is false in every model. – Carl Mummert Jul 11 '13 at 17:11
  • How does it differ from being false? – SasQ Jul 12 '13 at 07:18
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    @SasQ: saying that a statement is "true" or "false" is a claim about some specific model. Saying that a statement is identically false indicates that the statement is disprovable, so it is false in all models. – Carl Mummert Jul 12 '13 at 11:53
  • I had "false" as a noun, not a verb, on my mind. What's the difference between this special symbol and just "false" (as a noun)? Is there some special fancy symbol for truth too? – SasQ Jul 12 '13 at 23:26
  • Yes, $\top$ is an identically true statement. – Carl Mummert Jul 13 '13 at 00:10
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    Note: "identically false/true" is often called "valid/invalid" instead. – Noldorin Apr 21 '14 at 14:50
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    An excellent. It's great to see a positive characterisation of intuitionistic and classical logic, rather than the usual meaningless 'intuitionistic logic is classical logic without the law of excluded middle'. – Miles Rout Sep 12 '16 at 22:43
  • Do you have an example of a statement that can only be proved by contradiction? – Cronus Dec 23 '17 at 18:55
  • @Cronus: the principle LPO is easy to prove by contradiction, and is not provable in any constructive theory I know of, even ones that include intuitionistic logic. It is not fully equivalent to the law of the excluded middle, but the most natural way to prove it, starting with intuitionistic logic, is to assume excluded middle. https://en.wikipedia.org/wiki/Limited_principle_of_omniscience – Carl Mummert Dec 23 '17 at 22:30
  • @Carl Wow, okay. Can't imagine how anyone can do mathematics without these things! – Cronus Dec 24 '17 at 10:03

If a statements says "not $X$" then it is perfectly fine to assume $X$, arrive at a contradiction and conclude "not $X$". However, in many occasions a proof by contradiction is presented while it is really not used (let alone necessary). The reasoning then goes as follows:

Proof of $X$: Suppose not $X$. Then ... complete proof of $X$ follows here... This is a contradiction and therefore $X$.

A famous example is Euclid's proof of the infinitude of primes. It is often stated as follows (not by Euclid by the way):

Suppose there is only a finite number of primes. Then ... construction of new prime follows ... This is a contradiction so there are infinitely many primes.

Without the contradiction part, you'd be left with a perfectly fine argument. Namely, given a finite set of primes, a new prime can be constructed.

This kind of presentation is really something that you should learn to avoid. Once you're aware of this pattern its amazing how often you'll encounter it, including here on math.se.

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    Another example of this is the common presentation of the proof that there is no surjection from $\Bbb N$ to $(0,1)$. The usual presentation begins "suppose we have such a surjection…" and concludes "therefore it is not a surjection, and we have a contradiction". A simpler presentation simply lets $f$ be an arbitrary function $\Bbb N\to (0,1)$, follows the same argument, and concludes "…therefore, $f$ is not surjective". – MJD Nov 24 '12 at 22:08
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    Could you detail the part "construction of new prime follows" in your Euclid example please? I'd bet your version of the proof is either flawed, or uses contradiction. The key point being that "construction of new prime follows" normally relies on the assumption that only a finite number of primes exist. – Axel Nov 24 '12 at 23:16
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    Could you give another example or elaborate on this one? I don't really get how you want to finish the proof. – heinrich5991 Nov 25 '12 at 00:29
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    @Axel: The construction is: given a (possibly empty) finite set of primes, we can construct a number larger than one which is not divisible by any prime in the set. Therefore a prime exists which is not in the set. Therefore the set of all primes must be larger than any finite set, i.e., infinite. – Dietrich Epp Nov 25 '12 at 01:34
  • So are you saying that contradiction is often used when an argument by contrapositive would do just fine? – asmeurer Nov 25 '12 at 01:58
  • I would say that saying "the set of all primes must be larger than any finite set, i.e., infinite" implies the contradiction. – tst Nov 25 '12 at 02:23
  • @DietrichEpp: So what I get is: "Given a finite set of primes, you show that it doesn't contain all primes. Thus the set of all primes must be infinite." This IMHO is still prove by contradiction, even if in poor disguise. – Axel Nov 25 '12 at 07:51
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    @Axel: Where is the contradiction? The point illustrated in this answer is that the contradiction comes from saying "suppose that that the set of all primes is finite", but this is unnecessary for the proof. What I said is "suppose **a** set of primes is finite". No contradiction there. – Dietrich Epp Nov 25 '12 at 07:54
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    @Axel and Dietrich: My answer is a personal remark and not intended to start debates. I can see both your views. For me the point here is form. *Starting* with the assumption that there are finitely many primes suggests that the construction argument that follows depends on it, while it does not. (That remains valid even now that we know that there are infinitely many primes.) The contradiction part only comes after that: If I have a procedure that keeps churning out unique numbers, how do you prove that it will produce infinitely many numbers? – WimC Nov 25 '12 at 08:19
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    @DietrichEpp: You're simply not there yet. You have shown that any finite set of primes does not contain all primes. Thus the set of all primes cannot be a finite set. Fine. Now show that the set of all primes is infinite. The key part is still missing. – Axel Nov 25 '12 at 08:19
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    @tst ... and you would be right. Off topic: Another interesting view is that the construction of more primes produces an injection of $\mathbb{N}$ into the set of primes, which is therefore infinite. Would you then still demand a proof that $\mathbb{N}$ is infinite? – WimC Nov 25 '12 at 08:37
  • @WimC, well actually I am not sure. I don't know how many theorems of set theory require contradiction for their proof. So the theorem $\exists f:A\to B$ injective $\implies|A|\le|B|$, can be proved without contradiction? Also the theorem that $\mathbb{N}$ is infinite, can be proved without contradiction? – tst Nov 25 '12 at 10:57
  • @Axel As WimC pointes, you can show by induction that for any positive $n$ there exists at least $n$ prime numbers... Then, to conclude the proof without contradiction, you can construct an injective function from the prime numbers to themselves, which is not surjective... – N. S. Nov 26 '12 at 04:26
  • @tst I think that the formal definition of an infinite set is the following: S is infinite if and only if you can find $f: S \to S$ which is injective but not surjective... It is trivial to see that $\mathbb N$ does satisfy this definition, and it is easy to prove that any finite set doesn't satisfy this definition......Proving that this definition is equivalent to the intuitive understanding might require contradiction though.... – N. S. Nov 26 '12 at 04:29
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    I usually see the proof stated as the following: `Suppose there is only a finite number of primes and ALLPRIME contains all primes in existence. Then ... proof that there exist another number p which is a prime but not in ALLPRIME ... This is a contradiction so there are infinitely many primes.` which is a proof by contradiction. I've never seen it presented the way you present it. – Lie Ryan Nov 26 '12 at 04:32
  • @LieRyan we mean the same proof. I left out the ALLPRIME part. Sorry for the misunderstanding. – WimC Nov 26 '12 at 06:05
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    @N.S.: Yes, that's possible.But now if you put it all together into a complete proof, I think: 1. It's neither more compact nor easier to understand than the one using contradiction. That's the key thing I wanted to point out. Simple test: I can explain the proof to my boy of 7 years using contradiction. I don't manage to explain the other one to him (at least so that he does understand it) 2. It doesn't match the scheme presented above ("complete proof ox X follows"), since that's not needed here. – Axel Nov 26 '12 at 08:10
  • @heinrich5991: See asmeurer's answer and my comment. – WimC Nov 27 '12 at 21:34
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    It seems that in cases like the infinitude of primes, a complete proof requires *either* an argument by contradiction, or by induction. I find the indirect proof more elegant than the inductive one, but of course that's a matter of taste. – user7530 Dec 17 '12 at 21:49
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    @Axel: BTW, this is the complete proof (also closer to Euclid's): take any list of primes $p_1,p_2,\dots, p_n$. (Doesn't have to be first few primes, or all the primes. E.g. could be $\{2,5,11\}$.) Consider the new number $p_1p_2\cdots p_n+1$. This isn't divisible by any prime in the list, so any prime factor of it is a new prime that wasn't in our list. So any finite list of primes can be extended by adding new primes. Note that this actually gave us a construction that works without assumptions. See also http://math.stackexchange.com/questions/240/ and its linked questions. – ShreevatsaR Jan 30 '14 at 20:21
  • @ShreevatsaR: Your proof is alright, but it is nevertheless based on assumptions... ;-) – Axel Jan 30 '14 at 20:34
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    @Axel: I don't understand what you mean. Every proof is based on assumptions/axioms. That is not the question here; the question is whether it is necessary or better to use a proof by contradiction, specifically. Take some time to read the answers at the linked question and other questions linked to it, to see the issues involved. (Oh, if you were referring to my statement that "this actually gave us a construction that works without assumptions" — I meant *false* assumptions (so we got an unconditional useful result), but couldn't fit it in the 600 char limit.) – ShreevatsaR Jan 30 '14 at 20:40
  • To prove that a topological space is connected... take open disjoint non empty A and B. Show that B is empty using NOWHERE that B is not empty. And conclude by a useless contradiction that the space is connected. – André Caldas Oct 12 '16 at 23:36
  • @DietrichEpp how about "Every finite set of primes has a successor" and therefore the primes are surjective into the natural numbers. Definitely no contradiction there. – samerivertwice Mar 28 '17 at 17:26

It somewhat depends on whether you are intuitionist or not (or both? or neither? Who knows without the law of excluded middle). According to the Wikipedia article even intuitionists accept some versions of what one could call indirect proof, but reject most. In that sense, a direct proof would be preferable (and is often even a bit more elegant).

An example:

Theorem. There exist irrational numbers $a,b$ such that $a^b$ is rational.

Proof: Assume that $a,b\notin \mathbb Q$ always implies $a^b\notin \mathbb Q$. Then $u:=\sqrt 2^{\sqrt 2}\notin \mathbb Q$ and $u^\sqrt 2=\sqrt 2^{\sqrt 2\cdot\sqrt 2}=\sqrt 2^2=2\notin \mathbb Q$ - contradiction!

Indeed, an intuitionist would complain that we do not exhibit a pair $(a,b)$ with $a,b\notin \mathbb Q$ and $a^b\in \mathbb Q$. Instead, we only show that either $(\sqrt 2,\sqrt 2)$ or $(u,\sqrt 2)$ is such a pair. Converting the proof given above to a direct and constructive proof would in fact require you to actually prove one of the two possible options $u\in \mathbb Q$ or $u\notin\mathbb Q$.

Hagen von Eitzen
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    Of course you are correct the intuitionist would complain that the proof does not exhibit a specific pair. But an intuitionist would not agree that the proof produces two pairs such that either the first pair is an example or the second is an example. The intuitionistic reading of that claimed conclusion would say that, to prove the "or", we would have to prove which pair is the example, which is exactly what the proof does not do. The intuitionist would accept the proof shown here as showing "it is not the case that for all $a,b \not \in \mathbb{Q}$, $a^b \not \in \mathbb{Q}$". – Carl Mummert Nov 24 '12 at 21:20
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    By [Gelfond–Schneider](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem) $\sqrt2^{\sqrt2}$ is transcendental. Voilà! No more proof by contradiction ;-) – kahen Nov 24 '12 at 23:11
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    @kahen Well, the [proof of Gelfond Schneider](http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes8.pdf) referenced in the [wikipedia article](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem) you referenced seems to contain a step which is proven by contradiction... (if Delta not 0 ... conclude that Delta is 0) – mkl Nov 25 '12 at 22:54

See this post: Are proofs by contradiction weaker than other proofs?.
There are some wonderful answers related to your question - and addresses, directly, your "aside": See, in particular, what JDH writes.

One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. That is, direct proofs help clarify the necessary and sufficient conditions that make a theorem true, and provide a structure demonstrating how these conditions relate, and how the chain of implications imply the conclusion.

Indirect proofs, on the other hand (aka "proofs by contradiction") only tell us that supposing a proposition to be otherwise leads to a contradiction at some point. But such a proof doesn't really provide the sort of insight that can be gained from direct proofs.

That is not to say that indirect proofs don't have their place (e.g., they come in handy when asked to prove propositions during a time-limited exam!). They often help "rule out" certain propositions on the basis that they contradict well established axioms or theorems. Also, indirect proofs are sometimes more intuitive than direct proofs. For example, proving that $\sqrt{2}$ is not rational using a proof by contradiction is clean, and intuitive.

Sometimes an indirect proof will emerge first, after which one can seek to proceed with trying to construct a direct proof to prove the same proposition. That is, providing an indirect proof of a proposition often motivates the construction of direct proofs.


I found this blog entry (Gowers's Weblog) When is a proof by contradiction necessary. from which I'll quote an introductory remark:

It seems to be possible to classify theorems into three types: ones where it would be ridiculous to use contradiction, ones where there are equally sensible proofs using contradiction or not using contradiction, and ones where contradiction seems forced. But what is it that puts a theorem into one of these three categories?

The post follows immediately with a nice reply from Terence Tao.

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  • Is proving $\sqrt 2$ is not a rational possible by direct proof? – sonicboom Nov 24 '12 at 16:30
  • Somebody else posted that they had never encountered one either so I am wondering if this is a case where proof by contradiction is the only way of proving the fact. – sonicboom Nov 24 '12 at 16:41
  • I think your question (in the title) is good, and probably not yet fully addressed by any of the answers yet. – amWhy Nov 24 '12 at 16:43
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    The definition of "irrational" is "not rational", so the statement "$\sqrt{2}$ is irrational" is inherently a negative one. As such there is no "direct" proof (unless one somehow comes up with a definition of "irrational" that does not invoke "rational"). – Zhen Lin Nov 24 '12 at 16:48
  • @ZhenLin I agree. – amWhy Nov 24 '12 at 16:53
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    What if you prove that for $n,m\in\mathbb Z$ and $m\ne0$, the distance between $n/m$ and $\sqrt{2}$ is at least $1/(3m^2)$? Could that qualify as a "direct" proof of irrationality? – Michael Hardy Nov 24 '12 at 17:34
  • Well, that just shifts the contradiction elsewhere, no? Consider how the rest of the proof would go: we then assume $\sqrt{2} = n / m$, in which case $0 \ge 1 / (3m^2) > 0$, a contradiction. – Zhen Lin Nov 24 '12 at 18:03
  • @sonicboom: the usual proof that $\sqrt{2}$ is irrational is a direct proof that "$\sqrt{2}$ is rational" implies $\bot$, and so the usual proof is a direct proof that $\sqrt{2}$ is not rational. – Carl Mummert Nov 24 '12 at 18:12
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    @Michael Hardy: in fact some constuctivists actually take that to be the definition of *irrational*, so that this is stronger to them than just "not rational". But the classical definition of "irrational" is just "not rational", and that is the perspective of my previous comment in this thread. – Carl Mummert Nov 24 '12 at 18:28
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    @sonicboom http://mathoverflow.net/questions/32011/direct-proof-of-irrationality http://math.stackexchange.com/questions/20567/irrationality-proofs-not-by-contradiction – MJD Nov 24 '12 at 22:10
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    @ZhenLin Don't you need something more specific than "not rational" to define "irrational"? Are imaginary numbers, matrixes, dogs, and mattresses rational? Are they irrational? – Peter Olson Nov 25 '12 at 04:15
  • @amWhy: You say *"One of the advantageous to constructing direct proofs of propositions, when this is feasible, is that one can discover other useful propositions in the process. "* I agree but proofs by contradiction (or rather, attempts to proofs) can also lead to wonderful discoveries (non-euclidean geometry comes to mind). – ypercubeᵀᴹ Nov 29 '12 at 20:24

A few points from my (limited) experience:

  • I love proof by contradiction and I have used it in graduate level classes and no one seemed to mind so long as the logic was infallible.
  • For me, it's much easier to think about a proposition in terms of "What if this wasn't true?". That is usually my first instinct, this makes proof by contradiction the natural first choice. For instance, if I were to be asked to prove something like "Prove that a non-singular matrix has a unique inverse". My first instinct would be "What if a non-singular matrix had 2 inverses?" and from then on, the proof follows cleanly.
  • Sometimes, however, contradictions don't come cleanly and proof by simple logical deductions would probably take 5 lines whereas contradiction will take millions. I could point you to specific proofs but I'll have to do some digging. Further, if you look at every proof and try using Proof by Contradiction, another problem you will face is that sometimes, you will state your intended contradiction but never use it. In other words, solve using direct proof.
  • Another aspect about Proof by Contradiction (IMHO) is that you really must know all definitions and their equivalent statements fairly well to come up with a nice contradiction. Else, you will end up proving several lemmas on the way which looks clean in a direct proof but not so much in a Proof by Contradiction, but again, this might be a personal choice.

In summary, if you find it easier to think in terms of "What if not" then go ahead, use it but make sure your proof skills using other strategies are as good because $\exists$ nail that you cannot hit with the PbC hammer that you'll carry.

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What is a proof by contradiction? This is actually quite difficult to answer in a satisfactory way, but usually what people mean is something like this: given a statement $\phi$, a proof of $\phi$ by contradiction is a derivation of a contradiction from the assumption $\lnot \phi$. In order to analyse this, it is very important to distinguish between the statement $\phi$ and the statement $\lnot \lnot \phi$; the two statements are formally distinct (as obvious from the fact that their written forms are different!) even though they always have the same truth value in classical logic.

Let $\bot$ denote contradiction. When we show a contradiction assuming $\lnot \phi$, what we have is a conditional proof of $\bot$ from $\lnot \phi$. This can then be transformed into a proof of the statement $\lnot \phi \to \bot$, which is the long form of $\lnot \lnot \phi$ – in other words, we have a proof that "it is not the case that $\lnot \phi$". This, strictly speaking, is not a complete proof of $\phi$: we must still write down the last step deducing $\phi$ from $\lnot \lnot \phi$. This is the point of contention between constructivists and non-constructivists: in the constructive interpretation of logic, $\lnot \lnot \phi$ is not only formally distinct from $\phi$ but also semantically distinct; in particular, constructivists reject the principle that $\phi$ can be deduced from $\lnot \lnot \phi$ (though they may accept some limited instances of this rule).

There is one case where proof by contradiction is always acceptable to constructivists (or at least intuitionists): this is when the statement $\phi$ to be proven is itself of the form $\lnot \psi$. This is because it is a theorem of intuitionistic logic that $\lnot \lnot \lnot \psi$ holds if and only if $\lnot \psi$. On the other hand, it is also in principle possible to give a "direct" proof of $\lnot \psi$ in the following sense: we simply have to derive a contradiction by assuming $\psi$. Any proof of $\lnot \psi$ by contradiction can thus be transformed into a "direct" proof because one can always derive $\lnot \lnot \psi$ from $\psi$; so if we can obtain a contradiction by assuming $\lnot \lnot \psi$, we can certainly derive a contradiction by assuming $\psi$.

Ultimately, both of the above methods involve making a counterfactual assumption and deriving a contradiction. However, it is sometimes possible to "push" the negation inward and even eliminate it. For example, if $\phi$ is the statement "there exists an $x$ such that $\theta (x)$ holds", then $\lnot \phi$ can be deduced from the statement "$\theta (x)$ does not hold for any $x$". In particular, if $\theta (x)$ is itself a negative statement, say $\lnot \sigma (x)$, then $\lnot \phi$ can be deduced from the statement "$\sigma (x)$ holds for all $x$". Thus, proving "there does not exist an $x$ such that $\sigma (x)$ does not hold" by showing "$\sigma (x)$ holds for all $x$" might be considered a more "direct" proof than either of the two previously-mentioned approaches.

Can all proofs by contradiction be transformed into direct proofs? In some sense the answer has to be no: intuitionistic logic is known to be weaker than classical logic, i.e. there are statements have proofs in classical logic but not intuitionistic logic. The only difference between classical logic and intuitionistic logic is the principle that $\phi$ is deducible from $\lnot \lnot \phi$, so this (in some sense) implies that there are theorems that can only be proven by contradiction.

So what are the advantages of proof by contradiction? Well, it makes proofs easier. So much so that one algorithm for automatically proving theorems in propositional logic is based on it. But it also has its disadvantages: a proof by contradiction can be more confusing (because it has counterfactual assumptions floating around!), and in a precise technical sense it is less satisfactory because it generally cannot be (re)used in constructive contexts. But most mathematicians don't worry about the latter problem.

Zhen Lin
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As "Inquest"'s answer mentions, it's often easier to find a proof by contradiction than a direct proof. But after you do that, you can often make the proof simpler by rearranging it into a direct proof. It is not good to make a proof appear more complicated than it really is.

To see another disadvantage of some proofs by contradiction, consider this:

Proof: To prove $A$, assume not $A$. [insert 50 pages of argument here] We have reached a contradiction. Therefore $A$. End of proof

Now ask yourself: Which of the propositions proved in those 50 pages are erroneous and could be proved only because one relied on the false assumption that not $A$, and which are validly proved, and which are true but not validly proved because the assumption that not $A$ was relied on? It's not so easy to tell without a lot more work. And if you remember a proof of one of those propositions, you might just mistakenly think that it's been proved and is therefore known to be true. So it might be far better to limit the use of proof by contradiction to some portions of those 50 pages where no other method works.

Perhaps proofs of non-existence can be done only by contradiction. Here I might offer as an example the various proofs of the irrationality of $\sqrt{2}$, but for the fact that I've seen it asserted that if $m$, $n$ are integers, than $m/n$ differs from $\sqrt{2}$ by at least an amount that depends on $n$ --- I think it might have been $1/(3n^2)$. Here's another example: How would one prove the non-existence of a non-trivial (i.e. $>1$) common divisor of $n$ and $n+1$?

I've seen a book on logic asserting that a proof by contradiction of a non-existence assertion does not constitute an "indirect proof", since the assertion is inherently negative. I don't know how conventional that is.

Michael Hardy
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    Formally, a negative statement such as $\lnot \phi$ is an abbreviation for $\phi \to \bot$, i.e. it is the statement that $\phi$ leads to an absurdity; accordingly, to prove $\lnot \phi$, one must show that assuming $\phi$ leads to contradiction. This is completely orthodox logic. – Zhen Lin Nov 24 '12 at 16:46
  • But when must a statement be written in the form $\lnot\varphi$ ? – Michael Hardy Nov 24 '12 at 16:50
  • There isn't any "must" about it. If $\phi$ is a compound formula such as $\psi \lor \theta$ you can just push the $\lnot$ inward and get $(\lnot \psi) \land (\lnot \theta)$. But that just causes the number of negative statements to multiply... – Zhen Lin Nov 24 '12 at 17:01
  • So when _must_ it be written in a form that has negative statements? Does the answer depend on which formal language you write it in? – Michael Hardy Nov 24 '12 at 17:31
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    Of course. And it depends on what things you take as primitive. (Is $=$ more primitive than $\ne$? Sometimes constructivists take an "apartness" relation $\mathrel{\#}$ to be primitive, in which case $=$ becomes the negation of $\mathrel{\#}$!) – Zhen Lin Nov 24 '12 at 18:05

Another example of a contradiction proof that provides no idea on a constructive proof is the strategy-stealing argument. For certain symmetric games, the second player cannot have a winning strategy. If he did, the first player could "pretend" to be the second player and steal his winning strategy, stealing it from him, a contradiction.

An interesting example is the game Hex. It is easy to show that Hex cannot end in a tie, and the strategy-stealing argument does apply to it. Therefore, it is a first player win. But for symmetric $n$ x $n$, the actual winning strategy is still not known. Thus, this is an example of something that has been proven using contradiction and not constructively (yet).

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    I would formulate it as: Let $S$ be any strategy for the second player. Then by symmetry the first player can apply $S$ too. Since not both players can win and both are applying $S$, $S$ is not a winning strategy. – WimC Nov 27 '12 at 21:31

There is nothing wrong with proof by contradiction. You can show that they work using a truth table. In the end, that's all that really matters, right?

As far as I know, you can't know for certain that something is not provable by a direct proof. However, a proof by contradiction might be an easier way to prove some things, like the irrationality of certain numbers. For example, I have never seen a direct proof of the irrationality of $\sqrt{2}$.

EDIT: As Carl Mummert said in his answer, the above part in italics is not true. There are propositions which are only provable by contradiction.

A proof by contradiction can be also be formulated as a proof by contrapositive. If we know $Q$ is false, if we can show $P\Rightarrow Q$ then we have proved that $P$ is false. Whether you view this as "proof without contradiction" or not is up to you. In any case, they are logically equivalent.

Espen Nielsen
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    As I explained in my answer, we do know that there are things that are not provable directly modulo the usual formalization of proofs. The usual proof that $\sqrt{2}$ is not rational is a direct proof when it is formalized in the usual way, although it is a direct proof of a negative statement, which can be deceptive at first. – Carl Mummert Nov 24 '12 at 19:10
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    I found the link I had lost, which explains some of this: http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/ – Carl Mummert Nov 24 '12 at 22:43
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    Thank you for pointing this out, and for the reference. Both your comment and the article were very informative and enlightening! – Espen Nielsen Nov 24 '12 at 23:12

First of all, this is not an answer to the title but to the aside question and is just an example of why you would prefer a constructive proof to a proof by contradiction. Consider the example below,

Prove that $x^2 = 1$ has a root.

Proof by contradiction: Assume that $x^2 = 1$ has no root. Let $f(x) = x^2 - 1 $ then $x^2 = 1$ has a root if and only if $f(x_0) = 0$ for some $x_0$. By assumption $x^2 =1$ has no root and thus, $f(x) \neq 0$ for every $x$. Note that $f$ is continuous and $f(0) = -1$ and $f(2) = 3$. Hence, by the intermediate value theorem, $\exists x_0$ such that $f(x_0) = 0$ which is a contradiction. Therefore, $x^2=1$ has a root.

Constructive proof: For $x^2=1$ if and only if $x^2-1=0$ iff $(x-1)(x+1)=0$. Hence, for $x=\pm 1$ the equation is satisfied, namely the roots are $-1$ and $1$.

The difference is not about the length of the proofs but the information you have. In constructive proof, you know what the roots are but not in the proof by contradiction. Of course, in proof by contradiction, you could have said "let $x_0 = 1$, then $x_0^2=1$ which is a contradiction since $1$ is a root." but then, it is not a clear distinction between the two types of proofs.


I do believe there are some proofs that are only demonstrable through contradiction, and I'm going to attempt to describe them logically:

Let X be a logical statement such that: X $\rightarrow$ y, where y is a known contradiction (such as 2+2=5 in the normal arithmetic structure). Without knowing anything else of X, $\neg$X implies nothing and nothing implies $\neg$X (and hence is not provable). But, of course, assuming X implies a contradiction, and thus, $\neg$X.

This form of statement X is isolated, in that it only relates to itself and the contradiction. I do believe they can be constructed though, for it seems it they can be described.

With that said, in real math and logic, or in general real world scenarios, I don't believe any statements of this form exist, except possibly ones that are constructed to meet this criteria and otherwise meaningless. The proof of the primes was eventually proved without contradiction, to my understanding; until math had been more developed, I think that the statement "the number of primes is $\infty$" was basically an isolated logical statement at Euclid's time and for many years after probably, in that there were no other things known to imply it and it didn't imply anything else useful towards its proof.

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My non-mathematical response.

A == B equals !(A != B)

You always end up with a binary decision, is or is not. And in any language is = !(is not).

But I guess it is too simple to be ok.

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Bart Calixto
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    `is = !(is not)` does not hold in minimal and intuitionistic logic. See [Carl's answer](http://math.stackexchange.com/a/243866/107671). – user26486 Feb 24 '15 at 18:03

An interesting example of this is the entire study of Smooth Infinitesimal Analysis. It relies on not having the law of the excluded middle (i.e. no proof by contradictions are accepted) in order to be valid. Thus if everything provable by contradiction was also able to be proven directly, then there could not be smooth infinitesimal analysis! Look at Bell's book for more details, though the wiki gives a good example.

Chris Rackauckas
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Whether a proof is "by contradiction" really just depends on the statement you started with. If your inital statement is $P \rightarrow Q$, then showing the equivalent $\neg Q \rightarrow \neg P$ is "proof by contradiction". But in reality, the "direct" proof for $ P \rightarrow Q$ is just a proof "by contradiction" for $\neg Q \rightarrow \neg P$. The only reason why we started with $P \rightarrow Q$ instead of $\neg Q \rightarrow \neg P$ is our intuition.

This is just my opinion, but also remember that sometimes, it is also very valuable to know what holds if $Q$ does not hold.

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  • It really depends by what you accept as proof. Every logical system has certain logical axioms, nobody is forcing you to accept them. $\neg \neg A \leftrightarrow A$ is in fact not provable in intuitionistic logic. – Panayiotis Karabassis Nov 25 '12 at 00:41
  • ok fine. while i don't know what intuitionistic logic is, i was assuming that the law of excluded middle would hold. – kutschkem Nov 27 '12 at 11:21

Following Carl Mummert in considering the three main systems of propositional logic, let's re-interpret the question once again as

Does there exist a proof by contradiction that is valid in Classical Logic, yet invalid in Minimal Logic (resp. Intuitionistic Logic)?

The systems of Minimal Logic, Intuitionistic Logic and Classical Logic are three systems of propositional logic of strictly increasing strength. (I will be using the textbook 'Foundations of logic and mathematics' by Nievergelt as a reference, especially Sections 1.1 and 4.1.) To begin to answer this question, we first need to formalise what 'proof by contradiction' is, as a logical principle. Let us look at two examples.

Take first the usual proof of the infinitude of primes: Suppose $p_1, \ldots, p_N$ is the list of all primes. Then the smallest prime factor of $p_1 \cdots p_N + 1$ is larger than $p_N$. Thus there are infinitely many primes. The underlying logical principle applied at the word 'thus' is the so-called Law of Reductio Ad Absurdum: $$(P \to Q) \to ((P \to \neg Q) \to \neg P),$$ where $P$ is '$p_1, \ldots, p_N$ is the list of all primes' and Q is 'the smallest prime factor of $p_1 \cdots p_N + 1$ is a prime larger than $p_N$'. So if the Law of Reductio Ad Absurdum is valid, then the proper conclusion of the above proof is that $p_1, \ldots, p_N$ is not the list of all primes, which is reasonable as a possible definition of the infinitude of primes (where the prefix 'in-' means 'not', so that 'in-finite' means 'not-listable').

There is another kind of proof of contradiction, namely of the Pigeonhole Principle: Given $n$ holes, if $n + 1$ pigeons are put into them, then there must be some hole with at least two pigeons. So the proof goes: Were there no hole having at least two pigeons, then at most $n$ pigeons were put into the $n$ holes. Thus, if $n + 1$ pigeons were put into the $n$ holes, then there is some hole with at least two pigeons. And the underlying logical principle at the word 'thus' is now the so-called Converse Law of Contraposition: $$(\neg P \to \neg Q) \to (Q \to P),$$
where $P$ is 'there is some hole with at least two pigeons' and $Q$ is '$n + 1$ pigeons are put into the holes'.

By these two examples, I hope that the reader sees and is convinced that what is generically regarded as 'proof by contradiction' is formalisable as either the the Law of Reductio Ad Absurdum or the Converse Law of Contraposition, which are two separate laws distinct from each other.

The subtlety now arises that in fact

  1. the Law of Reductio Ad Absurdum is valid in Minimal Logic, in Intuitionistic Logic and in Classical Logic.
  2. The Converse Law of Contraposition is not valid in Minimal Logic (resp. Intuitionistic Logic). However, adding the Converse Law of Contraposition to Minimal Logic (resp. Intuitionistic Logic) gives a logic equivalent to the full Classical Logic (see Appendix).

So finally, we can arrive at an answer to the re-interpreted question in the yellow box. For our first example, the proof of the infinitude of primes uses 'proof by contradiction' in the sense of the Law of Reductio Ad Absurdum. This proof is valid in Classical Logic, but by (1), is also valid in Minimal Logic and in Intuitionistic Logic. However, for our second example, the proof of the Pigeonhole Principle uses 'proof by contradiction' in the sense of the Converse Law of Contraposition. Although this proof is valid in Classical Logic, by (2), it is not valid in Minimal Logic nor in Intuitionistic Logic. So we must be careful not to reject as non-intuitionistic or non-minimalistic those proofs in classical mathematics that uses the Law of Reductio Ad Absurdum, and inspect carefully whether it is this law or the Converse Law of Contraposition that is being employed.


For the convenience of the reader, I write down the axioms of these three systems of logic, as taken from Nievergelt's book. One of the purposes of writing this down is that in @Carl Mummert's answer, he uses a constant symbol $\bot$ to denote the falsum. However, it is possible avoid the falsum and to write down the axioms of Minimal Logic, Intuitionistic Logic and Classical Logic completely over the language $\{\neg, \to, \vee, \wedge\}$, with the symbol $\neg$ for negation, the symbol $\to$ for implication, the symbol $\vee$ for disjunction and the symbol $\wedge$ for conjunction. In this language, the use of a constant symbol $\bot$ for the falsum is avoided.

To give the details, let $CL^-$ be the system consists of the following two axiom schemas:

  1. $P \to (Q \to P)$
  2. $(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$

Then Classical Logic (CL) is $CL^-$ together with the Converse Law of Contraposition (p.58).

Let $T$ denote $CL^{-}$ together with the additional five axiom schemas:

  1. $(P \wedge Q) \to P$
  2. $(P \wedge Q) \to Q$
  3. $P \to (Q \to (P \wedge Q)$
  4. $P \to (P \vee Q)$
  5. $Q \to (P \vee Q)$
  6. $(P \to R) \to ((Q \to R) \to ((P\vee Q) \to R))$

Then Minimal Logic (ML) is $T$ plus the Law Of Reductio Ad Absurdum (p.228). And, Intuitionistic Logic (IL) is $T$ plus the Special Law of Reductio Ad Absurdum -- $$(P \to \neg P) \to \neg P$$ and also plus the Law of Denial Of the Antecedent $$\neg P \to (P \to Q).$$

The facts are $ML + \text{Law of Denial Of the Antecedent} \Leftrightarrow IL$ (Exercise 755, p.231), $ML + \text{Law of Double Negation} \Leftrightarrow CL$ (Theorem 653, p.229) and $IL + \text{Law of Double Negation} \Leftrightarrow CL$ (Exercise 754, p.231). Since the Law of Reductio Ad Absurdum is an axiom of ML, hence it is valid in both $IL$ and $CL$. Next, $ML$ is strictly weaker than $IL$ since the Law of Denial of the Antecedent is not valid in $ML$. And also $IL$ is strictly weaker than $CL$ since the Law of Double Negation is not valid in $IL$. Hence, the Law of Double Negation being the special case of the Converse Law of Contraposition by taking $Q = \top$ as the verum, the Converse Law of Contraposition is not valid in $ML$ nor in $IL$.

Colin Tan
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