In a proof by contradiction, how do we know the assumption is the cause of the contradiction? And not just the result of some other property more fundamental to numbers?

In other words, how can we be sure we arrived at the contradiction because of the proposition $P$ we assumed, and not because of some other proposition $Q$ we are not even aware of?

The answer that comes in mind is "Well, if there were some proposition $Q$ causing the contradiction, then it should also manifest in the event of $ \neg P$."

Is it this straightforward or there's more to it? What if the proposition $Q$ can only manifest in the event of $P$, and not at all in the event of $\neg P$?

Sorry if this is nonsense - just trying to clear out some confusions.

M. Winter
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    If you don't truly know if the the proposition Q is true, maybe you shouldn't be using it in your proof in the first place. – WeakestTopology Oct 06 '17 at 22:35
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    If you read Wikipedia [Proof by contradiction](https://en.wikipedia.org/wiki/Proof_by_contradiction) you see that the proof method depends on the Law of the excluded middle and that is one reason it is objectionable by some mathematicians. Your question is a very important one which is easily "swept under the rug". You are justified in questioning the method. – Somos Oct 07 '17 at 01:00
  • @Somos Glad to read I was not completely confused. I'll look up the Law of the excluded middle in hopes of more clarity. Thanks. – Stephen Oct 07 '17 at 01:31
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    @Somos: Proof by contradiction does not rely on the law of the excluded middle: it only requires the law of noncontradiction. The invocation of the law of the excluded middle that is often associated with a proof by contradiction is about proving a proposition from its double negation. –  Oct 07 '17 at 04:44
  • Possible duplicate of [Are proofs by contradiction really logical?](https://math.stackexchange.com/questions/1667884/are-proofs-by-contradiction-really-logical) – user21820 Oct 07 '17 at 07:32
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    Athough what Hurkyl said above is true for **some** definitions of "proof by contradiction", the typical variant that mathematics students encounter in textbooks and classes are actually of the form "$\neg A \to \bot \vdash A$", which **does** require LEM (the law of excluded middle) to justify fully. Hence I think the link I've provided will give a fuller answer to your question. – user21820 Oct 07 '17 at 07:37
  • @somos It is also worth pointing out that the problem Stephen is asking about has nothing to do with the excluded middle, but rather with there not being contradictions in the axioms. Of course for (I want to say most) a fair amount of proofs by contradiction you don't actually need to prove the result that way it's just easier. – DRF Oct 07 '17 at 15:23
  • It is also worth noting that people sometme make mistakes and so if you end up in a contradiction one possibility is that a mistake in reasoning is the cause. How can you decide between this possibility and/or that the proposition P is false? – Somos Oct 07 '17 at 15:39
  • @Morgan Rodgers Because in a direct proof you can check each of the statements for correctness. In a proof by contradicition you can't. – Somos Oct 07 '17 at 17:20
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    Checking a statement is not the same as "proving" it. – Somos Oct 07 '17 at 22:34
  • @DoughnutPump I now see what you mean. – Stephen Oct 07 '17 at 23:10
  • $Q$ cannot originate from outside of our system of axioms. So, either it was *unknowingly*/*accidentally* assumed in the proof (in which case the proof is not a valid one - but not because of the method of a PbC), or it has already been proven in the system in which we are working, in which case, it cannot be the cause of the contradiction (unless some of our axioms are inconsistent - which is by definition false). So indeed, back to proof-101 - be sure of the veracity of every statement! To avoid introducing a ''$Q$'' in it. – Stephen Oct 07 '17 at 23:19
  • @DoughnutPump: Your comment is inaccurate. See my answer for an explicit example (at the end) where we can have some assumption that contributes to the contradiction but is totally part of a valid proof. In fact, in many proof systems (especially Fitch-style) you necessarily have to consider assumptions in order to prove certain statements (such as conditional statements), **regardless of** whether you know it's true or not. – user21820 Oct 08 '17 at 11:21
  • @Somos: Your comment raises an interesting issue, but perhaps it may not be so clear to other readers. What you are saying is that in a direct proof (namely not using LEM or DNE or proof by contradiction) every statement that you deduce is empirically verifiable, whereas in an indirect proof, since you can deduce anything from a contradiction, it's entirely possible that every statement you have deduced under a false assumption is also false, and so one cannot 'double-check' each by testing it independently. That's indeed true, and you may want to look up the BHK interpretation for more. – user21820 Oct 08 '17 at 11:26
  • @user21820 Thanks, you got it right. That is what I am claiming. Every statement in a direct proof can be spot checked for validity, In a contradiction proof, statments can be true or false so you have to fall back on making sure that each step follows logically from previous steps. My concern is more for validity of statements than validity of proof methods, although proofs should use valid inferences but it's harder to check. I knew about BHK type interpretations, but I am not into proofs as object of manipulation. – Somos Oct 08 '17 at 13:08
  • @Somos: By the way, you're welcome to the [logic chat-room](https://chat.stackexchange.com/rooms/44058/logic) if you'd like further discussion, including about the posts I've linked. =) – user21820 Oct 09 '17 at 05:40
  • How about a simple case: Where one of the prior assumptions (a1,a2,a3..) was really an assumption, not a logic deduction Then you might have just injected a proposition that was an alternative to a2. Now your argument doesn't imply new mathematics but that you inserted a proposition that had a different branch than the internally assumed a2. Neither is correct or incorrect logically; just inconsistent with each other. For example, Euclid's parallel line "axiom". If I have misinterpreted your question, let me know and I will delete this! – rrogers Oct 10 '17 at 21:13
  • You might have a look at this : http://math.mit.edu/~cohn/Thoughts/contradiction.html – Watson Sep 26 '19 at 14:16

5 Answers5


In mathematics, we deduce theorems from axioms, so we end up knowing that the theorems are true in any situation where the axioms are true. We get no information about situations where some of the axioms are false.

Now suppose we prove a theorem T by contradiction; we assume not-T and, by means of this assumption plus our axioms, we arrive at a contradiction. Then we know that either not-T is false or (at least) one of our axioms is false. That knowledge is exactly the same as knowing that if the axioms are true, then not-T must be false and therefore T must be true. If, on the other hand, some of the axioms are false, then we get no information.

So the information we get from the proof by contradiction is the same as what we'd get from a direct proof: As long as the axioms are true, we know T; if some axioms are false, all bets are off.

Andreas Blass
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  • I'm not fully sure I understand what you mean by 'false' axioms. Isn't an axiom by definition true, and so all statements deduced from them are also true in the system formed by the axioms? I'm not sure I understand what is a 'false' axiom. – Stephen Oct 06 '17 at 22:56
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    @Stephen One of the usual axioms about the natural numbers is that $0$ is not the immediate successor of anything. If someone were to take theorems about the natural numbers and claim that they're also true about the real numbers, I'd tell him that he's wrong because that axiom, true of the natural numbers, is false of the real numbers. More generally, axioms are (barring errors) true in the situation that they're intended to describe, but they can be false in other situations. – Andreas Blass Oct 06 '17 at 23:28
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    Would it be more correct to say that the axioms are *inconsistent* than to say that they're *false*? – templatetypedef Oct 07 '17 at 01:14
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    @templatetypedef No. Consistent axioms, like the standard axioms about $\mathbb N$, can be false when one attempts to apply them to other contexts like $\mathbb R$. If the axioms were inconsistent, they'd be false in **all** situations. – Andreas Blass Oct 07 '17 at 01:32
  • Even properties, can become false if applied in the wrong context. That's partially why, this site is so hung up on specifics for problems. without conventions and context if I say ab without defining with context what a or b are, it could be string concatenation, or multiplication,etc. and a could be a real number, and maybe so is b so what's the value of ab if they are real is this true of when they aren't ? –  Oct 07 '17 at 01:47
  • [We do not know that the axioms we use are non-contradictory (whence it could be the case that they cannot be true all at once)!](https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem) – yo' Oct 07 '17 at 11:26
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    @yo' I assume that by "we do not know" you mean something like "we cannot prove, on the basis of those same axioms," so that your comment is justified by Gödel's second incompleteness theorem. If we consider other means of knowing things, the justification is a lot weaker. For example, a clear intuition of the cumulative hierarchy may well allow us to know that ZFC is consistent. – Andreas Blass Oct 07 '17 at 13:36
  • @Stephen I think your confusion about "aren't axioms true" might go away if you called them "hypotheses" and "definitions" relevant in your probem, leaving "axioms" to refer to the usual rules of logic and proof which is really OK since you're not asking about fundamentals that tangle with things like Godel's theorem. – Ethan Bolker Oct 07 '17 at 13:42
  • @AndreasBlass I don't know. Given some of the work I've seen on pulling back theorems to minimal axiomatic structures, ZFC might just be way too strong to be consistent. I'm joking a bit, but it really is a huge sledge hammer for 99% of mathematics. I still really like it of course. – DRF Oct 07 '17 at 15:27
  • @AndreasBlass Yes, that's what I mean. I should have said "we cannot know" of course! – yo' Oct 07 '17 at 15:32
  • @DRF I agree that ZFC is more than needed for most of mathematics. Nevertheless, the intuition of the cumulative hierarchy of sets shows that it's consistent. Once I know that ZFC is consistent, I can, of course, deduce the consistency of weaker theories, like those used in reverse mathematics. – Andreas Blass Oct 07 '17 at 15:34
  • @DRF: Incidentally, I think ZFC is elegant and consistent, but I'm not sure whether it's arithmetically sound. And I believe that nearly all the mathematics needed to develop real-world applications can be done in ACA. What do you think? – user21820 Oct 07 '17 at 15:35
  • @AndreasBlass: I'm personally not convinced that one can have an intuition of the cumulative hierarchy without assuming something that is more or less equivalent to the replacement schema from ZFC. In fact, a number of logicians in history have pointed out this exact defect in the iterative conception, including Boolos. It seems we **can** reasonably justify countable stages, but do not seem to be able to justify a stage after all of them, without assuming the existence of an uncountable well-ordering of labels for all countable stages. Any thoughts about that? – user21820 Oct 07 '17 at 15:38
  • @user21820 By Gödel's theorem, any intuition adequate for showing that ZFC is consistent (like the intuition of the cumulative hierarchy) cannot be formalized without using some assumption that goes beyond ZFC. But I was not talking about formalizing the intuition, just about using it in order to see that ZFC is consistent. – Andreas Blass Oct 07 '17 at 15:44
  • @AndreasBlass: Yes I know that, but like with PA, it's not necessary that one needs something strictly stronger, so the question or hope is that someone can give a justification for ZF[C] or perhaps even just Con(ZF[C]) that requires less philosophical commitment in some aspects and more in some other aspects, that at least might seem plausibly more reasonable. More precisely, can you give a reason why you think it is more justifiable to believe that ZF is arithmetically sound than to believe that ZF is arithmetically unsound? – user21820 Oct 07 '17 at 16:43
  • @user21820 There is some evidence that ZF is arithmetically sound. (1) Its axioms are evidently true in the cumulative hierarchy. (2) No one has yet deduced from ZFC an arithmetical statement that we know (or even suspect) to be false. (Of course neither (1) nor (2) is as convincing as a mathematical proof would be, but they are nevertheless reasonable evidence.) In contrast, I don't know of any evidence that ZF is arithmetically unsound. – Andreas Blass Oct 07 '17 at 17:48
  • @AndreasBlass I think I understand what you mean in your original answer. Basically, all axioms are assumed to be consistent in a specific system. But, they may not be consistent in other systems. Is this what you mean by false axioms? That is, they are ‘false’ in the system in which they cause contradictions? – Stephen Oct 07 '17 at 23:08
  • I also understand why under this, a PbC is just as valid as a direct proof. Still PbC seems to be more heavily dependent on LEM than a direct proof. Why exactly is there a controversy about LEM? Is it more of a philosophy of mathematics thing (a bit like Gödel's theorems) or there are also solid grounds for questioning it in mainstream mathematics? – Stephen Oct 07 '17 at 23:09
  • @Stephen I was using "false" in the semantic sense of "false in a structure" (like $\mathbb R$), but similar things could be said about the syntactic notion of "inconsistent with some system of axioms." – Andreas Blass Oct 08 '17 at 00:00
  • @Stephen LEM is used if you prove $P$ by deducing a contradiction from $\neg P$. But LEM is not involved in proving $\neg P$ by deducing a contradiction from $P$. Logical systems that reject LEM are usually based not on "true vs. false" like classical logic but on "known vs. unknown" or "proved vs. unprovable". For more details, you should look up "intuitionism" and "constructive logic". – Andreas Blass Oct 08 '17 at 00:05
  • @Stephen: As explained in my linked post, there is little philosophical ground to question LEM for statements about reality. Mathematicians mostly also believe that statements about natural numbers (such as "Every positive integer is the sum of four squares.") are statements about reality in an abstract sense, and so would have no reason to question LEM for these statements. There are many other kinds of statements, and one may be justified to question LEM for them. – user21820 Oct 08 '17 at 11:07
  • @Stephen: Also, Godel's theorems are **not** a philosophy thing at all, and are absolutely true of **any** useful formal system that can reason about natural numbers or finite strings. Of course, the incompleteness theorems themselves can only be proven or justified in some formal system or logical framework that permits the required deductive steps, but the required assumptions are so mild that anyone who doubts the truth of the incompleteness theorems will also doubt the vast majority of concrete mathematics, not to say modern mathematics. – user21820 Oct 08 '17 at 11:11

This post explains the intuitive justification that the technique of proof by contradiction is valid. Hence I think it addresses one aspect of your question. But I wish to point out an additional aspect that closely resembles your question. Suppose we have the following proof structure:

  If $P$:

    If $Q$:


      Contradiction.   [Suppose we can deduce this. Is it 'because of' $Q$ or $P$?]

    $\neg Q$.   [Is this a valid deduction? I will explain below why it is!]

It is valid to deduce $\neg Q$ under the outer assumption of $P$, as shown above. Why? As explained in the linked post, all our deductive rules are designed to be truth-preserving, namely that every sentence we can deduce is true in its context (including the assumptions it is under). Therefore we cannot deduce a contradiction except in an impossible context. Now in the above proof structure it must be that it is impossible for both $P$ and $Q$ to be true, but from the given structure we cannot determine which is not true. This is exactly like what you are asking in your question! Nevertheless, the last deduction above is valid because under assumption that $P$ is true, it is impossible for $Q$ to be true too and hence $Q$ must be false.

It could very well be that $Q$ is actually true but $P$ is the false nonsense that allows you to deduce a contradiction under the assumptions of both $P$ and $Q$. Despite that, whatever we said above is still correct! Why? Well if $P$ is false, then what we can deduce under assumption that $P$ is true is irrelevant, because we are not deducing a sentence that is false under no assumptions.

For programmers, you can think of each sentence as an assert statement, and each "If" context-header as an if-construct. The very fact that our deductive rules are truth-preserving implies that every proof we write following our deductive rules is one that does not make any assertion error. Thus if we manage to prove a contradiction somewhere, it implies that when you run the proof as a program it will never reach that assert statement! So let us look at the above proof structure again. The only way you can reach the last statement is if $P$ is true, in which case $Q$ must be false since we cannot reach the "Contradiction" assertion. So when we reach $\neg Q$ we are indeed correct to assert it. The other possible situation is that $P$ is false, in which case we never even run all the statements under it, so it does not matter that we have made some assertions under it!

Finally, I want to just point out that it is not always the case that only one of our assumptions is the 'cause' when we can deduce a contradiction. Consider the following valid proof structure:

  If $P$:

    If $\neg P$:

      Contradiction.   [Who 'caused' this?]

    $\neg \neg P$.

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  • This was helpful. My knowledge in logic is quite limited though, so a quick one; isn't ¬¬$P$ $ $ just $ $ $P$? – Stephen Oct 07 '17 at 23:25
  • @Stephen: In classical logic, yes. But in other logics, like minimal logic or intuitionistic logic, that might or might not be the case. – hoffmale Oct 08 '17 at 05:33
  • @Stephen For a quick example of why it's not instantly obvious that $\neg \neg P = P$: "I'm not unhappy to see you" isn't the same thing at all as "I'm happy to see you". – Patrick Stevens Oct 08 '17 at 07:09
  • @Stephen: As hoffmale said, they are equivalent in classical logic, since negation simply flips the truth value, so double negation does nothing. In any case, that last example was merely to show that a contradiction can be 'caused' by the current assumptions being impossible **in conjunction**, rather than necessarily being due to some single wrong assumption. That's why I think my answer best addresses your question, since we often do proof by contradiction within some context under some other assumptions, yet it's valid to conclude the negation of the innermost assumption. – user21820 Oct 08 '17 at 11:01
  • @Stephen: You may be interested in [this post about some statements that can't have a boolean truth value](http://math.stackexchange.com/a/1888389/21820) (scroll to the part about Quine's paradox). In most modern foundations for mathematics, such statements are impossible to make, and so that is why we can still afford to have LEM in modern mathematics. But if we want to permit this kind of statements, then we can't have LEM for them. – user21820 Oct 08 '17 at 11:14

A proof by contradiction relies on the fact that ($P$ and not-$Q$) is the negation of ($P$ implies $Q$). The law of the excluded middle, along with the law of noncontradiction, gives that exactly one of these two statements is true, so if ($P$ and not-$Q$) leads to a contradiction it must be a false statement, and therefore ($P$ implies $Q$) is true. Note that if you obtain a contradiction that is unrelated to your assumption of ($P$ and not-$Q$) then you have shown that the universe you are working in is inherently contradictory, we usually like to think that that is not the case.

I would like to mention that the idea that the "law of the excluded middle" is controversial is... perhaps overrepresented in some parts of the internet compared to the opinions you would encounter among mathematicians. I would challenge you to try to, on your own, come up with a mathematical statement such that neither the statement nor its negation is true (or where both are true). I won't hold my breath.

Morgan Rodgers
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  • You misrepresent the most common view against the law of the excluded middle, which is not that it is not true, but rather that it should not be considered provable. Good arguments for this include that (1) LEM is usually not necessary (most theorems can be proven constructively); (2) even if that fails, any classical statement using LEM can be translated into a constructive statement that is proven without LEM, or simply proven with an additional hypothesis that LEM holds; (3) LEM is undesirable in logic because, unlike the other laws of inference, it is not simply "true by definition". – 6005 Oct 07 '17 at 16:24
  • I still like your first paragraph and the second one correctly points out that mathematicians are not bothered by such considerations about the underlying logic. (+1) – 6005 Oct 07 '17 at 16:26
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    @6005 I'm not intending to represent that as the common argument, though I see how it comes across that way. But yeah, really just trying to say that whether using LEM is the logically appropriate way to approach a proof is more a question for philosophers and logicians. (I would also say that the other answers support this). Mathematicians (in most areas of mathematics) are happy to take the fact that it is true a sufficient justification for using it. – Morgan Rodgers Oct 07 '17 at 17:12
  • @MorganRodgers You are assuming that every "mathematical statement" is either true or false or both. What if it is neither true or false? How do you eliminate that possibility? – Somos Oct 08 '17 at 13:23
  • @Somos Give me an example of a statement where that could be a possibility. I'm not interested in naval gazing. "For every $x$ satisfying P, we have that Q." Yes, I'm assuming something like this is either true for every $x$, or else there is an $x$ for which it is not true. – Morgan Rodgers Oct 08 '17 at 14:09

In other words, how can we be sure we arrived at the contradiction because of the proposition $P$ we assumed, and not because of some other proposition $Q$ we are not even aware of?

Let's consider the possibilities for $Q.$ I will assume the Law of the Excluded Middle, because I do not know enough about logic systems without it to say whether proof by contradiction works in any of them.

I also assume that the proof is set in some axiomatic system $A,$ since I do not know how to do proofs outside of any axiomatic system.

So we either have that $Q$ is always true under $A,$ or $Q$ is not always true under $A.$

Case 1

If $Q$ is always true under $A,$ and $Q$ leads to a contradiction, then $A$ is an inconsistent system. It really is not suitable for any kind of proof, including proof by contradiction.

Case 2

If $Q$ is not always true under $A,$ then either $Q$ is entailed by the axioms $A$ together with the assumption $P$ (but not by $A$ alone) or $Q$ is not entailed by the axioms $A$ together with the assumption $P.$ Let's split this case into two sub-cases.

Case 2a

If $Q$ is not entailed by the axioms $A$ together with the assumption $P,$ then $Q$ represents an unwarranted assumption that we made. This makes the proof invalid even if its conclusion is true. Our human fallibility makes it possible for such things to happen. It is possible to introduce such an assumption into a direct proof as well, making it equally invalid.

Case 2b

If $Q$ is entailed by the axioms $A$ together with the assumption $P$ but not by $A$ alone, then $P$ indeed is necessary in order to arrive at the contradiction, and the theorem is provable. But if $Q$ really is instrumental in making some of the steps in the proof, and we were not aware of this fact, then I think again we have an invalid proof of a true conclusion. Again, this is a possible result of human fallibility that can also happen in direct proofs.


Taking all the cases together, there really are only two ways for a proof by contradiction to go wrong. One is if we are working with an inconsistent set of axioms, which is very bad news altogether. The other is if we have an unjustified step somewhere in the proof.

An example of an "unjustified step" occurred famously in Andrew Wiles's first announcement that he had proved Fermat's Last Theorem. Someone (actually, I think multiple people) found a mistake in the proof he presented. After he made a considerable additional effort, he was finally able to present a proof without that mistake, and this proof was accepted.

Some comments under the original question raised the issue of how we check the intermediate steps of a proof by contradiction, claiming that it is easier to check the steps in a direct proof since their conclusions are all true.

Things that we want to prove typically have the form $S\implies T,$ for which a direct proof typically involves assuming $S$ and then showing that $T$ follows. In the intermediate steps of the proof, we have some facts that depend on $S,$ which we cannot "check" by simply observing that they are true; we can check them by verifying the logic in every step leading up to that part of the proof, or we can check them by coming up with an alternative proof showing that they follow from $S.$

We may also introduce some known facts (which do not depend on $S$) in the course of the proof, which we can check simply by verifying that they are true facts.

A third possibility is that we derive something from $S$ that we could have known to be true without assuming $S.$ This is wasteful; we could improve the proof by simply introducing these facts as known without showing a logical derivation from $S.$

The same things happen in proof by contradiction. We will have some steps that we can check only by checking every step in the logic leading up to them or by devising an alternative proof, we may have known facts that we can check more easily, and we may even have wasted effort by deriving something from our (false) assumption that we could have simply brought in as a known fact.

David K
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  • In your particular sense, yes a conditional statement may be difficult to check if the condition is rarely (or never) satisfied. But if you interpret [proofs as games](http://math.stackexchange.com/a/1782071/21820) (see the comments under it), every use of LEM invokes an oracle. If the proof is purely intuitionistic, then the Refuter can be constructively satisfied by the Prover's justification. If LEM is used, then it falls apart because the Prover is claiming a disjunction without having justified either. In this precise sense a direct proof is interactively checkable. See also BHK. =) – user21820 Oct 09 '17 at 04:38

It is not that by $P\Rightarrow \bot$ (the symbol $\bot$ denotes contradiction) we assume $\neg P$ just because it seems there is something "wrong" with $P$. It is more that we cannot avoid it anyway.

Either the contradiction arises from $P$ and it would be reasonable to assume $\neg P$, or the contradiction remains when removing $P$, but then the already existing inconsistent axiom system suffices to prove $\neg P$ (as well as $P$) using the principle of explosion. So in either way, adding $\neg P$ is reasonable $-$ or at least not more inconsistent as the axiom system itself.

M. Winter
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