In formal logic, it seems that a contradiction only arises between two statements. Is it possible to have a set of three statements that together are a contradiction, but where any two of the statements do not contradict each other?

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    A > B, B > C, C > A. – hide_in_plain_sight Feb 10 '20 at 17:29
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    @bob {A1, A2, …, An, ¬(A1 & A2 & … & An)}. – user76284 Feb 11 '20 at 1:33
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    @hide_in_plain_sight That should be an answer. It's the most simple to understand one from everything that I read here. – Fabian Röling Feb 11 '20 at 1:36
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    @hide_in_plain_sight, why would that be a contradiction? It's only a contradiction if the transitivity of the comparison operator is an axiom. For example it is a perfectly valid statement regarding the rules of a rock-paper-scissors game. – Boluc Papuccuoglu Feb 11 '20 at 19:40
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    @BolucPapuccuoglu And the commas could mean something else, too. Why assume people are actually trying to communicate when they say things? – hide_in_plain_sight Feb 11 '20 at 20:00

To answer specifically the question in your last sentence, the answer is yes. The set {P, P→Q, and ¬Q} is one such set. Any two formulas from the set do not contradict, but all together they do.

{P, P→Q} is consistent

{P→Q, ¬Q} is consistent

{P, ¬Q} is consistent

{P, P→Q, and ¬Q} is inconsistent. P and P→Q imply that Q, so that Q Λ ¬Q, which is a contradiction.

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    The set is certainly a contradiction. However, I'm not clear that this is a case in which 'three statements contradict each other'. This is because I'm not sure what 'each other' means precisely. But +1 for an adept answer. – Geoffrey Thomas Feb 10 '20 at 10:06
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    I'll note that propositional logic proofs have a tendency to only use two statements per deduction. So, in a formal proof you'd end up with two deductions, each using only two statements: {P, P→Q} ∴ Q, {Q, ~Q} ∴ ⊥ . This tendency to only operate on two statements is probably why OP only saw contradictions which arise from two statements. In real-world papers, often one or more formal steps in the proof will be abbreviated into one compound step. So, you might say {P, P→Q, and ¬Q} ∴ ⊥, with the understanding that your audience will be expand this out into explicit steps. – Brian Feb 10 '20 at 15:47
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    As a completely irrelevant aside, I'm sort of amazed that my answer has gotten so many upvotes... Sometimes I'll pour my heart and soul into a question/answer and be lucky to get +3... Here I probably spent about 5 minutes (the hardest party was copying and pasting the "→" and "¬" symbols lol...). – Adam Sharpe Feb 10 '20 at 16:33
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    @AdamSharpe This is quite common when questions are shown in the Hot Network Questions bar networkwide. You happened to offer a simple enough, sufficient answer early on. That way, many visitors who were intrigued by the title brought upvotes, much more than our usual traffic could possibly provide. And as soon as there are many upvotes, this becomes a self-reinforcing effect. – Philip Klöcking Feb 10 '20 at 20:27


  1. This sandwich has ham
  2. This sandwich has butter
  3. This sandwich does not have ham OR it does not have butter

Or less yummingly, consider (P, Q, not P OR not Q).

The nice property of our sandwich-making example is that it is easily extensible to n: just add more ingredients.

I'm hungry.

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    I think the demorgan'sed version is better: "This sandwitch does not have both ham and butter" is more clearly a contradiction with 1 and 2 – Cruncher Feb 11 '20 at 15:17
  • @Cruncher I see what you mean, I ended up wording it like this because I felt it was the least ambiguous, especially when extended to n. I wouldn't use "both" also for the sake of extending it to n easily. But feel free to edit if you have a way to improve the understanding of the example. – Julien Lopez Feb 11 '20 at 17:29

It depends on the precise definition of Contradiction.

If we define it "syntactically" as a pair made of a statement and its negation, the answer is obvious.

But we may define it "semantically",equating "contardictory" with unsatisfiable, then we have unsatisfiable sets with more than two formulas: P, Q, not-P or not-Q.


This is a multi sentence version of the Liar's paradox:

i) Statement iii) is false

ii) Statement i) is false

iii) Statement ii) is false

Any two of these sentences do not contradict each other until you define and consider the final statement.

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    "Any two of these sentences do not contradict each other" – That's not true. Statements i) and ii) contradict each other because statement ii) asserts that statement i) is false. Likewise, i) and iii) contradict each other, and ii) and iii) contradict each other. However, the three sentences together form a liar's paradox, whereas any two of them merely form a contradiction, not a liar's paradox. – Tanner Swett Feb 10 '20 at 17:35
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    "Statements i) and ii) contradict each other because statement ii) asserts that statement i) is false." Why is that a contradiction at all? $\lnot \phi_3 \land \lnot \phi_1" certainly isn't contradictory. "Being false" is not a contradiction. "5 < 3" isn't a contradiction, just false. – Polygnome Feb 10 '20 at 22:28
  • I would have build the 3 statements "circulatory" 1 -> 2 is false , 2-> 3 is false , 3 -> 1 is false ... but even in the given example just build a Bool's logic table .. and you see that it only becomes a contradiction once you demand all 3 statements at the same time ... – eagle275 Feb 11 '20 at 12:50
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    @Polygnome Two statements form a contradiction if they cannot both be true at the same time. Statement i) and statement ii) cannot both be true at the same time. Therefore, statement i) and statement ii) form a contradiction. – Tanner Swett Feb 11 '20 at 13:11
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    @TannerSwett There is not contradiction until you consider what statement iii) is. Consider two further statements iv) and v), there is no contradiction between the following: iv) Statement v) is false, and v) statement vi) is false – Matt Feb 11 '20 at 16:14

As pointed by @hide_in_plain_sight, A>B, B>C, C>A is certainly "... a set of three statements that together are a contradiction, but where any two of the statements do not contradict each other".

But the title question is different: A>B, B>C, C>A are NOT "three statements that contradict each other". For any value of A, B and C, there will always be a case where two of this set do not contradict each other. For example, if the values are 1>2, 2>3, 3>1, the first two statements do not contradict each other. A case of three statements contradicting each other would be {A>B, B>A, A=B}. For such question, most answers fall into the same mistake.


I suspect the simplest is:

1) P is true
2) Q is true
3) P and Q are not both true

Any two of these can be true with no contradiction.

Equally good:

1) P
2) not Q
3) P != Q

  • use predicate logic instead... true/false is not enough. – Dee Feb 17 '20 at 15:40
  • @Dee Why do you say that? – David Schwartz Feb 17 '20 at 19:57

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