Why does the capacity of the formula a -> ~a to be true seem so counterintuitive?

Can you give me some ordinary language examples of this case?

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    – Philip Klöcking
    Commented Mar 28 at 17:10
  • 1
    In what context are you asking us about (A → ¬A)? Are you using a truth table definition, a english dictionary definition? Intuitionist logic, Classical Logic, Common Sense, Propositional Calculus, Aristotelian Syllogism? etc... Please provide more context. Commented Mar 30 at 22:30
  • "If you think you understand quantum mechanics, you don't understand quantum mechanics." Richard Feynman (strictly not applicable, but a nice example).
    – RodolfoAP
    Commented Apr 2 at 7:22

9 Answers 9


It might help to understand the conditional hypothetically. "If I accepted the supposition that A is true, it would follow that A is actually false. Since it is also the case that if A is false then A is false, I may conclude that A is false either way." It does indeed sound rather odd, but it is a form of reasoning by reductio.

An example might be the classic proof that the square root of two is irrational. You start with the supposition that there exists a pair of natural numbers n, m with no common divisor (other than 1) such that n/m equals the square root of two and then show that this supposition entails its own falsehood because n and m must both be divisible by two.

It is worth bearing in mind that the status of the sentence A → ¬A depends on the conditional you are using and on the underlying logic.

  1. In the case of the material conditional from classical logic, A → ¬A is a logically contingent sentence with the same truth value as ¬A. Both (A → ¬A) → ¬A and (¬A → A) → A are theorems of classical logic. The material conditional is simply a truth function and A → ¬A can be understood as: were A true, ¬A would also be true, and since this is contradictory, A is false.

  2. In intuitionistic logic and minimal logic, (A → ¬A) → ¬A is a theorem, but not (¬A → A) → A. We can understand the conditional to mean something like: I can manipulate a proof of A into a proof of ¬A and this proves ¬A. But if I manipulate a proof of ¬A into a proof of A, this only proves ¬¬A rather than A.

  3. In the basic relevance logic B, (A → ¬A) → ¬A does not hold, but it is a feature of the stronger systems R and E. As with classical logic, this represents a form of reasoning by reductio.

  4. With strict implication within normal modal logics, □(A → ¬A) holds if and only if □¬A. The strict conditional □(A → ¬A) means that A → ¬A holds in all possible worlds.

  5. With Stalnaker's C2 conditional, A → ¬A is false. In Stalnaker's semantics, the consequent holds in the world as it would be if the antecedent were true. In the closest possible world in which A holds, ¬A cannot also hold, so the conditional is false. The only exception is when A is necessarily false, in which case the conditional is taken to be vacuously true.

  6. With David Lewis' VC logic, A □→ ¬A is false, unless A is necessarily false, in which case the conditional is taken to be vacuously true.

  7. In connexive logic, A → ¬A is always false. Connexive logic attempts to express what is meant by a non-trivial implication relation. A proposition that is true cannot imply its own falsehood.

  8. If we think of logical entailment as a kind of meta level conditional, A ⊢ ¬A never holds, provided A is logically contingent.

  9. With probabilistic conditionals, P(¬A | A) = 0. (Or undefined, if P(A) = 0). The fact that P(¬A | A) = 0 can be understood as stating that not-A can never hold when A is assumed true.


I think part of the reason why it seems so counterintuitive is the same reason a statement like "if the sky is green, then the moon is made of cheese" seems counterintuitive. Namely, that when we use this kind of construction in everyday language, it's usually for cases where

  1. The antecedent can be true, and
  2. When the antecedent is true, this "causes" the consequent to be true in some way.

Neither of these hold for the example above, and similarly they don't hold for A → ¬A, since the only case where this implication holds is when A is not true to begin with. So even if it can be true logically, our intuition tells us it's false because it doesn't match the conditions above.

  • "the only case where this implication holds is when A is not true to begin with" But the implication is just false, so there is no case where it "holds". - 2. "this implication holds is when A is not true" So Why don't you offer a logical reasoning to prove that the implication is true? Commented Mar 29 at 17:04
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    @Speakpigeon Judging by your other replies, you seem to reject the classical definition of implication. Which is not "bad" per se, different logical formalisms exist. But let me ask you then, if you don't accept that A → B is true when A is false, how would you interpret the sentence "If I lose this bet I will eat my shoe"? If the speaker wins the bet and doesn't eat their shoe, would you consider them to have broken their promise?
    – BackusNaur
    Commented Mar 30 at 18:39
  • "you seem to reject the classical definition of implication" I don't. I don't reject the classical definition. The definition I reject is not "classical". The only classical definition we know of is that of Aristotle's. - 2. "how would you interpret the sentence "If I lose this bet I will eat my shoe"?" Not telling you that, but it's really easy. Standard logic. Why are you unable to see it? - 3. "If the speaker wins the bet and doesn't eat their shoe, would you consider them to have broken their promise?" His promise to eat his shoe? Nobody takes it as a true promise. Commented Mar 31 at 15:50
  • Ok, fine. What about "If I lose this bet I will give you $10"? The speaker then wins the bet and doesn't pay $10. Have they broken their promise?
    – BackusNaur
    Commented Apr 1 at 16:13
  • "Have they broken their promise?" You just repeat Suber's 1997 fallacious argument equating a conditional to a promise. It's not even logic. It is arguing from a silly example. You really think this helps understand the conditional?! 10% of children age 5 understand the conditional. 50% at age 8. Why do you suppose grown-ups need a silly analogy to understand the conditional? Commented Apr 1 at 16:40

Why does the capacity of the formula 'a -> ~a' to be true seem so counterintuitive?

Because such a conditional is true only in the case where the assertion of (a) is a contradiction, to make this clear, let us translate the variables into English.

(a) = it is the case that (a)

(~a) = it is not the case that (a)

With the above translation of the variables given, a common misunderstanding is assuming that (a) is actually true, which would result in a straightforward contradiction as can be demonstrated with the following syllogism:

(P1) a -> ~a

(P2) a

(P3) ~a [derived from P1 and P2]

Conclusion: a ∧ ~a

If this understanding were to be correct, that the assertion of [a -> ~a] necessitates the assertion [a ∧ ~a], then [a -> ~a] would always be false, but the mere assertion of [a -> ~a] is not the same as stating that the syllogism above is true;[a -> ~a] does not assert or entail (a), the second premise above must be introduced, it can not be deduced from [a -> ~a] alone.

But this is not the case, rather, the implication (->) symbol is a conditional symbol that makes the sentence take the English form "If (a) then (~a)", or to put it in a more fleshed-out form:

"IF it is the case that (a) THEN it is not the case that (a)"

That is to say, if you assume that (a) is the case then you will find (a) not to be a case. Hence the sentence is asserting in essence that (a) is a contradiction, not that (a) is the case.

Can you give me some ordinary language examples of this case?

Sure, here is a clear example. If we assume that "indescribable" is a descriptive word (which it is) then you get the following:

If [Tom is indescribable] then [Tom is describable]

(a) = it is the case that [Tom is indescribable] == [Tom is indescribable]

(~a) = it is not the case that [Tom is indescribable] == it is the case that [Tom is describable] == [Tom is describable]

Here we can see that the statement "If [Tom is indescribable] then [Tom is describable]" takes the form "a -> ~a" and is true. Note how asserting that it is the case in reality that Tom is indescribable, is different from asserting that "if we were to claim that Tom is indescribable, then we would be forced to conclude that he is describable"; the former is false whilst the latter is clearly true.

  • Feel free to let me know if you need any further clarification
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    The material conditional A → B is logically equivalent to ¬A ∨ B. So A → ¬A is logically equivalent to ¬A ∨ ¬A, which is logically equivalent to ¬A. So A → ¬A is a logically contingent proposition that has the same truth value as ¬A. It does not require A to be a contradiction.
    – Bumble
    Commented Mar 28 at 19:56
  • @Bumble If A → ¬A is logically equivalent to ¬A, then to assert A → ¬A is to assert ¬A must be the case. And to assert that ¬A must be the case is to assert that A being the case would be a contradiction...... I don't see where I erred
    – Max Maxman
    Commented Mar 28 at 20:07
  • No. To assert A → ¬A is equivalent (at least in its truth conditions) to asserting ¬A. There is no must about it.
    – Bumble
    Commented Mar 28 at 21:55
  • @Bumble The must is to emphasize the fact that you are continuously asserting ¬A (it's not being used in a modal sense). If we are speaking purely in terms of truth conditionals, then asserting ¬A would mean that asserting A would be a contradiction, meaning A would entail a contradiction, and would hence be the logical equivalent of a contradiction (A would entail a contradiction and the contradiction would entail A). If we are speaking in terms of the semantics of A → ¬A, then it seems obvious that the individual who asserts such is asserting that A is a contradiction....
    – Max Maxman
    Commented Mar 28 at 22:10
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    Let us continue this discussion in chat.
    – Bumble
    Commented Mar 29 at 1:10

A simple answer is that A→~A just means that A is false. However, if we mean that A⊢~A, then we are saying something tantamount to “A is self-defeating.”

Take for example the set comprehension schema from Frege’s set theory. In even relatively weak logics, this schema gives rise to a contradiction, i.e. the Russell ‘Set’.

An example that is slightly less rigorous is the claim “s knows that P precisely if s believes P, is justified in believing P, and P is true,” as shown by Gettier. I say it’s slightly less rigorous since Gettier relies on an intuitive notion of what constitutes an instance of knowledge to show that JTB is insufficient for that intuition about knowledge.

Either way, showing that a philosophical claim is self-defeating is an important tool for demonstrating that a philosophical claim is dubious at best, and inconsistent at worst.

As a side-note, over the positive portion of Intuitionistic Logic we can axiomatize the rest with

  • (A→~A)→~A and
  • ~A→(A→B).

That is to say, the ability to constructively refute a claim and show that all contradictions are absurd requires that self-defeating claims are false and self-defeating concepts have no possible instances.


In classical logic, the implication A → B is equivalent to ¬A ∨ B. So, when B is ¬A, the implication A → ¬A can be represented as ¬A ∨ ¬A, which is equivalent to ¬A. This statement holds true when A is false, as ¬False = True.

Understanding that this reduces to ¬A, an ordinary language example would be the claim "it is false that the sun doesn't exist", which is true because the internal claim "the sun doesn't exist" is false (purportedly, assuming we are not brains in vats, etc.)

  • "In classical logic, the implication A → B is equivalent to ¬A ∨ B" Classical logic is Aristotelian logic and nothing else, so, no, the implication A → B is not equivalent to ¬A ∨ B. - 2. "because the internal claim "the sun doesn't exist" is false" You don't prove an implication true by assuming a particular case. The implication we are discussing is A → ¬A. If you assume ¬A, the implication is no longer A → ¬A, it is ¬A → (A → ¬A). Commented Mar 29 at 17:12
  • @Speakpigeon See en.wikipedia.org/wiki/Material_conditional, also see math.stackexchange.com/a/630399
    – Mark
    Commented Mar 29 at 17:33
  • See the scholarship on Aristotle's texts. Commented Mar 30 at 11:44

This conditional can only be true for statements which are always false

The statement "A → ¬A" arises (after some thought) when contemplating any proposition A which is inconsistent with some larger system.  

Knights and Knaves

When understanding logic, it is sometimes helpful to visit the land of Knights and Knaves. Knights always tell the truth, knaves always lie, and all of them know each other (so they know whether any other resident of the land is a Knight or a Knave).

Suppose that you find two people in this land, and one (we'll call them Sam) says, "At least one of us is a knave."

Now, consider the statement "Sam is a knave." Call this statement "A."

If Sam is a knave, then their statement must be false, which means that there are no knaves among them, therefore Sam can't be a knave. A → ¬A. We can conclude, therefore, that Sam cannot be a Knave, because Sam being a knave is inconsistent with the system of Knights and Knaves. A Knave couldn't make that statement.

This is not a paradox. A knight can make that statement (if the other person in the group is a knave). "A → ¬A" but not "¬A → A."

Proofs by Contradiction

These sorts of contradictions frequently in mathematics. It is one form of a common technique in proofs called a "Proof by Contradiction," where the writer assumes the opposite of the statement they're trying to prove (for example, when trying to prove that a solution does not exist, they could assume that a solution exists), and then demonstrate that assuming so leads to a contradiction.

Often, proofs by contradiction create a contradiction on some other fact, e.g. "(A ∧ B) → ¬B" where B is something that is already known to be true from prior works.

Example: No Largest Prime Number

Here is a practical example of this form - the proof that there is no largest prime number.

  1. Suppose that there is a largest prime number, and call it m.
  2. Let P be the set of all prime numbers m. {2, 3, 5, ..., m}. Because m is a finite number and all primes are whole numbers, P must have a finite number of members.
  3. Let n be the product of all members of P. 2 * 3 * 5 * ... * m
  4. Because P has a finite number of members, n has a non-infinite value.
  5. Because all members of P are whole numbers greater than one, n cannot be smaller than m.
  6. Consider the number n+1
  7. Because n+1 is exactly 1 more than a multiple of every member of P, it will have a remainder of 1 when divided by any prime number.
  8. Since no prime number divides n+1, it must itself be prime.
  9. Because n ≥ m, we know that n+1 > m.
  10. Because n+1 > m and n+1 is prime, m is not the largest prime number.

Here we have taken the statement, "There is a largest prime number," and used it (along with our existing system or arithmetic) to prove, "There is a prime number larger than that number." This creates an "A → ¬A" contradiction. From that, we can conclude that there must not be a largest prime number.

As with all proofs of this form, the contradiction doesn't purely arrive from A itself. It didn't show that A is impossible - only that it is inconsistent with our existing system of arithmetic. For example, if we change to a different system which defines multiplication over whole numbers such that A*B isn't guaranteed to be ≥ A (for example, modulo arithmetic), then it would be possible for there to be largest prime (or for some systems, we might be able to even disprove the existence of prime numbers entirely).


This answer elaborates that for the proposition P: A ⇒ B

  • A is false (ie. a contradiction) gives the case of vacuous truth of P
  • B is tautology (ie. is true) gives the case of trivial truth of P

So for P: A ⇒ ¬A,
A is false makes P both trivially and vacuously true!

The above follows from the meanings of trivially/vacuously true but may not contribute much intuition!

However note that the above only address the case: A is false.

If A is true, we immediately get the contradiction True ⇒ False
And so by reductio ad absurdum A cannot be true.

That covers all cases though the second case is likely easier on the intuition

  • What do you mean by that A is tautologically true/false? That A is a tautology/contradiction? Commented Apr 2 at 5:06
  • @ayylien Ok that was sloppy language. Corrected
    – Rushi
    Commented Apr 2 at 6:03
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    Btw, A doesn’t need to be a contradiction for A -> ~A to be true. It just need to be false Commented Apr 2 at 13:08

The only time a -> ~a is when a is false. When a is false, a can imply anything. This is known as the principle of explosion.


(Latin: ex falso [sequitur] quodlibet, 'from falsehood, anything [follows]'; or ex contradictione [sequitur] quodlibet, 'from contradiction, anything [follows]')

As conifold pointed out, this can come up in an argument from contradiction.

edit. My answer seems... controversial, but there's another answer on this stackexchange making the same connection that seems more well-received - perhaps it's explained better there: https://philosophy.stackexchange.com/a/10528/49717

And here: https://math.stackexchange.com/questions/1212338/why-if-the-antecedent-p-is-false-and-the-consequence-q-true-then-the-implicati


The idea that the implication A → ¬A could be true in some cases is a theoretic construct. And, sometimes, theoretic constructs are just false. That is, they do not match up with the real world. They are only theoretic constructs.

It helps sometimes in order to see through these theoretic constructs to focus on reality by looking at an actual example.

Let us suppose, then, that A stands for the statement “God exists”.

There are only two cases to consider:

(1) God exists

(2) God doesn't exist

That is, cases A and ¬A.


A is true. If A is true, then God exists, so we can say reasonably that A implies that God exists. This is just the implication A → A. And it is obviously true.


A is false. If A is false, God doesn't exist, and it is ¬A which is true. Thus, if ¬A is true, then God doesn't exist. That is: If ¬A, then ¬A. This just means that the implication ¬A → ¬A is true, which again is obviously true.

And there is no other case to consider. We have exhausted all logical possibilities.

In particular, we find no case where the implication A → ¬A would be true! Not in the real world, anyway.

There is nothing unintuitive about this.

NOTA: This is a "politically correct" rewording of Baby_philosopher's answer, which has been deleted.

I tried to respect the philosophical substance.

Alleged instance of A → ¬A:

Let A be "B might be false" so Not-A is "B is true", I would argue that A implies that not-A is possibly true, and so it is possible that A implies not-A.

Clearly, "B might be false" does not imply "B is true", so this example does not show that it is possible that A implies not-A.

However, this is an example of A where A does not imply not-A, which proves that it is possible that A does not imply not-A, if this is not self-evident to you.

Another alleged instance of A → ¬A:

If Tom is indescribable then Tom is describable.

This is just good old sophistry!

Either you are equivocating on the word "indescribable", or you are contradicting yourself when you suppose that it is true that Tom is indescribable; this even before you could arrive at the conclusion that Tom is describable.

Either way, the example is just nonsense, as most people seem to realise this.

This reminds me of the idea that for us to be able to enjoy reading fiction, we have to suspend disbelief. Here, to "enjoy" the paradox, we have to suspend logical acumen.

  • What if A were “one apple is the same as two apples” or “it’s not the case that ‘if you agree with me, then you agree with me,’”?
    – PW_246
    Commented Mar 29 at 15:58
  • @PW_246 "What if A were "one apple is the same as two apples" or “it’s not the case that ‘if you agree with me, then you agree with me’” OMG, A is false in both cases, so not-A is true, but this only implies that you are going to say "Ah-ah, gotcha!" Go on, explain to me why ¬A → (A → ¬A) is tau-to-lo-gi-cal! I can't seem to get enough of that stuff! You are wasting your time, you know? Commented Mar 29 at 16:48
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    I’m ultimately trying to see what your understanding of implication is. You say, “ In particular, we find no case where the implication A→¬A would be true! Not in the real world, anyway.” So no examples of (A→¬A)→¬A are applicable to the real world? I’m just trying to see what your actual position is, since it seems to be either a very nuanced position, or a position that is not that well thought-out. P.S., I don’t think ¬A→(A→¬A) is true when P→Q means “whenever P holds, Q also holds ” and ¬A means “A is currently falsifiable.”
    – PW_246
    Commented Mar 29 at 17:25
  • @PW_246 "your understanding of implication" It is like that of everyone else. Common sense. As opposed to the purely theoretic and ultimately false axiom that the implication is logically equivalent to the horseshoe. Common says that A → ¬A is false. - 2. "a very nuanced position or . . . not that well thought-out" It all comes down to common sense and our logical intuition. You learned some theoretic construct at school and you think the stuff is "well thought out". Because? Because you haven't really thought about it, so you are unable to fault the theoretic stuff. Commented Mar 30 at 11:23
  • @PW_246 "P.S., I don’t think ¬A→(A→¬A) is true when P→Q means “whenever P holds, Q also holds ” and ¬A means “A is currently falsifiable.”" Are you admitting that when you say that A → ¬A is true when A is false, you don't mean that the implication A → ¬A is true but that the horseshoe A ⊃ ¬A is true if A is false? Please explain to me how mathematicians are not wasting everybody's time when they call the horseshoe A ⊃ B not a horseshoe but "the implication A → B"? - 2. "and ¬A means “A is currently falsifiable." ¬A just means "A is false". Falsifiability is something else. Commented Mar 30 at 11:43

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