How can “a implies not a” be true?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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.)

Answer 5

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.

https://en.wikipedia.org/wiki/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

Answer 6

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.

First:

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.

Second:

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.