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If supposing that a statement is false gives rise to a paradox, does this prove that the statement is true?

Let me attempt to be a little more precise:

Suppose you have a proposition. Furthermore, suppose that assuming the proposition is false leads to a paradox. Does this imply the proposition is true? In other words, can I replace the "contradiction" in "proof by contradiction" with "paradox."

This question might still be somewhat ambiguous; I'm reluctant to attempt to precisely define "paradox" here. As a (somewhat loose) example however, consider some proposition whose negation leads to, for example, Russell's paradox. Would this prove that the proposition is true?

Edit: Define paradox as follows: a situation is a paradox if and only if it presents a logical inconsistency somehow "equivalent" to the one presented in Russell's paradox.

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    If you're not going to define, or at least bound, what you mean by paradox, how can you expect good answers? – Dave Sep 19 '14 at 17:48
  • If you won't define paradox, it's not possible to give a sensible answer. Do you mean paradox as in logical contradiction, P and not-P? Or paradox as in counterintuitive result, like the Banach-Tarski paradox? – user4894 Sep 19 '14 at 17:49
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    And why the negation of a prop? Can't you replace -P by -(-P) and just say a proposition that leads to a paradox? Can you formulate a clearer question? – user4894 Sep 19 '14 at 17:51
  • @user4894: I agree this question is somewhat ambiguous. Obviously, however, I do not mean "paradox" as in a counter-intuitive result -- the question would be vacuous if I did -- but rather some form of logical inconsistency. I call Russell's paradox a "paradox" because it is logically inconsistent for an object to both be and not be an element of a set. – user9151 Sep 19 '14 at 18:08
  • @user9151 Russell's paradox shows that allowing unrestricted set formation leads to a contradiction; therefore we must reject unrestricted set formation. Surely you are not objecting to proof by contradiction. I do not understand your point. Or, are you trying to understand proof by contradiction? – user4894 Sep 19 '14 at 18:20
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No, because the statement being true could also lead to paradox:

A: "This sentence is false."

If you presume A to be false, then A is true, and we have a paradox.

But this doesn't prove that A is true, because if A is true, then A is false, and we also have a paradox.

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Classical logic includes three laws. They are :

  1. The Law of Identity ( whatever is, is ) ( in other words, x = x )
  2. The Law of Non-Contradiction ( nothing can both be and not be )
  3. The Law of the Excluded Middle ( everything must either be or not be )

The validity of the method of proof which you have outlined follows from a combination of the second and the third law. The method is referred to as "Proof by contradiction", or "The reductio ad absurdum method". This method is a valid method of proof in any application of classical logic - specifically, mainstream mathematics.

In the context of a mathematical proof, the term paradox is equivalent to contradiction. That is to say, a statement of the form "P and not P".

Here is an outline of the proof by contradiction method as applied to proving a proposition A:

  • Assume not A.
  • Show that this assumption leads to a contradiction, e.g. B and not B.
  • This violates law 2, so we know that our assumption not A cannot be true (since it necessarily leads to a contradiction).
  • Finally, by law 3, if not A cannot be true, then A must be true.

Obviously the validity of this method requires that you accept the validity of the stated three laws of thought. People who study logic also consider logical systems where laws 2 and 3 are not included. Some may even reject law 1, though one can only wonder why.

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The converse is true, i.e.: "If assuming that a premise is true leads to a paradox, that shows that the premise isn't true."

There are mathematical proofs that work like this: Wikipedia calls it proof by contradiction.

For example, "How can we prove that the square root of 2 is irrational? Well, assume that it's rational. Then, (see Pythagorean Theorem Proof which shows that if it is rational then a paradox exists). Therefore (to avoid the paradox) the premise (that it is rational) cannot be true: therefore Q. E. D."

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In first-order logic, you can define 'p implies q' as 'q or not p' then 'p implies not p' means 'not p or not p' which means 'not p'. This reduces paradox to contradiction.

But it does not fit well with the ordinary interpretation of self-referential statements. So there is recursive function theory, which allows propositions as variables. In this form of logic p itself has to take the form of a function, which falls into infinite recursion if p is a paradox.

So you cannot necessarily perform the straightforward kind of reductio-ad-absurdum when your assumption involves self-reference. In a more flexible mathematics, the Cretan paradox, and also Russel's, remain undefined, rather than false.

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