4 ms·

You might say "but wait, haven't we just proven that it's true? So isn't that also a contradiction?" This would be a disaster, because it would prove that the a

by qnleigh 7mo ago

You might say "but wait, haven't we just proven that it's true? So isn't that also a contradiction?" This would be a disaster, because it would prove that the axioms of arithmetic are inconsistent! Now 1+1=3 for all we know. The catch is that when we proved that the sentence is not false, we used proof by contradiction, and for proof by contradiction to be a valid method of proof, we need to assume that the axioms we are working with are consistent (and therefore can't produce a contradiction). So really all we have proved is that either: - the sentence is true Or - the axioms of arithmetic are inconsistent We can't prove that the axioms of arithmetic are consistent, so we haven't actually proven that the sentence is true. Contradiction avoided. This issue is actually a major part of Gödel's theorem; we can only avoid a paradox of the axioms of arithmetic can't prove their own consistency. These theorems apply to any system of axioms that are rich enough to state the liar's paradox.

eis 7mo ago
> These theorems apply to any system of axioms that are rich enough to state the liar's paradox. Isn't that circular reasoning or tautological though? Rephrased: any system that can state something that these theorems apply to, can have the theorems applied to. I think the word "rich" is too inaccurate in this context. It is not clear why there can't be a more "rich" system which does not suffer from this issue and can't state the liars paradox.
- qnleigh 7mo ago
  Yeah rich is a vague word here. Really we're trying to say 'if a set of axioms can express provability, and can get a sentence to refer to itself, then it can state the liar sentence. And once it can state the liar sentence, then the rest of Gödel's argument follows.' There's no circularity.
discarded1023 7mo ago
> We can't prove that the axioms of arithmetic are consistent [...] Sure we can! [1] ... but it requires (logically) stronger axioms. Assessing the relative strength of axioms along these (Gentzen's) lines goes by the name "ordinal analysis". It's not clear to me that stronger axioms are always less plausible than weaker ones (as axioms). An alternative is to abandon your insistence on consistency. Another thread points to an article by Graham Priest but not to one of his main research interests: paraconsistency. This line of work aims to route around these issues (paradox in general) by making inconsistencies less explosive. A quick google turned up some relevant discussion [2]. I have it on good authority that the wheels fall off at some point. [1] https://en.wikipedia.org/wiki/Gentzen%27s_consistency_proof https://en.wikipedia.org/wiki/Gentzen%27s_consistency_proof [2] https://math.stackexchange.com/questions/1524715/how-do-inconsistent-arithmetics-get-around-g%C3%B6del-incompleteness https://math.stackexchange.com/questions/1524715/how-do-inco...
- qnleigh 7mo ago
  > We can't prove that the axioms of arithmetic are consistent using the axioms themselves. We can prove consistency using a stronger set of axioms, but those axioms have their own liar sentence, and so they can't prove their own consistency. And without knowing if the stronger set of axioms is consistent, we can't be sure that we have really proved the consistency of arithmetic.

by qnleigh 7mo ago

eis 7mo ago

> These theorems apply to any system of axioms that are rich enough to state the liar's paradox. Isn't that circular reasoning or tautological though? Rephrased: any system that can state something that these theorems apply to, can have the theorems applied to. I think the word "rich" is too inaccurate in this context. It is not clear why there can't be a more "rich" system which does not suffer from this issue and can't state the liars paradox.

qnleigh 7mo ago
Yeah rich is a vague word here. Really we're trying to say 'if a set of axioms can express provability, and can get a sentence to refer to itself, then it can state the liar sentence. And once it can state the liar sentence, then the rest of Gödel's argument follows.' There's no circularity.

discarded1023 7mo ago

> We can't prove that the axioms of arithmetic are consistent [...] Sure we can! [1] ... but it requires (logically) stronger axioms. Assessing the relative strength of axioms along these (Gentzen's) lines goes by the name "ordinal analysis". It's not clear to me that stronger axioms are always less plausible than weaker ones (as axioms). An alternative is to abandon your insistence on consistency. Another thread points to an article by Graham Priest but not to one of his main research interests: paraconsistency. This line of work aims to route around these issues (paradox in general) by making inconsistencies less explosive. A quick google turned up some relevant discussion [2]. I have it on good authority that the wheels fall off at some point. [1] https://en.wikipedia.org/wiki/Gentzen%27s_consistency_proof https://en.wikipedia.org/wiki/Gentzen%27s_consistency_proof [2] https://math.stackexchange.com/questions/1524715/how-do-inconsistent-arithmetics-get-around-g%C3%B6del-incompleteness https://math.stackexchange.com/questions/1524715/how-do-inco...

qnleigh 7mo ago
> We can't prove that the axioms of arithmetic are consistent using the axioms themselves. We can prove consistency using a stronger set of axioms, but those axioms have their own liar sentence, and so they can't prove their own consistency. And without knowing if the stronger set of axioms is consistent, we can't be sure that we have really proved the consistency of arithmetic.