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He mentions those in relation to his own pet theories but in no way does he address GP's argument. What his theorem boils down to is: 1. Let A be a formal sys
by zepolud 13y ago
He mentions those in relation to his own pet theories but in no way does he address GP's argument.
What his theorem boils down to is:
1. Let A be a formal system capable of describing the natural numbers.
2. Assume A is inconsistent. It follows that there exists such Φ that A ⊢ Φ and A ⊢ ¬Φ
3. ?????
4. A is consistent!
At step 3. he appears to conflate A⊢Φ ∧ A⊢¬Φ with Φ∧¬Φ and use it to refute the assumption in 2. But Φ∧¬Φ is not generally true outside the context of A and therefore the refutation does not follow.
- stiff 13y agoMy understanding is that if proof by contradiction is allowed in A, Φ ∨ ¬Φ is a true sentence in A, so if you try to include the sentence "A is inconsistent" in A, it follows that A ⊢ Φ and A ⊢ ¬Φ, therefore in A Φ ∧ ¬Φ is true, but this contradicts Φ ∨ ¬Φ, therefore A is consistent. In other words, it is impossible to show the inconsistency of any system that includes the law of excluded middle. This is really trivial in the end. Inconsistency is "Φ ∧ ¬Φ", law of excluded middle is "Φ ∨ ¬Φ", so in the assumption of law of excluded middle is hidden the assumption of consistency. That's what he means by The above theorem means that the assumption of consistency is deeply embedded in the structure of classical mathematics Edit: I agree with you in the end, the proof is valid, but the conclusion does not hold in A, but one "level" above, so it does not contradict Goedels theorem. Am I reading this right?