Perplexed comments on Rationality Quotes: January 2011 - Less Wrong
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Comments (268)
Sure you can.
Edit:
(P->Q) -> (Q->P) is not a fallacy. ∀P,Q: (P->Q) -> (Q->P) is a fallacy, and its negation is ∃P,Q: (P->Q)^~(Q->P) which is indeed a theorem in first order logic.
Huh?? If you allow quantification over propositions, you are no longer using first order logic.
I think you were closer to being on track before your edit. The first thing to realize is that a fallacy is not a false statement. It is an invalid inference scheme or rule of inference.
So, with P and Q taken to be schematic variables (to be instantiated as propositions), the following is a fallacy (affirming the consequent):
P -> Q |- Q -> P
Or, you could have simply corrected the words "additional axiom" in the quoted claim to "additional axiom scheme".