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billswift comments on Rationality Quotes: January 2011 - Less Wrong
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Comments (268)
Re-writing Descartes's "I think, therefore I am" as ("If I think, then I am" and "I think"); therefore "I am". Then the joke's "I think not" would be denying the antecedent, which is still a fallacy, of course.
I seem to represent P -> Q and ~Q -> ~P the same way in my mind, but giving the resulting fallacies different names reduces ambiguity, so I guess this is a useful distinction.
P→Q is not logically equivalent to ~Q→P. Perhaps you meant P→Q and ~Q→~P.
Fixed, thanks.
Affirming the consequent is a totally different fallacy -
"If P, then Q and Q is true; therefore P".
Denying the antecedent with P and Q:
P -> Q
~P
Therefore ~Q
Affirming the consequent with ~Q and ~P
~Q -> ~P
~P
Therefore ~Q
Wow, I feel kind of bad just writing those chains of "deduction". Anyways, the same result was concluded from the same minor premise, the only difference is the major premise, and P -> Q and ~Q -> ~P are equivalent.
edit: formatting
By the principle of explosion, all fallacies are the same.
(∀P,Q ((P->Q) ^ Q) -> P) <-> (∀P,Q ((P->Q) ^ ~P) -> ~Q)
That depends on the definition of same. All fallacies imply each other, but the premises and conclusions in these two should be represented identically by a computer.
No. This is not the case. Just because something is a fallacy doesn't make its negation true. Thus for example (P->Q) -> (Q->P) is a fallacy. But ~((P->Q) ->(Q->P) ) is not a theorem of first order logic. So even if I throw (P->Q) -> (Q->P) as an additional axiom in I can't get a general explosion in first order logic. Contradictions lead to explosion, but fallacies do not necessarily do so.
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".
Er, sorry. Meant propositional calculus not first order logic. I think my statement works in that context.
What's specifically going on here is that (P=>Q) => (Q=> P) is false whenever P is false and Q is true.
Adding it as an axiom schema to propositional calculus results in a contradiction. It cannot be added as a single axiom to first-order logic.
Yes, you are correct. I was confused in a very stupid way.