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endoself comments on Rationality Quotes: January 2011 - Less Wrong
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Comments (268)
That's not even rational, it's affirming the consequent.
It's a joke about rationality. Why should all rationality quotes need to be direct and inspiring?
I would in general approve of a joke about rationality. I just don't think that this was particularly related to rationality or particularly funny, and it contained a logical fallacy without the fallacy being the butt of the joke, so it was not very rational.
However, on the assumption that at least once in René's life he did indeed walk into a bar and refuse his regular drink in such manner (it seems possible), the premises and the conclusion are all true.
I think the implication that the cessation of existence was immediate was at least as strongly implied as that the exchange took place in French.
Of course. One of the things that learning logic from philosophy teaches you is to be nitpicky about deriving causation from conditionals. A truth table, for better or worse, contains no field for "strong implication contradicted".
I don't think I derived this implication from the `I think, therefore I am,' I think I got it from how it happened right after, though I can't be sure about that specific instance of causation in my brain.
Best summary of the justification for Bayesian AI I've ever heard.
How many Less Wrongers does it take to ruin a joke?
None. Any general intelligence should be able to do it.
Two, judging by the number of people between your comment and the top one, with the top comment being excluded because the ruining of a joke is defined not to include the initial statement of the joke and, due to the way in which you mentioned the ruining of the joke, you presumably were commenting on an existing situation, rather than one which you had just completed ;)
Hmm, it was only one that time.
Really? What if they totally mess up the punchline? Or accidentally use a synonym of the word that was being set up for a pun?
Good point, I did assume that the joke was told correctly.
AUGH
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.