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Challenge for logic nuts
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7/23/2014 7:04:59 PM Posted: 2 years ago Decode this for me:
Proof of Theorem: Say that an event E is continuous provided that for any 0X04;xX04;P(E), there is an event FX38;E with P(F)=x. By Axiom 5, without loss of generality {XW12;A} and {YW12;A} are continuous for any (Borel measurable) A. (Proof: If necessary, enrich the probability space that X is defined on to introduce a random variable U uniformly distributed on [0,1] and independent of X. The enrichment will not change any gamble orderings by Axiom 5. Then if 0X04;xX04;P(XW12;A), just choose aW12;[0,1] such that aP(XW12;A)=x and let F={XW12;A&UX04;a}. Ditto for Y.) Now, given an event A and a random variable X, let AX be the random variable equal to X on A and equal to zero outside of A. Let A={X=W22;W34;} and B={Y=W22;W34;}. Define the random variables X1 and Y1 on [0,1] with uniform distribution by X1(x)=W22;W34; if xX04;P(A) and X1(x)=0 otherwise, and Y1(x)=W22;W34; if xX04;P(B) and Y1(x)=0 otherwise. Since P(A)X05;P(B) by (7), it follows that X1(x)X04;Y1(x) everywhere and so X1X04;Y1 by Axiom 3. But AX and BY are probabilistically equivalent to X1 and Y1 respectively, so by Axiom 5 we have AXX04;BY. If we can show that AcX<BcY then the conclusion of our Theorem will follow from the second part of Axiom 4. Let X2=AcX and Y2=BcY. Then P(X2=+W34;)<P(Y2=+W34;), X2* and Y2* have finite expected values and X2 and Y2 never have the value W22;W34;. We must show that X2X04;Y2. Let C={X2=+W34;}. By subdivisibility, let D be a subset of {Y2=+W34;} with P(D)=P(C). Then CX2 and DY2 are probabilistically equivalent, so CX2X04;DY2 by Axiom 5. Let X3=CcX2 and Y3=DcY3. Observe that X3 is everywhere finite. Furthermore P(Y3=+W34;)=P(Y2=+W34;)W22;P(X2=+W34;)>0. Choose a finite N sufficiently large that NP(Y3=+W34;)>E(X3)W22;E(Y3*) (the finiteness of the right hand side follows from our integrability assumptions). Let Y4 be a random variable that agrees with Y3 everywhere where Y3 is finite, but equals N where Y3 is infinite. Then E(Y4)=NP(Y3=+W34;)+E(Y3*)>E(X3). Thus, Y4>X3 by Axiom 2. But Y3 is greater than or equal to Y4 everywhere, so Y3X05;Y4. By Axiom 1 it follows that Y3>X3. but DY2X05;CX2 and X2=CX2+X3 and Y2=DY2+Y3, so by Axiom 4 we have Y2>X2, which was what we wanted to prove. Thanks 
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7/23/2014 7:08:14 PM Posted: 2 years ago Darn. Forgot that logic symbols don't show up. Here's the link: http://alexanderpruss.blogspot.com.br...

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7/23/2014 7:27:56 PM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: F That. 
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7/23/2014 8:24:39 PM Posted: 2 years ago I would, BUT I CAN'T WORK UNDER THESE CONDITIONS. *throws paper at wall*

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7/23/2014 10:13:32 PM Posted: 2 years ago At 7/23/2014 7:27:56 PM, Envisage wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: 404 coherent debate topic not found. Please restart the debate with clear resolution. Uphold MarxistLeninistMaoistSargonistn7ism. 
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7/31/2014 1:30:49 PM Posted: 2 years ago At 7/23/2014 7:27:56 PM, Envisage wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: Did I just read right, Well now you know how I feel when you get all technical' LMAO 
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7/31/2014 7:01:30 PM Posted: 2 years ago Logic dictates that this will never serve a useful purpose to solve it.
I am not what you think I am. You ARE what you think I am 
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7/31/2014 8:57:20 PM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: Yes. 
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8/1/2014 9:46:48 AM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: 42 "It is one of the commonest of mistakes to consider that the limit of our power of perception is also the limit of all there is to perceive." " C. W. Leadbeater 
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8/1/2014 10:10:52 AM Posted: 2 years ago At 7/23/2014 7:04:59 PM, Toviyah wrote: It's basically saying that if there's a nonzero probability for an infinite payoff with only the risk of a finite loss, then the reward from taking the gamble is worth the risk. Similarly, if both options offer a chance at an infinite payoff but there's a greater probability of an infinite payoff from taking the gamble and no higher risk of a negative infinite payoff then you should take the bet. So therefore it's better to choose to believe in God because there's a better chance of an infinite payoff (heaven). 
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9/8/2014 6:57:27 PM Posted: 2 years ago At 8/1/2014 10:10:52 AM, Enji wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: ....so that huge, impressivelooking chunk of logickese is nothing more than an elaborate version of pascal's wager!? _____ 
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9/8/2014 7:23:21 PM Posted: 2 years ago At 9/8/2014 6:57:27 PM, UchihaMadara wrote:At 8/1/2014 10:10:52 AM, Enji wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: That's exactly what it is. Toviyah cited the source for it above: [http://alexanderpruss.blogspot.com.br...] 