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# Challenge for logic nuts

 Posts: 90 Add as FriendChallenge to a DebateSend a Message 7/23/2014 7:04:59 PMPosted: 3 years agoDecode 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 AcX0.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
 Posts: 90 Add as FriendChallenge to a DebateSend a Message 7/23/2014 7:08:14 PMPosted: 3 years agoDarn. Forgot that logic symbols don't show up. Here's the link: http://alexanderpruss.blogspot.com.br...
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 7/23/2014 7:27:56 PMPosted: 3 years agoAt 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksF That.
 Posts: 13,775 Add as FriendChallenge to a DebateSend a Message 7/23/2014 8:24:39 PMPosted: 3 years agoI would, BUT I CAN'T WORK UNDER THESE CONDITIONS. *throws paper at wall*
 Posts: 1,465 Add as FriendChallenge to a DebateSend a Message 7/23/2014 10:13:32 PMPosted: 3 years agoAt 7/23/2014 7:27:56 PM, Envisage wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksF That.404 coherent debate topic not found. Please restart the debate with clear resolution. Uphold Marxist-Leninist-Maoist-Sargonist-n7ism.
 Posts: 2,919 Add as FriendChallenge to a DebateSend a Message 7/31/2014 1:30:49 PMPosted: 3 years agoAt 7/23/2014 7:27:56 PM, Envisage wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksF That.Did I just read right, Well now you know how I feel when you get all technical'LMAO
 Posts: 10,078 Add as FriendChallenge to a DebateSend a Message 7/31/2014 7:01:30 PMPosted: 3 years agoLogic dictates that this will never serve a useful purpose to solve it.Beware of the people who are in your circle but are not in your corner. And with the stroke of a pen people 18 to 21 who own a gun became criminals and public enemy #1 having committed no crime and having said nothing. Just like the Jews in Germany during WW2. Must be a weird feeling. When I hear people crying and whining about their first world problems I think about the universe with everything in it and people in wheelchairs and all of their problems go away.
 Posts: 2,109 Add as FriendChallenge to a DebateSend a Message 7/31/2014 8:57:20 PMPosted: 3 years agoAt 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksYes.
 Posts: 3,749 Add as FriendChallenge to a DebateSend a Message 8/1/2014 9:46:48 AMPosted: 3 years agoAt 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.Thanks42"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
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 8/1/2014 10:10:52 AMPosted: 3 years agoAt 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksIt's basically saying that if there's a non-zero 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).
 Posts: 1,049 Add as FriendChallenge to a DebateSend a Message 9/8/2014 6:57:27 PMPosted: 3 years agoAt 8/1/2014 10:10:52 AM, Enji wrote:At 7/23/2014 7:04:59 PM, Toviyah wrote: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 AcX0.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.ThanksIt's basically saying that if there's a non-zero 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).....so that huge, impressive-looking chunk of logickese is nothing more than an elaborate version of pascal's wager!? -_____-
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 9/8/2014 7:23:21 PMPosted: 3 years agoAt 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: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 AcX0.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.ThanksIt's basically saying that if there's a non-zero 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).....so that huge, impressive-looking chunk of logickese is nothing more than an elaborate version of pascal's wager!? -_____-That's exactly what it is. Toviyah cited the source for it above: [http://alexanderpruss.blogspot.com.br...]