First person plural
On 23 Jul 2015, at 21:44, John Clark wrote: On Thu, Jul 23, 2015 at 2:41 PM, Quentin Anciaux wrote: >> Yes, after the duplication but before the door of the duplicating chamber is opened John Clark may have a hunch that he (at this point the personal pronoun is not ambiguous because although there are 2 bodies they are identical so there is still just one John Clark) will see Moscow when the door is opened and make a bet. One of the John Clarks will win the bet and one will not; > Same thing with MWI, No not the same, both involve duplication but after that the similarities end. Not in a relevant way as far as the probabilities on the first person experience is concerned. > I remind you however you are that you are duplicated along the measurement apparatus. Yes. > So who's you who make the bet? In MWI "You" is the only thing that the laws of physics allow Quentin Anciaux to observe that is organized in a Johnkclarkian way; that is the thing that will give Quentin Anciaux money if the bet is lost and that is the thing Quentin Anciaux will have to give money to if the bet is won. With duplicating chamber stuff if the bet was "you will see Moscow" I don't know how to resolve the bet because I don't know who "you" is; maybe Quentin would have to give the Moscow Man $5 and the Washington Man would have to give Quentin $5, but that seems rather silly. What would be the point of Quentin Anciaux making such a bet? To make the bet, the one who bet must accompany the experiencer. This illustrate well the notion of first person plural, and show that both in Comp and in Everett we can use Dutsch-book definition of probability (as opposed to a frequency interpretation). You and I are in Helsinki, and we will both enter the annihilation- duplication box. Your bet is P(W & M) = 1, and so put your money on "W & M". I bet on "W or M". We are both annihilated and reconstituted in both cities. In W, you and me, realize that we live "W and ~M", which implies "W or M", I win, you loose. In M, you and me, realize that we live "M and ~W", which implies "W or M", I win, you loose. Conclusion: P(W or M) = P(W xor M) = 1. P(W and M) = 0. No ambiguity in pronouns at all, you and me remain the guys who remember our lives, including our visit of Helsinki and the button pushing. You can use this to explain how Everett-QM saves, empirically, the idealism of computationalism from solipism. Bruno John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On 23 Jul 2015, at 19:33, John Clark wrote: On Thu, Jul 23, 2015 at 12:55 PM, Quentin Anciaux wrote: > Actually John Clark "pretends" that half the John Clark's who say "I bet Quentin Anciaux will see spin up when the electron is measured" will win the bet. > Actually John Clark can say the same thing betting that after duplication he will find moscow behind the door, Yes, after the duplication but before the door of the duplicating chamber is opened John Clark may have a hunch that he (at this point the personal pronoun is not ambiguous because although there are 2 bodies they are identical so there is still just one John Clark) will see Moscow when the door is opened and make a bet. One of the John Clarks will win the bet and one will not; it can never be determined if "he" won the bet because as soon as the door was opened the 2 bodies were no longer identical, they had different memories, so that personal pronoun becomes ambiguous. That contradict the fact that you have agreed that both copies are the Helsinki guy. There is no ambiguity, you are both guys. But both guys have incompatible first person experiences, and that explains the prediction "W v M", as explained with all detail before (reread the last posts). It looks you change your mind on this, but then it looks you say whatever needed to deny the obvious. Bruno John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On 23 Jul 2015, at 18:58, John Clark wrote: On Thu, Jul 23, 2015 at 7:58 AM, Bruno Marchal wrote: >> That is neither right nor wrong because it is not clear what "the probability" refers to; the probability of *who* seeing spin up? > Oh, You said us that in the MWI there were no problem as the copies cannot met, and so the use of probability makes sense in QM. Quentin asked if the probability of "you" seeing spin up and seeing spin down is both 1. John Clark doesn't know how to answer that except to say if MWI is correct then the probability of John Clark seeing spin up is 1 and the probability of John Clark seeing spin down is 1. But Chris Peck, if I understood him correctly, seems to agree that in QM-Everett, we keep the usual probability. Indeed Everett justifies those probabilities, notably with MWI+Gleason theorem. And this uses the comp FPI. John Clark knows that's not exactly what was asked but if a better definition of "you" is given a better answer will be provided. It has been given, and we have agreed on it. We don't need a better definition of "you", we need only to take into account that the question is about the first person experience content, that is, the first person experience from the first person experience pov itself. As the experience W and M are incompatible, as you have agreed also, "W & M" is directly ruled out. Nobody will experience from that 1-1 view being in the two cities at once. That follows easily from the fact that the two brain copies are disconnected and cannot be aware of each other in any 1p direct view (unless magical telepathy of course, but we can't have it with the computationalist hypothesis and this protocol. > OK you did change your mind I change my mind all the time, but not in this case. > and I guess this is to hide the fact that your argument against the FPI and Chris Peck's argument would contradict each other. There may come a time when I disagree with Chris Peck, if and when that day comes I will not hesitate to say so. You may have noticed that I'm not particularly shy in that regard. I don't think Chris Peck is saying that P(up) = P(down) = 1 in QM (Everett or not). Of course this is as much ridiculous than to predict "W and M" in step 3, as the subjective experience of seeing simultaneously UP and DOWN, like W and M, are incompatible. To be oneself in superposition does not lead to a blurred experience of the two outcomes. Sorry, but I have no clue how you can maintain "W and M", except by confusing 1-you and 3-1-you. In the math part, it is the confusion between []p and []p & p, which has a long history in science and philosophy. Bruno John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
Le 23 juil. 2015 21:44, "John Clark" a écrit : > > On Thu, Jul 23, 2015 at 2:41 PM, Quentin Anciaux wrote: > >>> >>> >> >>> Yes, after the duplication but before the door of the duplicating chamber is opened John Clark may have a hunch that he (at this point the personal pronoun is not ambiguous because although there are 2 bodies they are identical so there is still just one John Clark) will see Moscow when the door is opened and make a bet. One of the John Clarks will win the bet and one will not; >> >> > >> Same thing with MWI, > > No not the same, both > > involve duplication but after that the similarities end. >> >> > >> I remind you however you are that you are duplicated along the measurement apparatus. > > Yes. >> >> > >> So who's you who make the bet? > > In MWI "You" is the only thing that the laws of physics allow Quentin Anciaux to observe that is organized in a Johnkclarkian way; Quentin Anciaux is in no way involved in that, only the matter who's unfortunately organized in a johnclarkian way is involved in both experiments and questions, and unfortunately again in both experiments this matter is duplicated... that is the thing that will give Quentin Anciaux money if the bet is lost and that is the thing Quentin Anciaux will have to give money to if the bet is won. With duplicating chamber stuff if the bet was "you will see Moscow" I don't know how to resolve the bet because I don't know who "you" is; maybe Quentin would have to give the Moscow Man $5 and the Washington Man would have to give Quentin $5, but that seems rather silly. What would be the point of Quentin Anciaux making such a bet? > > John K Clark > > > -- > You received this message because you are subscribed to the Google Groups "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. > > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On Thu, Jul 23, 2015 at 2:41 PM, Quentin Anciaux wrote: > >> >> Yes, after the duplication but before the door of the duplicating chamber >> is opened John Clark may have a hunch that he (at this point the personal >> pronoun is not ambiguous because although there are 2 bodies they are >> identical so there is still just one John Clark) will see Moscow when the >> door is opened and make a bet. One of the John Clarks will win the bet and >> one will not; > > > > Same thing with MWI, > > No not the same, both involve duplication but after that the similarities end. > > > I remind you however you are that you are duplicated along the measurement > apparatus. > > Yes. > > > So who's you who make the bet? > > In MWI "You" is the only thing that the laws of physics allow Quentin Anciaux to observe that is organized in a Johnkclarkian way; that is the thing that will give Quentin Anciaux money if the bet is lost and that is the thing Quentin Anciaux will have to give money to if the bet is won. With duplicating chamber stuff if the bet was "you will see Moscow" I don't know how to resolve the bet because I don't know who "you" is; maybe Quentin would have to give the Moscow Man $5 and the Washington Man would have to give Quentin $5, but that seems rather silly. What would be the point of Quentin Anciaux making such a bet? John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
Le 23 juil. 2015 19:33, "John Clark" a écrit : > > > On Thu, Jul 23, 2015 at 12:55 PM, Quentin Anciaux wrote: >>> >>> > Actually John Clark "pretends" that half the John Clark's who say "I bet Quentin Anciaux will see spin up when the electron is measured" will win the bet. >> >> > >> Actually John Clark can say the same thing betting that after duplication he will find moscow behind the door, > > Yes, after the duplication but before the door of the duplicating chamber is opened John Clark may have a hunch that he (at this point the personal pronoun is not ambiguous because although there are 2 bodies they are identical so there is still just one John Clark) will see Moscow when the door is opened and make a bet. One of the John Clarks will win the bet and one will not; Same thing with MWI, I remind you however you are that you are duplicated along the measurement apparatus. So who's you who make the bet? it can never be determined if "he" won the bet because as soon as the door was opened the 2 bodies were no longer identical, they had different memories, so that personal pronoun becomes ambiguous. > > John K Clark > > -- > You received this message because you are subscribed to the Google Groups "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. > > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On Thu, Jul 23, 2015 at 12:55 PM, Quentin Anciaux wrote: > Actually John Clark "pretends" that half the John Clark's who say "I bet > Quentin Anciaux will see spin up when the electron is measured" will win > the bet. > > > Actually John Clark can say the same thing betting that after duplication > he will find moscow behind the door, > > Yes, after the duplication but before the door of the duplicating chamber is opened John Clark may have a hunch that he (at this point the personal pronoun is not ambiguous because although there are 2 bodies they are identical so there is still just one John Clark) will see Moscow when the door is opened and make a bet. One of the John Clarks will win the bet and one will not; it can never be determined if "he" won the bet because as soon as the door was opened the 2 bodies were no longer identical, they had different memories, so that personal pronoun becomes ambiguous. John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On Thu, Jul 23, 2015 at 7:58 AM, Bruno Marchal wrote: >> >> >> That is neither right nor wrong because it is not clear what "the >> probability" refers to; the probability of *who* seeing spin up? > > > > > Oh, You said us that in the MWI there were no problem as the copies cannot > met, and so the use of probability makes sense in QM. > Quentin asked if the probability of "you" seeing spin up and seeing spin down is both 1. John Clark doesn't know how to answer that except to say if MWI is correct then the probability of John Clark seeing spin up is 1 and the probability of John Clark seeing spin down is 1. John Clark knows that's not exactly what was asked but if a better definition of "you" is given a better answer will be provided. > > > OK you did change your mind > I change my mind all the time, but not in this case. > > > and I guess this is to hide the fact that your argument against the FPI > and Chris Peck's argument would contradict each other. > There may come a time when I disagree with Chris Peck, if and when that day comes I will not hesitate to say so. You may have noticed that I'm not particularly shy in that regard. John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
Le 23 juil. 2015 17:58, "John Clark" a écrit : > > > On Thu, Jul 23, 2015 Quentin Anciaux wrote: >> >> > >> No he did not. He >> [Quentin means John Clark. I think] >> pretends probabilities do have meaning in MWI. When he says 0.5 with his bet > > Actually John Clark "pretends" that half the John Clark's who say "I bet Quentin Anciaux will see spin up when the electron is measured" will win the bet. Actually John Clark can say the same thing betting that after duplication he will find moscow behind the door, and about half the time he will be right. And round and round we go. Granted there will be mathematical problems if there are an infinite number of John Clarks and not just 10^500^500 of them, but that is a difficulty that the many worlds theory has never entirely cleared up in my opinion. Yes it needs work but I still think the theory is on the right track. > > John K Clark > > > -- > You received this message because you are subscribed to the Google Groups "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. > > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On Thu, Jul 23, 2015 at 7:51 AM, Bruno Marchal wrote: > >>> >> Then under MWI, same thing you're garanteed to see all results > > >> > Yes, provided that "you" means somebody who remembers being > > Quentin Anciaux > > at this instant. MWI says everything that doesn't violate the laws of > physics will happen, so in one of those many worlds you have been elected > Pope > , > and in another you have graduated from > Ringling Brothers and Barnum & Bailey Clown College > , > and in yet another you have won the Nobel Prize. And all of them are > "you" because all of them remember reading this post at this instant. > > > Oh! You change your mind. Now computationalism is like MWI, > I have always thought that computationalism was compatible with MWI, if not I would have never come to believe that MWI was the least ridiculous of all known quantum interpretations. > > > you agree with Quentin > I was unaware that Quentin and I agreed on anything. > > > that if there is no computationalist FPI, > I have no idea what absurd logical contortions you underwent to form that conclusion, and please don't tell me, I just ate. > > you have a problem with how Everett justify the use of probability in QM A transactional approach is interesting but very few of even the most enthusiastic supporters of Everett (including me) think that all the mathematical difficulties of dealing with probability if infinity is involved have been ironed out. But if spacetime is quantized then Everett might not need to deal with infinities at all. John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On Thu, Jul 23, 2015 Quentin Anciaux wrote: > > > No he did not. He > [Quentin means John Clark. I think] > pretends probabilities do have meaning in MWI. When he says 0.5 with his > bet > > Actually John Clark "pretends" that half the John Clark's who say "I bet > Quentin Anciaux will see spin up when the electron is measured" will win the bet. Granted there will be mathematical problems if there are an infinite number of John Clarks and not just 10^500^500 of them, but that is a difficulty that the many worlds theory has never entirely cleared up in my opinion. Yes it needs work but I still think the theory is on the right track. John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: FPI & possible continuations
On 22 Jul 2015, at 20:59, Terren Suydam wrote: On Tue, Jul 21, 2015 at 9:27 AM, Bruno Marchal wrote: On 20 Jul 2015, at 21:40, Terren Suydam wrote: Question for Bruno or anyone else: Let's say I see a UFO. There are potentially many competing explanations for what I saw. Does FPI entail that all of them could be a continuation from my current state, so long as the explanation robustly produces the phenomena I experienced? FPI requires that all of them need to be taken into account to "evaluate" what happens next. But some explanation might correspond to very rare computations, other might be more numerous. So this is one way in which FPI differs from Many Worlds QM scenarios, because in those scenarios, the splitting is the result of divergence from a common physical reality, whereas FPI indeterminacy is the result of divergence from a common phenomenological reality. Well said. This does not change the fact that the Everett-QM indeterminacy is a particular case of the first person indeterminacy, as I think you agree, with Quentin and me and others. Eventually, if comp is true, the first one (the result of divergence from a common physical reality) must be explain through the second one (the result of phenomenological divergence, common for the guy before the split). All right? (this of course use step 7, and is not relevant at step 3) Bruno Terren In other words, so long as what I experience is identical, relative to each possible explanation of the phenomena (e.g. aliens / military prototype / atmospheric disturbance / holographic projection / etc), does that not entail computational equivalence among the potential continuations, even if the measure would differ among them? Is my ongoing experience the only thing that matters when it comes to the set of "the infinite computations going through my state"? If not, what principle could rule out a particular explanation despite it potentially being able to produce identically the phenomena in my experience (UFO)? Nothing is ruled out, but statistically, the computations which have a bigger measure will be more probable. If you see a UFO, may be "there is a UFO" in the normal physical reality. That means that in all normal computations an UFO is there. Then that UFO is multiplied along all things which multiply you. You will be (comp)-entangled to it. For example, there will be as much UFO than there are equivalent (from your 1p pov) position of electron possible in your body (already a continuum if we postulate classical QM (and thus that QM is the solution of the FPI). Or the UFO belongs to a normal dream, in which case you will wake up, in the normal histories. Or the UFO is based on more rare computations, and the probability that you belong to them will drop down. Naively, what you expect is determined by the "mass" of computations going through your state. Although the rare experience seems as much real than the normal one, they are relatively rare. Even if you find yourself in one, from there you should bet on the normal continuations starting from that non normal situation. Similarly, you should never bet on a non normal computation, unless you die or are on drugs. Basically it is like the lottery: you should not expect to win the biggest gain, despite you cannot rule out the possibility. In our case, all finite computations may be ruled out, as they have a (naive) measure null, compared to infinite computations multiplied by (dovetailing on) the real numbers. Empirically Nature used a random oracle to get that self-multiplication right, and we can expect this to be proved necessary for the comp-measure measure. Now, that reasoning is a bit naive, and it is virtually impossible to "count" the computations, or even to recognize them in some 3p way. It can be proved easily that most computations cannot have their semantic extracted mechanically from the code of the program doing them, and that is why I handle the math of the measure in an indirect way from the logic of self-reference. Bruno Terren -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything- l...@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://g
Re: A riddle for John Clark
On 23 Jul 2015, at 05:09, chris peck wrote: Quentin >> Then under MWI, same thing you're garanteed to see all results, so probability should also be one Deterministic branching leads to trouble rendering the idea of probability coherent. Go figure! Who would ever have guessed determinism and chance were difficult to marry... Computationalism, once the 1p/3p distinction is made clear, put transparent, 3p describable, light on this. (Determinism + ontology rich enough to duplicate oneself) ===> chance. Even Tegmark "rediscovered" this in his recent book, as Jason Resch quoted once. Then elementary arithmetic confirms the quantum probabilities logic(s) with the []p & <>t (and some others) views. That is, at the exact place(s) forced by the UD Argument. This is pure math and has been thoroughly verified. It is not well known because few physicists dare to think on Gödel's theorem (especially after Penrose), and few logicians knows about Everett. Well, there are other factors which are more contingent. The point is that computationalism explains that 3p-determinism entails 1p-indeterminism. Bruno Subject: Re: A riddle for John Clark To: everything-list@googlegroups.com From: meeke...@verizon.net Date: Wed, 22 Jul 2015 16:25:00 -0700 On 7/22/2015 12:08 AM, Bruno Marchal wrote: On 21 Jul 2015, at 19:42, meekerdb wrote: On 7/21/2015 10:30 AM, Bruno Marchal wrote: So maybe one could see W AND W the same way I can see my computer screen AND my dog - just by attending to one or the other. You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. It follows from physics. We don't know that. Then why did you assert the necessity of a physical connection: "You will need a long neck to attend a conference in Moscow, and a party in Washington." We just assume that the physics is rich enough to implement locally universal machine, so that comp make sense, but then we arrive at the computationalist difficulties. Physics assume a brain/mind link which has to be justified, and the UDA shows the change we have to introduce. But you have effectively asserted that the duplicate persons at different locations do not experience both locations - their minds are separate because their brains are. If that is more than just an assumption it is because it is relying on the physical basis of mind. If you reject the physical basis of mind then you might expect the duplicates to share one mind. Brent But does it follow from UD computations? It should, (at step 7 and 8) and the point is only that it is testable. Up to now, it is working well. But to explain this, we need to dig deeper in computer science. Are you OK with the steps 0-6? 0-7? From your other posts, I think you were OK. So we can perhaps come back on step 8. I think Bruce Kellet has also some problem there. That can only be more intersting than the nonsense about step 3 that we can hear those days. Bruno Brent -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it,
Re: A riddle for John Clark
On 23 Jul 2015, at 13:07, Quentin Anciaux wrote: Le 23 juil. 2015 09:24, "chris peck" a écrit : > > Quentin > > > >> Is measuring spin up under MWI has a probability of one or 0.5 under MWI? > > we've done this sketch before...and John Clarke just did the same sketch with you hours ago...Why do you need things repeated to you so much? No he did not. He pretends probabilities do have meaning in MWI. When he says 0.5 with his bet he ignores he is entangled with the measurement apparatus and duplicated with it, with one john winning and one losinf his bet. Exactly. The deny of the FPI has been shown now equivalent with the deny of the use of probability in QM (beyond having be shown inconsistent per se, or based on the 1-3 confusion). Case close. (Normally). Bruno > > David Wallace, a proponent of MWI at Oxford University, puts it this way with regards to Schrodinger's Cat: > > "We're not really sure how probability makes any sense in Many Worlds Theory. So the theory seems to be a theory which involves deterministic branching: if I ask what should I expect in the future the answer is I should with 100% certainty expect to be a version of David who sees the cat alive and in addition I should expect with 100% certainty to be a version of David who sees the cat dead." > > What Wallace does is tackle incoherence head on. Does he over come it? Im not brainy enough to say. But I am brainy enough to see that he doesn't take the Bruno-Quentin approach of praying the problem will go away by pretending it doesn't exist. > > > > Date: Thu, 23 Jul 2015 08:48:51 +0200 > Subject: RE: A riddle for John Clark > From: allco...@gmail.com > To: everything-list@googlegroups.com > > > > Le 23 juil. 2015 05:09, "chris peck" a écrit : > > > > Quentin > > > > > > >> Then under MWI, same thing you're garanteed to see all results, so probability should also be one > > > > Deterministic branching leads to trouble rendering the idea of probability coherent. Go figure! Who would ever have guessed determinism and chance were difficult to marry... > > Then you're refuting MWI as not being able to correctly renders the probabilities, right? > > Is measuring spin up under MWI has a probability of one or 0.5 under MWI? > > Quentin > > > > > > Subject: Re: A riddle for John Clark > > To: everything-list@googlegroups.com > > From: meeke...@verizon.net > > Date: Wed, 22 Jul 2015 16:25:00 -0700 > > > > > > On 7/22/2015 12:08 AM, Bruno Marchal wrote: > >> > >> > >> On 21 Jul 2015, at 19:42, meekerdb wrote: > >> > >>> On 7/21/2015 10:30 AM, Bruno Marchal wrote: > > > > So maybe one could see W AND W the same way I can see my computer screen AND my dog - just by attending to one or the other. > > > You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. > >>> > >>> > >>> It follows from physics. > >> > >> > >> We don't know that. > > > > > > Then why did you assert the necessity of a physical connection: "You will need a long neck to attend a conference in Moscow, and a party in Washington." > > > >> We just assume that the physics is rich enough to implement locally universal machine, so that comp make sense, but then we arrive at the computationalist difficulties. Physics assume a brain/ mind link which has to be justified, and the UDA shows the change we have to introduce. > > > > > > But you have effectively asserted that the duplicate persons at different locations do not experience both locations - their minds are separate because their brains are. If that is more than just an assumption it is because it is relying on the physical basis of mind. If you reject the physical basis of mind then you might expect the duplicates to share one mind. > > > > Brent > > > >> > >> > >> > >>> But does it follow from UD computations? > >> > >> > >> It should, (at step 7 and 8) and the point is only that it is testable. > >> Up to now, it is working well. But to explain this, we need to dig deeper in computer science. > >> > >> Are you OK with the steps 0-6? 0-7? From your other posts, I think you were OK. So we can perhaps come back on step 8. I think Bruce Kellet has also some problem there. That can only be more intersting than the nonsense about step 3 that we can hear those days. > >> > >> Bruno > >> > >> > >> > >> > >>> > >>> Brent > >>> > >>> -- > >>> You received this message be
Re: A riddle for John Clark
On 23 Jul 2015, at 00:19, John Clark wrote: On Wed, Jul 22, 2015 Quentin Anciaux wrote: > So you're claiming that the probability of seeing spin up while doing a measurement of the spin is one (likewise seeing spin down) right? That is neither right nor wrong because it is not clear what "the probability" refers to; the probability of *who* seeing spin up? Oh, You said us that in the MWI there were no problem as the copies cannot met, and so the use of probability makes sense in QM. OK you did change your mind, and I guess this is to hide the fact that your argument against the FPI and Chris Peck's argument would contradict each other. It looks a bit opportunistic to me, and it annihilates your previews post on the subject. What I am claiming is that if the MWI is correct and if Quentin Anciaux performs a spin measurement on a electron then Quentin Anciaux will see spin up with 100% probability and Quentin Anciaux will see spin down with 100% probability. In the description of the wave, yes, a typical 3-1 view. But n QM we use that to evaluate outcomes of future measurement, and we get probabilities. I am also claiming that if Quentin Anciaux measures the spin of a electron and I say "I bet Quentin Anciaux got spin up" I will win the bet 50% of the time. Again assuming that the MWI is correct. You get only probability 100% if you include yourself in the wave, by what you say above, or you get solipsisme, as you allow probabilities, but only for you, which is indeed a way to confuse the 3p and the 1p. Bruno > Then, for one if you agree, you're just saying MWI is false, so you lied to us for years saying the contrary, And likewise the probability of Quentin Anciaux fucking a horse is 100% and the probability of Quentin Anciaux fucking a mule is 100%; but if I say "I bet it was a mule not a horse that Quentin Anciaux fucked" I will win my bet 50% of the time. John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On 22 Jul 2015, at 22:15, John Clark wrote: On Wed, Jul 22, 2015 at 2:18 PM, Quentin Anciaux wrote: > Then under MWI, same thing you're garanteed to see all results Yes, provided that "you" means somebody who remembers being Quentin Anciaux at this instant. MWI says everything that doesn't violate the laws of physics will happen, so in one of those many worlds you have been elected Pope, and in another you have graduated from Ringling Brothers and Barnum & Bailey Clown College, and in yet another you have won the Nobel Prize. And all of them are "you" because all of them remember reading this post at this instant. Oh! You change your mind. Now computationalism is like MWI, you agree with Quentin that if there is no computationalist FPI, there is no probability in QM either. At least this makes you coherent, but then you have a problem with how Everett justify the use of probability in QM (and indeed it is a particular case) on the FPI. Bruno > you and Clark are in total disagreement contrary to your encouragement in trolling would let us believe. Trolling? Unlikely as it is do you think it is conceivable that in one of those many worlds there is somebody who sincerely disagrees with you and Bruno, or would such a thing violate the laws of physics? John K Clark -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: A riddle for John Clark
On 23 Jul 2015, at 09:24, chris peck wrote: Quentin >> Is measuring spin up under MWI has a probability of one or 0.5 under MWI? we've done this sketch before...and John Clarke just did the same sketch with you hours ago...Why do you need things repeated to you so much? Well Quentin points to the fact that your critics on the FPI is inconsistent with Clarks' critics. That so true that you succeed in making Clark changing his mind on the difference between the FPI used in comp and in QM. Nice, now he is a bit more coherent, and ... contradicted by anyone using QM, as it is a probabilistic theory. Bruno David Wallace, a proponent of MWI at Oxford University, puts it this way with regards to Schrodinger's Cat: "We're not really sure how probability makes any sense in Many Worlds Theory. So the theory seems to be a theory which involves deterministic branching: if I ask what should I expect in the future the answer is I should with 100% certainty expect to be a version of David who sees the cat alive and in addition I should expect with 100% certainty to be a version of David who sees the cat dead." What Wallace does is tackle incoherence head on. Does he over come it? Im not brainy enough to say. But I am brainy enough to see that he doesn't take the Bruno-Quentin approach of praying the problem will go away by pretending it doesn't exist. Date: Thu, 23 Jul 2015 08:48:51 +0200 Subject: RE: A riddle for John Clark From: allco...@gmail.com To: everything-list@googlegroups.com Le 23 juil. 2015 05:09, "chris peck" a écrit : > > Quentin > > > >> Then under MWI, same thing you're garanteed to see all results, so probability should also be one > > Deterministic branching leads to trouble rendering the idea of probability coherent. Go figure! Who would ever have guessed determinism and chance were difficult to marry... Then you're refuting MWI as not being able to correctly renders the probabilities, right? Is measuring spin up under MWI has a probability of one or 0.5 under MWI? Quentin > > > Subject: Re: A riddle for John Clark > To: everything-list@googlegroups.com > From: meeke...@verizon.net > Date: Wed, 22 Jul 2015 16:25:00 -0700 > > > On 7/22/2015 12:08 AM, Bruno Marchal wrote: >> >> >> On 21 Jul 2015, at 19:42, meekerdb wrote: >> >>> On 7/21/2015 10:30 AM, Bruno Marchal wrote: > > So maybe one could see W AND W the same way I can see my computer screen AND my dog - just by attending to one or the other. You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. >>> >>> >>> It follows from physics. >> >> >> We don't know that. > > > Then why did you assert the necessity of a physical connection: "You will need a long neck to attend a conference in Moscow, and a party in Washington." > >> We just assume that the physics is rich enough to implement locally universal machine, so that comp make sense, but then we arrive at the computationalist difficulties. Physics assume a brain/ mind link which has to be justified, and the UDA shows the change we have to introduce. > > > But you have effectively asserted that the duplicate persons at different locations do not experience both locations - their minds are separate because their brains are. If that is more than just an assumption it is because it is relying on the physical basis of mind. If you reject the physical basis of mind then you might expect the duplicates to share one mind. > > Brent > >> >> >> >>> But does it follow from UD computations? >> >> >> It should, (at step 7 and 8) and the point is only that it is testable. >> Up to now, it is working well. But to explain this, we need to dig deeper in computer science. >> >> Are you OK with the steps 0-6? 0-7? From your other posts, I think you were OK. So we can perhaps come back on step 8. I think Bruce Kellet has also some problem there. That can only be more intersting than the nonsense about step 3 that we can hear those days. >> >> Bruno >> >> >> >> >>> >>> Brent >>> >>> -- >>> You received this message because you are subscribed to the Google Groups "Everything List" group. >>> To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. >>> To post to this group, send email to everything-list@googlegroups.com . >>> Visit this group at http://groups.google.com/group/everything- list. >>> For more op
Re: A riddle for John Clark
On 23 Jul 2015, at 01:25, meekerdb wrote: On 7/22/2015 12:08 AM, Bruno Marchal wrote: On 21 Jul 2015, at 19:42, meekerdb wrote: On 7/21/2015 10:30 AM, Bruno Marchal wrote: So maybe one could see W AND W the same way I can see my computer screen AND my dog - just by attending to one or the other. You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. It follows from physics. We don't know that. Then why did you assert the necessity of a physical connection: "You will need a long neck to attend a conference in Moscow, and a party in Washington." Because we are at the step 3 protocol. The point is logical. Comp assumes a physical reality stable enough to have computer working deterministically, without anything non Turing emulable in them. Only later we will understood, from the reasoning, that such a physics needs to be extracted from arithmetic. We just assume that the physics is rich enough to implement locally universal machine, so that comp make sense, but then we arrive at the computationalist difficulties. Physics assume a brain/mind link which has to be justified, and the UDA shows the change we have to introduce. But you have effectively asserted that the duplicate persons at different locations do not experience both locations - their minds are separate because their brains are. Yes, as we assume computationalism (and thus some amount of physics needed to be able to say yes to a doctor). What is not assumed is that such a physical reality is primitive. Later, we get the proof that it cannot be primitive, and that such physics has to be derived from RA, or, if it contradicts it, we will refute computationalism. If that is more than just an assumption it is because it is relying on the physical basis of mind. I don't think so. At this stage it relies on some physics, to implement computer. It does not rely on the fact that such physics is primitive, and so it does not rely on the existence of a physical basis of mind. By definition of comp, it relies only on the fact that we are in a physical universe in which we can implement locally universal machine. No possibility to say "yes" to a doctor without it. If you reject the physical basis of mind then you might expect the duplicates to share one mind. Not at all, because we know, or have good reason to believe, that our brain are physical, and that our human consciousness needs it to manifest itself relatively to others and relatively to the physical universe. The conclusion of the UDA never put any doubt on this. It rejects only the idea that physicalism is true. No problem at all with physics, as long as the empirical world confirms the physics extracted from Robinson Arithmetic (and computationalism) as it does up to now. At some point we use only that (p -> ~p) -> ~p. (with p being for example the physical supervenience thesis. Bruno Brent But does it follow from UD computations? It should, (at step 7 and 8) and the point is only that it is testable. Up to now, it is working well. But to explain this, we need to dig deeper in computer science. Are you OK with the steps 0-6? 0-7? From your other posts, I think you were OK. So we can perhaps come back on step 8. I think Bruce Kellet has also some problem there. That can only be more intersting than the nonsense about step 3 that we can hear those days. Bruno Brent -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com . Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything- l...@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this gro
RE: A riddle for John Clark
Le 23 juil. 2015 09:24, "chris peck" a écrit : > > Quentin > > > >> Is measuring spin up under MWI has a probability of one or 0.5 under MWI? > > we've done this sketch before...and John Clarke just did the same sketch with you hours ago...Why do you need things repeated to you so much? No he did not. He pretends probabilities do have meaning in MWI. When he says 0.5 with his bet he ignores he is entangled with the measurement apparatus and duplicated with it, with one john winning and one losinf his bet. > > David Wallace, a proponent of MWI at Oxford University, puts it this way with regards to Schrodinger's Cat: > > "We're not really sure how probability makes any sense in Many Worlds Theory. So the theory seems to be a theory which involves deterministic branching: if I ask what should I expect in the future the answer is I should with 100% certainty expect to be a version of David who sees the cat alive and in addition I should expect with 100% certainty to be a version of David who sees the cat dead." > > What Wallace does is tackle incoherence head on. Does he over come it? Im not brainy enough to say. But I am brainy enough to see that he doesn't take the Bruno-Quentin approach of praying the problem will go away by pretending it doesn't exist. > > > > Date: Thu, 23 Jul 2015 08:48:51 +0200 > Subject: RE: A riddle for John Clark > From: allco...@gmail.com > To: everything-list@googlegroups.com > > > > Le 23 juil. 2015 05:09, "chris peck" a écrit : > > > > Quentin > > > > > > >> Then under MWI, same thing you're garanteed to see all results, so probability should also be one > > > > Deterministic branching leads to trouble rendering the idea of probability coherent. Go figure! Who would ever have guessed determinism and chance were difficult to marry... > > Then you're refuting MWI as not being able to correctly renders the probabilities, right? > > Is measuring spin up under MWI has a probability of one or 0.5 under MWI? > > Quentin > > > > > > Subject: Re: A riddle for John Clark > > To: everything-list@googlegroups.com > > From: meeke...@verizon.net > > Date: Wed, 22 Jul 2015 16:25:00 -0700 > > > > > > On 7/22/2015 12:08 AM, Bruno Marchal wrote: > >> > >> > >> On 21 Jul 2015, at 19:42, meekerdb wrote: > >> > >>> On 7/21/2015 10:30 AM, Bruno Marchal wrote: > > > > So maybe one could see W AND W the same way I can see my computer screen AND my dog - just by attending to one or the other. > > > You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. > >>> > >>> > >>> It follows from physics. > >> > >> > >> We don't know that. > > > > > > Then why did you assert the necessity of a physical connection: "You will need a long neck to attend a conference in Moscow, and a party in Washington." > > > >> We just assume that the physics is rich enough to implement locally universal machine, so that comp make sense, but then we arrive at the computationalist difficulties. Physics assume a brain/mind link which has to be justified, and the UDA shows the change we have to introduce. > > > > > > But you have effectively asserted that the duplicate persons at different locations do not experience both locations - their minds are separate because their brains are. If that is more than just an assumption it is because it is relying on the physical basis of mind. If you reject the physical basis of mind then you might expect the duplicates to share one mind. > > > > Brent > > > >> > >> > >> > >>> But does it follow from UD computations? > >> > >> > >> It should, (at step 7 and 8) and the point is only that it is testable. > >> Up to now, it is working well. But to explain this, we need to dig deeper in computer science. > >> > >> Are you OK with the steps 0-6? 0-7? From your other posts, I think you were OK. So we can perhaps come back on step 8. I think Bruce Kellet has also some problem there. That can only be more intersting than the nonsense about step 3 that we can hear those days. > >> > >> Bruno > >> > >> > >> > >> > >>> > >>> Brent > >>> > >>> -- > >>> You received this message because you are subscribed to the Google Groups "Everything List" group. > >>> To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. > >>> To post to this group, send email to everything-list@googlegroups.com. > >>> Visit this group at http://groups.google.com/group/everything-list. > >>> For more options, visit http
RE: A riddle for John Clark
Quentin >> Is measuring spin up under MWI has a probability of one or 0.5 under MWI? we've done this sketch before...and John Clarke just did the same sketch with you hours ago...Why do you need things repeated to you so much? David Wallace, a proponent of MWI at Oxford University, puts it this way with regards to Schrodinger's Cat: "We're not really sure how probability makes any sense in Many Worlds Theory. So the theory seems to be a theory which involves deterministic branching: if I ask what should I expect in the future the answer is I should with 100% certainty expect to be a version of David who sees the cat alive and in addition I should expect with 100% certainty to be a version of David who sees the cat dead." What Wallace does is tackle incoherence head on. Does he over come it? Im not brainy enough to say. But I am brainy enough to see that he doesn't take the Bruno-Quentin approach of praying the problem will go away by pretending it doesn't exist. Date: Thu, 23 Jul 2015 08:48:51 +0200 Subject: RE: A riddle for John Clark From: allco...@gmail.com To: everything-list@googlegroups.com Le 23 juil. 2015 05:09, "chris peck" a écrit : > > Quentin > > > >> Then under MWI, same thing you're garanteed to see all results, so > >> probability should also be one > > Deterministic branching leads to trouble rendering the idea of probability > coherent. Go figure! Who would ever have guessed determinism and chance were > difficult to marry... Then you're refuting MWI as not being able to correctly renders the probabilities, right? Is measuring spin up under MWI has a probability of one or 0.5 under MWI? Quentin > > > Subject: Re: A riddle for John Clark > To: everything-list@googlegroups.com > From: meeke...@verizon.net > Date: Wed, 22 Jul 2015 16:25:00 -0700 > > > On 7/22/2015 12:08 AM, Bruno Marchal wrote: >> >> >> On 21 Jul 2015, at 19:42, meekerdb wrote: >> >>> On 7/21/2015 10:30 AM, Bruno Marchal wrote: > > So maybe one could see W AND W the same way I can see my computer > screen AND my dog - just by attending to one or the other. You will need a long neck to attend a conference in Moscow, and a party in Washington. You can use a tele-vision system, and communicate by SMS, but unless you build a new corpus callosum between the two brains, and fuse the limbic system, by comp, the two "original" persons have become two persons, having each its unique experience. That follows from mechanism, and so P(W xor M) = 1, and P(W & M) = 0, as no one can open door in Moscow, and see some other city in the direct way of the first person experience. >>> >>> >>> It follows from physics. >> >> >> We don't know that. > > > Then why did you assert the necessity of a physical connection: "You will > need a long neck to attend a conference in Moscow, and a party in Washington." > >> We just assume that the physics is rich enough to implement locally >> universal machine, so that comp make sense, but then we arrive at the >> computationalist difficulties. Physics assume a brain/mind link which has to >> be justified, and the UDA shows the change we have to introduce. > > > But you have effectively asserted that the duplicate persons at different > locations do not experience both locations - their minds are separate because > their brains are. If that is more than just an assumption it is because it > is relying on the physical basis of mind. If you reject the physical basis > of mind then you might expect the duplicates to share one mind. > > Brent > >> >> >> >>> But does it follow from UD computations? >> >> >> It should, (at step 7 and 8) and the point is only that it is testable. >> Up to now, it is working well. But to explain this, we need to dig deeper in >> computer science. >> >> Are you OK with the steps 0-6? 0-7? From your other posts, I think you were >> OK. So we can perhaps come back on step 8. I think Bruce Kellet has also >> some problem there. That can only be more intersting than the nonsense about >> step 3 that we can hear those days. >> >> Bruno >> >> >> >> >>> >>> Brent >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Everything List" group. >>> To unsubscribe from this group and stop receiving emails from it, send an >>> email to everything-list+unsubscr...@googlegroups.com. >>> To post to this group, send email to everything-list@googlegroups.com. >>> Visit this group at http://groups.google.com/group/everything-list. >>> For more options, visit https://groups.google.com/d/optout. >> >> >> http://iridia.ulb.ac.be/~marchal/ >> >> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Everything List" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> e
