On 19 Mar 2012, at 04:23, Stephen P. King wrote:

Hi,

To add to John Mikes' point, it is being assumed that there is an invariance of consiousness under the copy and paste operations. What is the nature of this continuity?

The numerical identity of the copied digital states, and of course the way a computer is working.


It might help to keep in mind exactly what generates said continuity. This reminds me a lot of the notion of a connection that is used in presheaves and fiber bundles. The continuity in is induced to mappings to elements of a continuous index set. Does the UD provide such a base space?

yes, by the Z1 *and X1*, and S4Grz1 logics, we can say that it exists very plausibly. But this is irrelevant for the reversal result, which conclusion is that such a continuity has to exist. The continuity is brought by the consistency and soundness constraint on the machine. Keep in mind that comp is a problem, not a solution of a problem. The beginning of the solution is in AUDA, but let us be sure you get the UDA, in which case you can address your question with the math of AUDA, still assessing the UDA. If the Z and X logics does not work, you can still propose a different theory of knowledge, but it has to be compatible with the UDA steps, notably the six one (the dream-video argument), and it has to be defined through self-reference, to keep the distinction quanta/qualia.



The difficulty is that there are no unique paths in the base space (assuming that it is the UD) unless we arbitrarily introduce a measure on the UD. Where does it come from?

From the comp assumption.


We seem to be making a bunch of distracting arguments and hand waving to distract that we are basically putting in by hand a continuity and yet claiming that there is none occurring naturally.

Because nature has become secondary to the mind of machine/number. But in the discussion with Clark we are not yet there. We are still on the 1-indeterminacy point. The problem is only that Clark avoid the question, or fake to not understand it.



If we just assume the set of the natural numbers (via Borel sets), isn't this just an arbitrary bias toward a particular measure and not something natural? Where is the transitivity or ordering of the numbers coming from?

From the axiom of a tiny fragment of arithmetic.


We could we be operating on a set of p-adic numbers of a very large prime...

We use comp. the doctor will put the description of your state on an hard disk. It is a natural number, not a p-adic number. You point is meaningless: it is like I was saying that "2x^2 = y^2" has no solution, and you tell me "Did you have considered real numbers". By Church thesis, if you use a universal system based for the Turing universal base, then, you will have the same mind-body problem. The choise of the initial universal computing system is not relevant for addressing the question. It light be for the solution, but this has to be derived from any initial system. If not, you juts "copy nature", and this is exactly what is shown unsound if we want keep the qualia/ quanta distinction.



Put all the names of cities aside for a moment, we are really taking about movement in space-time.

No. That is not relevant at all for the goal of the thought experiment.



So my question is, why are we putting our selves through such convoluted abstractions to talk about the simple idea of moving though space-time?

Because we address the question of self-duplication, and of the existence of the 1-indeterminacy, which later will play a role in showing that the TOE cannot assume a given or primitive physical universe.

Stephen, you have already acknowledge that you grasped UDA, but now it looks like you have some doubt on the step 3, which, to be honest, might comfirm my feeling that you still miss the methodology I am using. I assume comp, and I use natural numbers because people are familiar with them. Then all what is done in UDA is a reduction of the mind-body problem to a "belief in body" problem in arithmetic (or recursively equivalent structure). The measure is shown existing indirectly, by comp. Which measure should be used is the subject matter of AUDA, the math part. And yes, it remains a lot of problems to extract physics. The main and only advantage of this is that we address the quanta/qualia question, and we show that the Aristotelians must abandon comp to do so.

Bruno



Onward!

Stephen

On 3/18/2012 6:45 PM, John Mikes wrote:

I think (?) it is Bruno's sentece:
"They can perceive the difference, not the duplication"
(if not, I apologize, but my remark is still on)
To perceive a difference goes with full knowledge of the comparison, knowing the 'previous' format (existence). They (and I am indeed not for the entire thought-play) can notice "a" state - irrespectively from any former history.

Sorry to embarge into this time- and energy wasting strawmanship.

John Mikes



On Sat, Mar 17, 2012 at 5:15 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:

On 17 Mar 2012, at 05:05, John Clark wrote:

Bruno Marchal <marc...@ulb.ac.be> wrote:

>>If he knew he was duplicated both would mention it, if he didn't neither would.

>The point is that he cannot perceive it. he can not known it by any personal observation,

So you're saying that neither the original nor the copy can feel the duplication, it does not enter their consciousness, it does not change their consciousness, and so far I agree with you completely; but then in the next breath you say it DOES change their consciousness and the change is about as dramatic as a change can get, it's so ENORMOUS that a new individual is created. So do you believe they can perceive the duplication or do you not?

They can perceive the difference, not the duplication.




> You misunderstand Everett. he said that we cannot feel the split ever after the differentiation occurred.

Everett said they would not feel the split but they would certainly feel other things,

Sure, me too.


and there would not even be a differentiation unless there was something different about them to differentiate. Everett would also say that talking about 2 absolutely identical points of view is silly, if there is no difference between them then there is only one point of view.

Me too.



> Now you come back to the idea that if I throw a dice, the notion of probability does not apply because the guy looking at the dice is not the same that the guy who threw it, which is straw man.

I know you like the phrase but when asked to calculate probabilities, or anything else for that matter, it is not a straw man to ask just what you want me to calculate; the probability that the guy who sees 12 on the dice will see 12 is 100%, the probability that the guy who does not see 12 on the dice will see 12 is 0%, the probability that right now John K Clark will see 12 when he throes the dice in his hand is 1 in 36.

> So it looks you can give us an algorithm to predict what you will feel with certainty the result of your future self- localization. But I have already explain why it does not work.

I know that there is one chance in 36 that my future self (I don't see the need of the word "localization") will be certain the dice gave him a 12, and the algorithm to calculate this has been well known for centuries.

I was illustrating a point. If the dices are medelt long enough the quantum uncertainties adds up and generates the 36 (* a continuum) possibilities, in which case quantum indeterminacy, which is different from the classical statistical one, and different from the comp 1-indeterminacy. The indeterminacies looks alike, but have different explanations, and different consequences.




> *in both cities* he will feel to survive *one and entire in only one city*.

Correct, therefore we can conclude that the Helsinki man will feel he has survived in both cities because HE HAS BEEN DUPLICATED and is now *in both cities*.

But he feels he is in only one city. He used your trick to predict that he will be in Moscow with 100%, but he woke in Washington. Ah! But you say he know that he has been duplicated and that he is in Washington AND in Moscow. But how could he *know* that? He can only *verifie* that. The presence, or not, or the other, the doppelganger, is like a scientific needing some confirmation. He can give a call to Moscow, to say hello to "himself", but bad luck, he just learned that the reconstitution machine failed in Moscow. This illustrates that each copies can know where they are, but can only believe the other copy is or not in the other city. They personal perspective are different, they knew this in advance, they perceive the difference, but they can only bet on the duplication, not experience them. The experiences they (can) get are only "I wake up in Moscow", OR "I wake up in Washington", and never "I wake up in washington and I wake in Moscow". The probability here on those future personal experiences.




> But the obvious point here is that he will not FEEL having survived in both cities.

Just ask them! Ask the Moscow man if he is the Helsinki man and if he is experiencing Moscow and he will answer "yes" to both questions. Ask the Washington man if he is the Helsinki man and if he is experiencing Washington and he will answer "yes" to both questions. Therefore it doesn't take rocket science to conclude that the Helsinki man experienced Moscow AND Washington.

Then, given that you and me are already the result of the many duplication since the first amoeba, we have all the life "at once". I love the idea, and I think we might have a very deep common first person indeed, but this is not relevant for the question of predicting, for example the "movie" you will feel to see in the multiplication-movie thought experience. Here the answer is "white noise", because it will be lived by the vast majority of the copies.




> Both copies will FEEL having survive in only one city,

Yes, but it doesn't matter because BOTH are the Helsinki man who just happens to be in another place, and we change our position all the time without loss of identity.

You are incredible. Of course it does matter, given that the question is explicitly about those personal feelings.

You look like "I don't want to talk about that".



> Each of them cannot know what the other feels.

True, so the Washington man is not the Moscow man, although both are the Helsinki man. For some things like the integers H, M and W if H=M and H= W then M=W, but that does not work for everything, for example a watermelon is green and a pea is green but a watermelon is not a pea; it doesn't work for personal identity either.

> You know perfectly well who you are, and the duplication will not change this.

Yes I will always know who I am, I will know I am in Moscow and only Moscow and I will know I am in Washington and only Washington and I will know I am in Helsinki and only Helsinki. Odd yes, contradictory no because there are 3 I's.

> You are back to the confusion between a 3-view on 1-views and the 1-views themselves.

One of us is very confused indeed over this point, but I don't believe its me.

This is equivalent with saying "I am right".



> Ask them if they have seen, from their own eyes, Washington AND Moscow. They will deny this,

Sure, but each has seen one of those cities and both are the Helsinki man (although they are not each other), therefore the Helsinki man saw Washington AND Moscow; the Washington man didn't and the Moscow man didn't but the Helsinki man did.

Lol




> unless you introduce magical telepathy between them.

Now THAT is a straw man! Telepathy has nothing to due with it.

Then you avoid the necessary ignorance of most copies, ignorance on which experience they will "actually" live and have lived. Think about the multiplication movie experience. You predict that you will see all movie, and I agree if "you" means the 1-you that you can attribute to those people, but I disagree if by "you" you mean each of those persons as they will experience. They discourse is simple, formally, because those experience are given by *each* movie (not *all* movie).




> You are just avoiding putting yourself at the place of each copies

I the Helsinki man walk into the duplicating chamber and walk right out and find that I the Helsinki man am now in Moscow, and I the Helsinki man walk into the duplicating chamber and walk right out and find that I the Helsinki man is now in Washington, and I the Helsinki man walk into the duplicating chamber and walk right out and find that I the Helsinki man am still right here in Helsinki and wonder if the duplicating chamber has malfunctioned. All three are me the Helsinki man and there is a 100% chance I will go to Moscow only and a 100% chance I will go to Washington only and a 100% chance I will remain in Helsinki. There is a 0% chance that I the Helsinki man in Moscow will see Washington and there is a 0% chance that I the Helsinki man in Washington will see Moscow. What have I avoided?

First, that in the protocol you are annihilated in Helsinki, so there is 0% you will wake up in Helsinki. Then you confuse an intellectual hybrid of your 1-you with the set {3-you in M, 3-you in W}, to be able to predict that you will be in both city. You avoid the question which concerns your present feeling as anticipated by your older Helsinki. If he predicted 100% for Moscow, then he was wrong for the guy who feel to be in Washington, and why not to listen to him?

The repetition of the experiences explains that the majority of W-M strings will be random, even incompressible. So the bet on the actual lived futures, in that multiplication-movie experience, is white noise. With this protocol, seeing Flying circus is a white rabbit phenomenon.




> You forget to say that neither the W-man nor the M-man could have guess in advance to be those one, from the complete protocol information he got in Helsinki.

Guessing is not necessary,

Well, this is what is asked!!!



the Helsinki man could have assigned a probability of 100% that if he sees Moscow then he will become the Moscow


But that is not the question!!!



man because that's what the Moscow man is, the Helsinki man who saw Moscow. And what is the probability that the Helsinki man will actually see Moscow? 100% of course.

By definition of first person (content or sequence of content of the diary), if you say 100% for each step of the duplication experience, then you are predicting that you will see "flying circus", and the 2^(16180 * 10000) * (90 * 60) * 24 other copies will laugh at you. Ok, you will have some neighbors who saw "Flying circus" with (correct, wrong) subtitles, and those who say the beginning, or the movie, or the trailer in Chinese with Korean subtitles, etc.
Don't count on it!

You avoid to answer the question which concerns the futures 1-view on the 1-view, by avoiding doing the experience, and defining an abstract notion of person distributed in the copies to avoid the simple fact that we will just look at the diaries which describe the experiences, and that with the movie-multiplication protocol, they almost all describes "white noise". The number of "senseful movie grows linearly", the number of white noise movie grows exponentially.

I said it precisely in the protocol, you have to bet which movie you will describe in the diary after the experience. Obviously after the experience they have all view ONE movie. OK, there is one "winner", having seen a perfect version of "flying circus", but the vast majority have not. In those thought experiments, you have to put yourself coldly at the place of some sample of those person.

With the quantum multiplication movie experience, the pixels are in quantum superposition which contagiate to the spectator, so that the quantum wave describes the spectator seeing all the movies, but again, the spectators does not feel the split nor the superposition, and see only *one* movie, and most of them will see white noise, for the same reason that beam splitters split the intensity into 1/2.

So logically, it is just plausible that the quantum indeterminacy might be an instance of the comp first person plural (with duplication of populations) indeterminacy. But we are not yet there.

You seem to continue to oscillate between there is no 1- indeterminacy, because ... 100% for Moscow, and there is an indeterminacy (but it is trivial, nothing new).

Let us assume you accept the 1-indeterminacy (trivial or not might be just another topic), might we move to step 4? Hint: revise step 0, 1, 2. Step zero is the definition of comp.

Bruno


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