Le 15-juin-06, à 14:40, Stathis Papaioannou a écrit :

<x-tad-bigger>I've always wondered (from a position of relative mathematical naivete, please understand)</x-tad-bigger>
<x-tad-bigger> about the process in arguments like this whereby reasoning about arithmetic comes to</x-tad-bigger>
<x-tad-bigger> include the labels applied to arithmetical statements, on the grounds that these labels happen</x-tad-bigger>
<x-tad-bigger> themselves to be numbers. The function g as defined below is based on a look-up table, and </x-tad-bigger>
<x-tad-bigger> the contradiction consists in that this table has the same label applied to more than one</x-tad-bigger>
<x-tad-bigger> distinct value, i.e. fk(k) and fk(k)+1. Doesn't this just mean that the  chosen labelling</x-tad-bigger>
<x-tad-bigger> scheme is ill-chosen? Is there any way to separate f and g formally, perhaps to call g a</x-tad-bigger>
<x-tad-bigger> meta-function (or something) rather than a function of f, and thus avoid the contradiction?</x-tad-bigger>
<x-tad-bigger> Or is this whole train of thinking basically flawed, there being no method in general to</x-tad-bigger>
<x-tad-bigger> banish g from the cosy world of functions like f(x)=x^2 or sin(x) or |x|?</x-tad-bigger>


Just two questions before I leave, the first concerns this current thread, the second refers to an older post of you.

Why do you think g is defined from a look-up table? G applied on n generates the Fi when needed. g would like generate the fi but can't. I am using the convention that the Fi are the programmable function (thus the partial computable one), and the fi are the total computable functions.

The second question is related to your assessment (in older posts) that in a self-duplication W M, with annihilation of the "original" in Brussels, the probability are equal to 1/2. I don't disagree with this.

But are you aware of the difficulty of composing such probabilities? What if in Moscow I am read and annihilated again, and duplicate into Sidney and Beijing? I mean what are in Brussels the (first person) probabilities to reach and stay in Washington, Sidney and Beijing respectively?

The most common answers are "P(W) = 1/2 P(S) = P(B) = 1/4" and "P(W) = P(S) = P(B) = 1/3".
The problem here is that you can have a continuum of thought experiments, where you introduce ranges of amnesia in Moscow going from a case where "clearly" it should be 1/2 1/4 1/4 and (in the case of a complete amnesia of anything happened in Moscow) 1/3 1/3 1/3.

What would you say? I mention this problem because it has motivated me for going more technical. I think it shows that amnesia makes possible the relative fusion of comp histories ... like the one of the quantum (empirical) world. (I think George also assessed a relation between fusing and amnesia if I remember correctly (?)).

Bruno



http://iridia.ulb.ac.be/~marchal/


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