Those tired of Clark's argument can skip up to the (more interesting)
Holiday Exercise below.
My be this could help Clark to try to find a new argument, as again,
he just brought his usual invalid trick, as I show one last times.
On 10 Jul 2016, at 19:29, John Clark wrote:
On Sun, Jul 10, 2016 Bruno Marchal <[email protected]> wrote:
<snip>
> Pleasen replaced "all copies" by "each copy", and call them
as you want.
OK.
1) Each copy saw only one city.
Excellent! That is the correct 1-view description. Now, you just need
to interview each copy about the prediction made in Helsinki and
written in the diary to evaluate the better one.
If you agree that "Each copy saw only one city", you have to agree
that "each copy realize that "W & M" is refuted, that half of the
copies refute all specific predictions, and that all verifies "W V M".
2) All the copies together saw 2 cities.
Correct 3p description of the experiences of all copies. That is the
3-1 view. We need it to get the correct "1)", but "all the copies" is
not a person, that is why you correctly add "together" (which is the
3-1 view, in which we are not interested).
3) All the copies have an equal right to call themselves "John
Clark".
Yes, but below you are using "3') All the copies has an equal right to
call themselves "John Clark", like if the set of reconstituted people
was a sort of super-entity. That is the 3-1 view. But we are asked
about the 1-views. Again, better to use "each" than "all", you get
back to what you were asked to avoid.
4) The statement "John Clark will see two cities" turned out to
be unambiguously true.
In the 3-1 view, sure. But we asked about the 1-views.
If you were talking about the 1-views, then that is directly refuted
by each copy, or each 1-view (or all of them, well understood). Given
the enunciation of the problem, you are just wrong, as the
verification procedure has shown again above.
5) The statements "you will see one city" and "you will see 2
cities" and "you will see no city" turned out to be neither true nor
false because in a world of people duplicating machines the personal
pronoun "you" is ambiguous.
> When interviewing each of them in the cities, each have no
problem to understand to whom I am asking the question, whatever
words are used for it.
But the interviewees do NOT agree among themselves to whom the
question was asked.
Utter non sense, again made possible by the confusion you introduce by
abstracting from the precision given.
If I don't give the precision, you say: vague, ambiguous. If I give
you the precision (like in the papers, books), you say "pee-pee
jargon", and refuse to use them. When eventually you use them, you say
the result is not original, but fail to say why you don't move on the
next step.
So which one was right?
Trivially both when in Helsinki the prediction written in the diary
was "W v M", and none for any other. There is no ambiguity, and the
prediction are simple and clear, and the criteria of verification
(interviewing all copies) is very clear too.
You might try to use the exercise below to try to find another
refutation. Your current attempt has been debunked many times, and as
Telmo and other said, it is a bit boring.
==================
Holiday Exercise:
A guy undergoes the Washington Moscow duplication, starting again from
Helsinki.
Then in Moscow, but not in Washington, he (the one in Moscow of
course) undergoes a similar Sidney-Beijing duplication.
I write P(H->M) the probability in H to get M.
In Helsinki, he tries to evaluate his chance to get Sidney.
With one reasoning, he (the H-guy) thinks that P(H-M) = 1/2, and that
P(M-S) = 1/2, and so conclude (multiplication of independent
probability) that P(H-S) = 1/2 * 1/2 = 1/4.
But with another reasoning, he thinks that the duplications give
globally a triplication, leading eventually to a copy in W, a copy in
S and a copy in B, and so, directly conclude P(H-S) = 1/3.
So, is it 1/4 or 1/3 ?
Can you modify a bit the protocol so that we get any of those results?
Bruno
http://iridia.ulb.ac.be/~marchal/
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