On 27 Mar 2015, at 12:53, Bruce Kellett wrote:
LizR wrote:
On 28 March 2015 at 00:06, Quentin Anciaux <[email protected] <mailto:[email protected]
>> wrote:
1- It is assumed you have a machinery/program that is conscious.
(a
real conscious AI)
2- You have (for example) a conversation with it.
3- While doing that conversation, you record all inputs fed to the
machine.
4- You replay those inputs to the machine.
To make sure I have this right - you reboot it, or whatever - this
is a machine that starts from the same starting state as the one
you talked to originally. It doesn't remember the first
conversation, and hence by hypothesis goes through the same states
as before.
5- Assuming in 3 the machine was conscious, replaying the same
inputs, the machine should still be conscious.
6- You remove from the machine all the transistor not in use
during
that particular run (given the recorded input)
7- You replay those inputs to the ("crippled") machine. 8-
Assuming in 3 and 5 the machine was conscious, replaying the same
inputs, the machine should still be conscious as in 5 (because
what
you removed wasn't in use anyway).
OK
9- You break one transistor, but you make a device (in the MGA
it's
the projection of the record on the graph) that permits (even if
the
transistor is broke) to mimic the output at the exact moment it
should have happen if the transistor wasn't broken (like the lucky
cosmic ray replacing the firing of a neuron).
OK
10- Assuming in 3,5 and 8 the machine was conscious, replaying
the
same inputs, the machine should still be conscious as the broken
transistor while not working did nonetheless gave the correct
output thanks to the lucky ray/devide/movie projection.
11- You do 9 for all the transistor, so as to leave only the
mimic...
Aha. Yes that makes sense. It's a slippery logical slope ...
12- Assuming in 3,5,8 and 10 the machine was conscious, then the
machine is still conscious while no computation occur anymore....
contradicting computationalism.
Yes, so you are finally playing just a recording because for every
component you have to know exactly what its outputs were, so you
have to record everything, not just the inputs. At this point you
have shown that either consciousness can supervene on playing back
a recording OR that consciousness doesn't supervene on the original
physical substrate that was supposed to be performing the
computation.
From that, either computationalism is false or physical
supervenience is false.
Hmmm....I'm not sure where I sit on that. I do feel like some
sleight of hand has been pulled - not intentionally, of course.
Perhaps the broken version might still be conscious, which means
that ... eek. That's like saying Klara's conscious despite being
inert, isn't it?
I think it's the "thinking about what it all means afterwards" part
that ties my brain in knots. I want to just throw my hands up and
say "well of course physical supervenience doesn't work! How can a
bunch of atoms do that, anyway?" But then they do seem to ...
Computers are just bunches of atoms......
That's a physical computer, if that exists.
When I use computation, I mean some (sigma_1) arithmetical relation
between numbers.
I think it is the all or nothing aspect of computationalism that is
a problem. I have no difficulty in accepting that a simulation run
on a computer can be conscious -- silicon brain rather than wetware.
But I have difficult in accepting that a computation can be run in
the required way without a physical substrate of some sort (computer
or whatever).
The running of a computation is just what a universal number do when
given a program. It defines sequences of states related in some way by
that universal number. The measure problem comes from that all
computations are run by infinities of universal numbers.
A computation is just calculating a function over the reals.
Not the digital one of computationalism. We have only function from N
to N. We can use some recursive equivalence.
Consciousness involves memory
I don't know that.
and I do not think that a memory can be formed in the abstract.
I can refute that. The axioms of RA entails already the existence of
all computations, including all those involving memories.
Memory is remembered -- laying down memories increases entropy, and
abstract ideas do not have entropy.
?
(Although thinking abstract ideas does increase the entropy
associated with your brain!) I don't think any of this works without
a physical substrate, so, in the end, all consciousness supervenes
on the physical, no matter how indirectly.
Then computationalism is wrong. That is all what I show. You should
read some book on computer science. logicians ate metaphysics, but it
is a simple facts for them that RA proves the existence of all
computations. if phi_i(j) = k, RA prove it, and computation are
realized in RA semantic like you might believe that computations are
realized in some physical realm, but with comp, the qubit-->bit is a
two way road.
Thanks to the work of Matiyazevitch, a universal dovetailer is already
given by diophantine equation. So we don't need to believe in anything
more than the solution of the equation given below(*). It 50 years to
find such polynomial. The Turing universality of the diophantine
equation was not easy to prove. I show this only to illustrate that by
computation I mean a precise mathematical notion, discovered by Post,
Kleene, Church, Turing.
Bruno
(*)
31 unknowns: A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S,
T, W, Z, U, Y, Al, Ga, Et, Th, La, Ta, Ph, and two parameters: Nu and
X.
X is in W_Nu iff phi_Nu(X) stop if and only if
BEGIN:
Nu = ((ZUY)^2 + U)^2 + Y
ELG^2 + Al = (B - XY)Q^2
Qu = B^(5^60)
La + Qu^4 = 1 + LaB^5
Th + 2Z = B^5
L = U + TTh
E = Y + MTh
N = Q^16
R = [G + EQ^3 + LQ^5 + (2(E - ZLa)(1 + XB^5 + G)^4 + LaB^5 + +
LaB^5Q^4)Q^4](N^2 -N)
+ [Q^3 -BL + L + ThLaQ^3 + (B^5 - 2)Q^5] (N^2 - 1)
P = 2W(S^2)(R^2)N^2
(P^2)K^2 - K^2 + 1 = Ta^2
4(c - KSN^2)^2 + Et = K^2
K = R + 1 + HP - H
A = (WN^2 + 1)RSN^2
C = 2R + 1 Ph
D = BW + CA -2C + 4AGa -5Ga
D^2 = (A^2 - 1)C^2 + 1
F^2 = (A^2 - 1)(I^2)C^4 + 1
(D + OF)^2 = ((A + F^2(D^2 - A^2))^2 - 1)(2R + 1 + JC)^2 + 1
END
==================
Bruce
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