Quentin Anciaux wrote:
2015-05-04 10:24 GMT+02:00 Bruce Kellett <bhkell...@optusnet.com.au
Quentin Anciaux wrote:
But then again... you reject step 0, so why bother saying
because of that step N is invalid... well ok, if step 0 is
invalid, any further deduction from it is also invalid. Either
just for the argument, you go as if step 0 is valid and try to
show an error, meaning to show that step N is not deductively
valid from the steps 0 to N-1 so as to demonstrate that the
argument as a whole is invalid (and so is not deductively valid
from step 0)... that's not what you do, you start with 0 is
invalid... so well, yes the rest doesn't follow, no need for you
to attack a special particular step after that. But if the
argument is deductively valid, then if step 0 is true, the rest
follows.
You haven't followed the twists and turns of the discussion. I
rejected step 0 (in this particular discussion) because substitution
at the quantum level is impossible. But comp probably only needs
classical digital substitution. In that case, one can go on and
criticize the rest of the argument.
Yes but not by saying step 3 is invalid because you can't make copy at
that level... because what you're saying is step 0 is false so step 3 is
false... If you want to critic step N, you have to say it is false
because it does not follow deductively step N-1... if not, then it was
step N-1 you should have critized. So I've lost your point, what's your
critic if you accept steps 0,1,2 against step 3 ?
The point I was making about step 3 was that if quantum level
substitution was necessary, then deliberate copying was out. People then
claimed that chance duplicates would do as well. I rejected this by
point out that it violated step 3 where the duplication is explicitly
deliberate, and known to the subject. Chance copies fail to fit these
criteria.
Bruce
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