On 8/24/2012 11:33 PM, meekerdb wrote:
On 8/24/2012 7:05 PM, Stephen P. King wrote:
"...due to the law of conjugate bisimulation identity:
A ~ A = A ~ B ~ C ~ B ~ A = A ~ B ~ A
this is "retractable path independence": path independence only over
retractable paths.
I don't understand this. You write A~(B~A) which implies that B~A is
a "system" (in this case one being simulated by A).
Dear Brent,
The symbol "~" represent simulate, so the symbols A~(B~A) would be
read as "A simulating B while it is simulating A". A and B and C and D
... are universal simulators ala David Deutsch. The can run on any
physical system capable of universality.
But then you write
A~B~A=A~A
These would read as: "A simulating B simulating A", which is
different from "A simulating B while it is simulating A", a subtle
difference. The former is simultaneous while the latter is not.
and also
A~B~C~A =/= A~C~B~A =/= A~A
This seems inconsistent, since A~B~C~A = A~D~A where D=B~C,
How do you get D=B~C from? That is inconsistent with the Woolsey
identity rule . For example C could be capable of simulating B in the
process of it simulating A, which is different in content from C
simulating A while A is simulating B. Simulators do not commute the way
numbers do. BTW, a simulation relation is not necessarily an identity
like "=".
but then A~D~A=A~A. And A~C~B~A = A~E~A where E=C~B, and then
A~E~A=A~A. But then A~B~C~A = A~C~B~A.
I seem to be assuming a natural ordering on the symbols A, B, C, D,
etc. and a notion of being at the same level in the ordering with the
"(..)" symbols. I should have made this clear. My apologies! Does the
comment about telescope property not make sense?
You drop the parentheses, implying the relation is associative, but
then you treat it as though it isn't??
Not having pointed out the ordering caused a confusion. My
apologies. Thank you for pointing this out! This idea still needs a lot
of work, that I do admit!
Brent
--
Onward!
Stephen
http://webpages.charter.net/stephenk1/Outlaw/Outlaw.html
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