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


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!





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