I am wondering, if one reverses a rule in a causally invariant Wolfram model (i.e. changes it from A->B to B->A), whether the model remains causally invariant. Does anyone know an answer to the question?
I recal lthe definition
Causal Invariance: A property of multiway graphs whereby all possible
paths yield the isomorphic causal graphs. When causal invariance
exists, every branch in the multiway system must eventually merge.
Causal invariance is a core property associated with relativistic
invariance, quantum objectivity, etc. In the theory of term rewriting,
a closely related property is confluence. In a terminating system,
causal invariance implies that whatever path is taken, the “answer”
will always be the same.
Consider the rules B-> A, C → A acting on a word composed exclusively of letters from the alphabet {A, B, C}. This system is evidently causally invariant, because all the brances will merge to the same word after the substitution of all B and all C by A.
On the other hand, the system obtained by reversing the rules A-> B, A → C will be non-causally invariant for any word containing at least one “A”. For example, applying these rules to A, we obtain two brances B and C, which cannot merge, since there is not rule to transform neither B nor C.
Ok, yes, very good. If we have at least two rules, then the reverse rules do not have to be causally invariant. I also came up with a single rule example:
Consider the rule AB->AA. It is causally invariant for all starting states (can prove). If we start from ABA, we end with AAA in a single step. If we reverse the rule, then AAA can go into ABA or AAB.
I suspect, if we start from a graph (as opposed to a string), we can construct something very similar, although it seems more difficult to prove