Um, AIXI is not computable. Relatedly, K(AIXI) is undefined, as AIXI is not a finite object.
Also, A can simulate B, even when K(B)>K(A). For example, one could easily define a computer program which, given sufficient computing resources, simulates all Turing machines on all inputs. This must obviously include those with much higher Kolmogorov complexity.
Yes, you run into issues of two Turing machines/agents/whatever simulating each other. (You could also get this from the recursion theorem.) What happens then? Simple: neither simulation ever halts.
Um, AIXI is not computable. Relatedly, K(AIXI) is undefined, as AIXI is not a finite object.
Also, A can simulate B, even when K(B)>K(A). For example, one could easily define a computer program which, given sufficient computing resources, simulates all Turing machines on all inputs. This must obviously include those with much higher Kolmogorov complexity.
Yes, you run into issues of two Turing machines/agents/whatever simulating each other. (You could also get this from the recursion theorem.) What happens then? Simple: neither simulation ever halts.