The reason why this wouldn’t work is that sometimes what you’re calling “U(n)” would fail to be well defined (because some computation doesn’t halt)
No; the utility function is stipulated to be computable.
What Manfred is calling U(n) here corresponds to what the paper would call U(phi_n(k)).
The utility function is defined as being computable over all possible input.
phi_n(k) may not halt.
No; the utility function is stipulated to be computable.
What Manfred is calling U(n) here corresponds to what the paper would call U(phi_n(k)).
The utility function is defined as being computable over all possible input.
phi_n(k) may not halt.