so long as the infinite bitstring that computes the halt function for Turing Machines is simulated with every other bitstring
I don’t know what you mean by this.
A Turing machine with a halting oracle can (trivially) solve the halting problem. A Turing machine with access to a closed timelike curve can solve the halting problem. A Turing machine on its own cannot. And we do not have access to either a halting oracle or a closed timelike curve.
The halting oracle can be encoded (by God) as an infinite bitstring. Every finite prefix of that string appears in a list of all finite strings. That list of all finite strings can be generated by a Turing machine. But none of that matters, because we do not know and cannot know which subset of all those finite strings encodes a halting oracle.
Here is a simpler example of the issue here. Consider two Turing machines. Each of them ignores its input tape. One prints “Yes” and halts, and the other prints “No” and halts. One of these machines answers the Riemann Hypothesis. Of course, we do not know which, which makes them useless for answering the Riemann Hypothesis.
I don’t know what you mean by this.
A Turing machine with a halting oracle can (trivially) solve the halting problem. A Turing machine with access to a closed timelike curve can solve the halting problem. A Turing machine on its own cannot. And we do not have access to either a halting oracle or a closed timelike curve.
The halting oracle can be encoded (by God) as an infinite bitstring. Every finite prefix of that string appears in a list of all finite strings. That list of all finite strings can be generated by a Turing machine. But none of that matters, because we do not know and cannot know which subset of all those finite strings encodes a halting oracle.
Here is a simpler example of the issue here. Consider two Turing machines. Each of them ignores its input tape. One prints “Yes” and halts, and the other prints “No” and halts. One of these machines answers the Riemann Hypothesis. Of course, we do not know which, which makes them useless for answering the Riemann Hypothesis.