Re: ChatGPT and claude.ai agree that I have refuted the conventional Halting Problem proof technique ---

Op 30.jun.2025 om 23:50 schreef olcott:

On 6/30/2025 3:40 PM, joes wrote:

Am Mon, 30 Jun 2025 11:33:40 -0500 schrieb olcott:

On 6/30/2025 2:38 AM, Fred. Zwarts wrote:

Op 29.jun.2025 om 15:29 schreef olcott:

On 6/29/2025 5:34 AM, Fred. Zwarts wrote:

Op 28.jun.2025 om 15:08 schreef olcott:

>

HHH reaches its "return" instruction final halt state.

Cool, that makes DDD loop forever.
>

DDD correctly simulated by HHH cannot possibly reach its own final
halt state no matter what HHH does.
You are getting confused over what is being measured. If we were
measuring whether or not HHH halts then you would be correct. *We
are not measuring that*
>

You are confused. If HHH has a bug that does not allow it to reach
the final halt state of its simulation, then that is not a measure
for the halting behaviour specified in the input.

>
If HHH had a bug then someone could have pointed out the exact bug on
the last two years that its full source has been available.
https://github.com/plolcott/x86utm

Another counter-factual claim.
The bug has been presented to you many many times. The bug is that it
forgets to count the conditional branch instructions when simulating
itself.

>

The conditional branch instructions in HHH cannot possibly have any
effect whatsoever on whether or not the simulated DDD reaches its own
"return" instruction final halt state.

 

But of course they do. Once as part of the outermost simulator, deter-
mining whether to abort, and once as part of DDD, determining whether
to return.
>

HHH simply simulates DDD with a pure simulator until it conclusively
proves that its outermost simulated DDD cannot possibly reach its own
simulated "return" statement final halt state. When it aborts this DDD
all recursive emulations immediately stop.

Right, HHH is not a pure simulator.

 

The simulations would have halted
if *only* the outermost HHH was pure and did not abort.

 Since you know that is impossible because
every instance of HHH has the exact same
machine code at the same machine address
why bring it up?
 

 *That* is what
what Sipser's twisted-by-you words meant, not simulating a different
DDD that calls a pure simulator. The code of DDD, including HHH, is
fixed and "describes" a partial simulator; you can't in a program
magically refer to "itself", you have to arrange for the explicit
mention to fall together with its name (except in Lisp maybe).
>

The input to HHH(DDD) never stops running unless aborted
is saying the same thing as DDD correctly simulated by HHH:
 HHH(DDD) simulates DDD that calls HHH(DDD)
simulates DDD that calls HHH(DDD)
simulates DDD that calls HHH(DDD)
simulates DDD that calls HHH(DDD)
simulates DDD that calls HHH(DDD)
simulates DDD that calls HHH(DDD)
 Cannot possibly reach its own simulated
"return" statement final halt state.
 

Why repeating this failure of HHH? The simulated HHH, as you yourself admitted a few lines above, has the same code as the simulating HHH. If the simulating HHH is programmed to abort after a few cycles, the input for the simulating HHH specifies the same aborting code. Indeed, HHH has a bug so that it cannot reach that part of the specification, but that does not change the specification, it only makes HHH blind for that part of the specification.
Another effect of you strange construction is, that the input changes as soon as you change HHH. This makes you blind for the fact that each aborting HHH aborts just one cycle too soon, reporting an incorrect result, where the not-aborting HHH never returns to report.
You would help yourself with a construction where the simulating HHH and the simulated HHH would not reside in the same memory, but where the simulated HHH is a copy of the simulating HHH. Then it becomes more evident what exactly is the behaviour of the program specified in the input and what is the behaviour of the simulator.