Re: Title: A Structural Analysis of the Standard Halting Problem Proof

Am Fri, 25 Jul 2025 09:15:52 -0500 schrieb olcott:

On 7/25/2025 2:53 AM, joes wrote:

Am Thu, 24 Jul 2025 16:41:26 -0500 schrieb olcott:

On 7/24/2025 4:24 PM, joes wrote:

Am Thu, 24 Jul 2025 09:32:45 -0500 schrieb olcott:

 

Aborting prematurely literally means that after N instructions of
DDD are correctly emulated by HHH that this emulated DDD would reach
its own emulated "ret" instruction final halt state.
What value of N are you proposing?

>
Let's see: the call to HHH is #4, [waves hands], then another 4
inside the next level of simulation, and after another 4 the first
simulated HHH (the one called by the input, not the outermost
simulator. We are now 3 levels in) decides that enough is enough and
aborts,

>
Thus immediate killing its simulated DDD and everything else that HHH
was simulating thus no simulated DDD or simulated HHH can possibly
ever return no matter how many or how few X86 instructions that the
executed HHH correctly emulates.
This is the part that you fail to understand or understand that I am
correct and disagree anyway.

 

You failed to understand I was talking about the first simulated HHH
aborting, not the outermost simulator.

 
*I am trying to get you to understand that is impossible*
The only HHH that can possibly abort is the outermost directly executed
one.

True if the input changes along with the simulator, but not if we
simulate the fixed input (that aborts after 4+4=8 instructions of DDD,
when we encounter the second nested call to HHH) without prematurely
aborting.
I get that if you change what "HHH" refers to in order do extend the
simulation you necessarily simulate a different input. You don't.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.