Re: DDD correctly emulated by H0 --- Why Lie?

On 6/23/24 7:34 PM, olcott wrote:

On 6/23/2024 5:58 PM, Richard Damon wrote:

On 6/23/24 6:45 PM, olcott wrote:

>
You know what the freak I was talking from prior
discussions unless your brain is so damaged that
you can't remember anything from one post to the next.
>
In the case that you affirm that your brain <is>
this damaged then I humbly apologize.
>

>
>
No, you don't know what you are talking about.
>

So you insist on lying about this verified fact?
 _DDD()
[00002172] 55               push ebp
[00002173] 8bec             mov ebp,esp
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 According to the semantics of the x86 programming language
when DDD correctly emulated by H0 calls H0(DDD) this call
cannot possibly return.
 

I won't say it can't be true, but it hasn't been proven, largely because it seems you don't know how to do a formal logic proof.
Also, with JUST what you have said, I can prove it FALSE with an unpure H0, but I won't push that as you have previous indicated your intentions, but that also means that you have previously indicated that H0 is supposed to be a Halt Decider, so we can say you definition of its requirements are just wrong, so again it is an incorrect question.
So, by that logic, you claim isn't correct (either being provably wrong or just the wrong questiion).
You also haven't shown that your H0 actually DOES the required correct simulation, so you can't claim your results either, even if it is established that on partial correct emulation by any version of H0 can reach the return in H0, because we don't know if your H0 actually does a correct emulation. (If it did, you should be able to show its trace)
Part of the problem, is with the input given, you can't get past the call 000015d2 instruciton as you don;t specify what is there, so we need to either assume that your problem is just an invalid one, or that we can use the implied input, which means that all of the code of H0 is included in the input, and thus any change to H0 implies a change to the input the way you do it (meaning we can't compare the results of different H0s).
So, in summary, you are just showing a total lack of understanding of what you are talking about.