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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: My reviewers think that halt deciders must report on the behavior
 of their caller --- Mike
Date: Wed, 9 Jul 2025 21:49:01 -0400
Organization: i2pn2 (i2pn.org)
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On 7/9/25 9:55 AM, olcott wrote:
> On 7/9/2025 6:44 AM, Richard Damon wrote:
>> On 7/8/25 3:49 PM, olcott wrote:
>>> On 7/8/2025 2:01 PM, Mike Terry wrote:
>>>>
>>>> This is one of PO's practiced tactics - he makes a claim, and 
>>>> regardless of how patently false that claim appears, he refuses to 
>>>> logically defend the claim beyond saying "the claim is true, and if 
>>>> you understood xxx you would realise it is true".
>>>>
>>>
>>> All of my claims are easily verified facts to those
>>> with the capacity to verify them.
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>>
>>> _DDD()
>>> [00002192] 55             push ebp
>>> [00002193] 8bec           mov ebp,esp
>>> [00002195] 6892210000     push 00002192  // push DDD
>>> [0000219a] e833f4ffff     call 000015d2  // call HHH
>>> [0000219f] 83c404         add esp,+04
>>> [000021a2] 5d             pop ebp
>>> [000021a3] c3             ret
>>> Size in bytes:(0018) [000021a3]
>>
>> Not a program, must include the code for HHH to be simulatable.
>>
> 
> You must have dementia.
> I have told you that HHH does emulate DDD
> then it emulates itself emulating DDD 500 times now.
> 

And thus you admit that you are lying.

IF the above is the input, then HHH can't "emulate itself" while 
"emulating the input" as it isn't part of the input.

And thus, you are admitting that HHH just isn't actually a "program" per 
computaiton theory, as those only look at their inputs.

If you include HHH in the input, then that defines what HHH actually is, 
and thus to "correctly emulate itself" that simulation need to do what 
HHH itself does.

Since HHH(DDD) returns 0, the call from DDD of HHH(DDD) must also 
emulate to it returning 0, and other "evaluation" is just incorrect.

Sorry, but you are just proving that you are just a stupid ignorant liar 
that doesn't know what he is talking about.

By your logic, computers don't work, as programs don't have to do what 
their code says they will do, and since simulations can not be 
resposible for the direct execution of the program their input 
represents, there are not UTMs, and thus not simulators, and thus no 
simulating halt deciders, so you whole world just vanishes in a puff of 
self-contradictions

>>>
>>> I am utterly shocked that you can't understand
>>> that DDD emulated by HHH according to the semantics
>>> of the x86 language cannot possibly reach past
>>> it own machine address [0000219a].
>>
>> But that DDD Can't be simulated by HHH without including the code that 
>> you refuse to accept is part of the input.
>>
> 
> HHH is the test program.
> DDD is the program under test.

Which to be a program, include the code of HHH.

> HHH is only determining whether or not DDD emulated
> by HHH can reach its own emulated final halt state.

Then it isn't a halt decider, and you are just admitting that youj whole 
arguement is based on the lie of a strawman.

> 
> The historical (official received view) answer to
> this same question is:
> 
> I have no idea. I give up. I will call it undecidable.
> 

Nope, The official answer is that the question of what the input does 
HAS an answer, it is just impossible to make a program that can compute 
that answer for all possible inputs.

In the case of your particular HHH that you have defined, DDD will halt, 
and thus HHH should return 1 to be correct, but it returns 0.

Nothing wrong with that, we understand that we will not know some things.

If you think all truth is knowable, do you accept that you will die?

And thus, if you know that to be true, you should also know WHEN that 
will happen.

Of course you don't know that (unless you make the fact self-determined) 
so you have to accept that some truth is not knowable.

And thus some mappings are uncomputable.

And you are just proved to be stupid and ignorant.