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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: Every D(D) simulated by H presents non-halting behavior to H ###
Date: Mon, 27 May 2024 17:34:28 -0400
Organization: i2pn2 (i2pn.org)
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On 5/27/24 3:52 PM, olcott wrote:
> On 5/27/2024 11:37 AM, Richard Damon wrote:
>> On 5/27/24 12:06 PM, olcott wrote:
>>> On 5/27/2024 10:56 AM, Richard Damon wrote:
>>>> On 5/27/24 11:43 AM, olcott wrote:
>>>>> On 5/27/2024 9:58 AM, Richard Damon wrote:
>>>>>> On 5/27/24 10:39 AM, olcott wrote:
>>>>>
>>>>> typedef int (*ptr)();  // ptr is pointer to int function in C
>>>>> 00       int H(ptr p, ptr i);
>>>>> 01       int D(ptr p)
>>>>> 02       {
>>>>> 03         int Halt_Status = H(p, p);
>>>>> 04         if (Halt_Status)
>>>>> 05           HERE: goto HERE;
>>>>> 06         return Halt_Status;
>>>>> 07       }
>>>>> 08
>>>>> 09       int main()
>>>>> 10       {
>>>>> 11         H(D,D);
>>>>> 12         return 0;
>>>>> 13       }
>>>>>
>>>>> The above template refers to an infinite set of H/D pairs where D is
>>>>> correctly simulated by either pure simulator H or pure function H. 
>>>>> This
>>>>> was done because many reviewers used the shell game ploy to endlessly
>>>>> switch which H/D pair was being referred to.
>>>>>
>>>>> Correct Simulation Defined
>>>>>     This is provided because many reviewers had a different notion of
>>>>>     correct simulation that diverges from this notion.
>>>>>
>>>>>     A simulator is an x86 emulator that correctly emulates 1 to N 
>>>>> of the
>>>>>     x86 instructions of D in the order specified by the x86 
>>>>> instructions
>>>>>     of D. This may include M recursive emulations of H emulating 
>>>>> itself
>>>>>     emulating D.
>>>>
>>>> And how do you apply that to a TEMPLATE that doesn't define what a 
>>>> call H means 
>>>
>>> *It is completely defined and you are just ignoring this definition*
>>
>> So, what instruction does the call H in D go to to be simulated?
>>
> 
> DISHONEST HEAD GAMES. WHEN WE APPLY THIS SAME REASONING TO THE
> LINZ TEMPLATE YOUR REASONING CALLS THE LINZ TEMPLATE NONSENSE.

Nope, because Linz doesn't try to pass a Template to H, but the machine 
built from the template H^.

As I said, if you assume the input is the machine built from the 
template you get the ability to define the simulation, you just now get 
every decider got a different input, so you can't just do the logic 
across them.

YOU are the one trying to do dishonest head games

(And who has been saying that insults are unprofessional?)

> 
>> As a template, there is no fixed H, so no instruction to look at.
>>
>>> H correctly simulates 1 to ∞ steps of D with either pure function H
>>> or pure simulator H. In none of these cases does the correctly simulated
>>> D ever reach its own simulated final state and halt.
>>>
>>> Do some of these instances of H play a game of poker with themselves
>>> before or after they simulate D? Yes they do because the H/D pairs
>>> are an infinite set.
>>>
>>
>> But, how do they correctly simulate something that isn't there?
>>
>> Either they are simulating an INSTANCE of the template, in which case 
>> each H is looking at a DIFFERENT instance, and you can't relate one 
>> result to the other, or they are trying to simulate the Template, at 
>> which point you have the problem that the code to be simulated hasn't 
>> been defined, and thus you can't do what you define to do.
> 
> I AM REFERRING TO THE EXACT SAME SORT OF INFINITE SET
> THAT THE LINZ TEMPLATE IS REFERRING TO AND YOU KNOW IT.

Nope, Linz choose A SPECIFIC H out of the set, and gives it a SPECIFIC 
H^ built from that SPECIFIC H, and then works with that set. There is 
ZERO logic about infinite sets in the part that shows that THIS H can't 
get the right answer to THIS input.

Only afterwords, by pointing out that the SPECIFIC H was abritrarily 
chosen, and the exact same steps can be done with ANY OTHER machine that 
might be considered to be a Halt Decider, does the proof go to show that 
no possible H can exsit.

You just don't seem to understand how to do proof with universal qualifiers.


> 
> WHEN EVERY ELEMENT OF AN INFINITE SET HAS THE SAME NON-HALTING
> PROPERTY THEN YOUR SHELL GAME SWITCHEROO IS OBVIOUSLY A HEAD GAME.

But they don't.

Every element D of the infinite set where H(D,D) answer non-halting, 
will halt even though the H is incapable of simulating the input to that 
point since it gives up to soon.


> 
> IF YOU WOULDN'T HAVE TRIED TO GET AWAY WITH THIS FOR TWO SOLID
> YEARS I WOULD NOT BE YELLING.
> 

But it is YOU who has been ignoring the definitions of the system, 
apparently because you never learned them.

Apparently in 20 year of studying the proof, you never say that H and H^ 
were refered to as "a Turing Machine", i.e. a singular, and not as an 
"infinite set" of such machines