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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DD correctly emulated by HHH --- Totally ignoring invalid
 rebuttals ---PSR---
Date: Sat, 8 Mar 2025 17:25:27 -0500
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On 3/8/2025 11:29 AM, olcott wrote:
> On 3/8/2025 9:00 AM, dbush wrote:
>> On 3/8/2025 9:03 AM, olcott wrote:
>>> On 3/8/2025 2:47 AM, Fred. Zwarts wrote:
>>>>
>>>> So, we agree that any simulator that tries to simulate *itself* 
>>>> cannot possibly reach the end of its simulation.
>>>
>>> Apparently you don't understand that inputs to a
>>> simulating termination analyzer specifying infinite
>>> recursion or recursive emulation cannot possibly
>>> reach their own final state and terminate normally.
>>
>> Apparently you don't understand that inputs to a termination analyzer, 
>> simulating or otherwise, are specified by the specification that is 
>> the halting function:
>>
>> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
>>
>> And HHH(DD)==0 fails to meet the above specification
> 
> Until you understand how and why this is necessary
> correct strongly held misconceptions will persist:
> 
> Replacing the code of HHH with an unconditional simulator and 
> subsequently running HHH(DD) cannot possibly reach
> its own "ret" instruction and terminate normally
> because DD calls HHH(DD) in recursive emulation.
> 
> 

Failing to understand requirements is not a rebuttal.

By the below stipulative definitions, your HHH does not meet the 
requirements to be classified as either a halt decider or a termination 
analyzer.

So it doesn't matter if HHH does what you says it does.  What matters is 
that it doesn't map DD to 1 / halting as required.


Stipulative definition 1:

Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
described as <X> with input Y:

A solution to the halting problem, sometimes referred to as a "halt 
decider" is an algorithm H that computes the following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed

Theorem 1:

No algorithm H exists that satisfies the above definition

Stipulative definition 2:

Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
described as <X>:

A termination analyzer is an algorithm H that computes the following 
mapping:

(<X>) maps to 1 if and only if X(Y) halts when executed directly for all Y
(<X>) maps to 0 if and only if X(Y) does not halt when executed directly 
for some Y

Theorem 2:

No algorithm H exists that satisfies the above definition