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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Defining a correct halt decider
Date: Tue, 3 Sep 2024 22:16:49 -0400
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On 9/3/24 9:17 AM, olcott wrote:
> On 9/3/2024 3:44 AM, Mikko wrote:
>> On 2024-09-02 16:06:11 +0000, olcott said:
>>
>>> A correct halt decider is a Turing machine T with one accept state 
>>> and one reject state such that:
>>>
>>> If T is executed with initial tape contents equal to an encoding of 
>>> Turing machine X and its initial tape contents Y, and execution of a 
>>> real machine X with initial tape contents Y eventually halts, the 
>>> execution of T eventually ends up in the accept state and then stops.
>>>
>>> If T is executed with initial tape contents equal to an encoding of 
>>> Turing machine X and its initial tape contents Y, and execution of a 
>>> real machine X with initial tape contents Y does not eventually halt, 
>>> the execution of T eventually ends up in the reject state and then 
>>> stops.
>>
>> Your "definition" fails to specify "encoding". There is no standard
>> encoding of Turing machines and tape contents.
>>
> 
> That is why I made the isomorphic x86utm system.
> By failing to have such a concrete system all kinds
> of false assumptions cannot be refuted.
> 
> The behavior of DDD emulated by HHH** <is> different
> than the behavior of the directly executed DDD**
> **according to the semantics of the x86 language

No it isn't, and you have accepted that conclusion by failing to 
correctly point out what was the first instruction that was correctly 
emulated by HHH that differed from the directly executed program.

Sorry, you "proof" has popped because you have been shown to have just 
lied with your claim that you can not support it.

> 
> HHH is required to report on the behavior tat its finite
> string input specifies even when this requires HHH
> to emulate itself emulating DDD.

Right, and it ignores it.

> 
> DDD never halts unless it reaches its own final
> halt state. The fact that the executed HHH halts
> has nothing to do with this.

except that it make the DDD that calls this HHH to halt.

There is just one HHH in the example, and if it aborts its simulation 
and halts, it makes the DDD that calls it, which is the DDD that it was 
given to decide on, halt, and thus HHH is incorrect to abort and say not 
halting.

> 
> HHH is not allowed to report on the computation that
> itself is contained within.

But it is allowed, and in fact is REQUIRED to report on computations 
that use copies of itself (even if they are code at the exact same 
address just executed in a different context).

Thus, since the HHH(DDD) that DDD calls is a DIFFERENT context from the 
HHH that is deciding on that DDD, HHH must, to be correct, correctly 
report on the behavior of that program, which by the mechanics of 
programs will do exactly what this instance of HHH(DDD) will do.

You just don't seem to understand how deterministic programs work.

> 
> Except for the case of pathological self-reference the
> behavior of the directly executed machine M is always
> the same as the correctly simulated finite string ⟨M⟩.

No, you have acccepted (by failing to refute) that this isn't true

> 
> That no one has noticed that they can differ does not
> create an axiom where they are not allowed to differ.
> 

Except it has actually be PROVEN, and isn't taken as an axiom.

As I have asked, what is the first instruction where they differ that 
was actually correctly emulated?

You failure to answer is just an admission that you are just a liar.

Your claim that the call instruction differs just proves you just don't 
know what you are talking about, because you explaination of what HHH 
sees in the call instruction is jst a LIE.

> No one noticed that they differ only because everyone
> rejected the idea of a simulating halt decider out-of-hand
> without review.
>

No, they no it has been proven that the are the same.

Sorry, you are just proving your ignorance, and deceptiveness, making 
you are pathological lying idiot.