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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Who here understands that the last paragraph is Necessarily true?
Date: Mon, 15 Jul 2024 22:19:57 -0400
Organization: i2pn2 (i2pn.org)
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On 7/15/24 8:34 AM, olcott wrote:
> On 7/15/2024 4:03 AM, joes wrote:
>> Am Sun, 14 Jul 2024 22:41:24 -0500 schrieb olcott:
>>> On 7/14/2024 9:04 PM, Richard Damon wrote:
>>>> On 7/14/24 9:27 PM, olcott wrote:
>>>>>
>>>>> Any input that must be aborted to prevent the non termination of
>>>>> simulating termination analyzer HHH necessarily specifies non-halting
>>>>> behavior or it would never need to be aborted.
>>>>
>>>> Excpet, as I have shown, it doesn't.
>>>> Your problem is you keep on ILEGALLY changing the input in your
>>>> argument because you have misdefined what the input is.
>> Specifically, the input HHH aborts.
>>
>>>> The input to HHH is ALL of the memory that it would be accessed in a
>>>> correct simulation of DDD, which includes all the codd of HHH, and
>>>> thus, if you change HHH you get a different input.
>>>>
>>>> If you want to try to claim the input is just the bytes of the function
>>>> DDD proper then you are just admitting that you are nothing more than a
>>>> lying idiot that doesn't understand the problem,
>>> Turing machines only operate on finite strings they do not operate on
>>> other Turing machines *dumbo*
>> Don't deflect. HHH as part of DDD (because it is called) needs to be
>> included in the input to the simulator.
>> Bedsides, TMs can be encoded as strings. Notwithstanding that HHH is not
>> and does not simulate a TM.
>>
> Richard insists that HHH report on the behavior of the TM that
> is *not* encoded as finite string. TM's are not allowed to report
> on the behavior of the computation that they are contained within.

Why ISN'T it encoded in the finite string?

Isn't that what HHH(DDD) MEANS? that the input is supposed to encode the 
program DDD?

So, since to be the TM equivalent, the input to HHH WOULD be the 
encoding of DDD, then HHH can surely be asked to report on its behavior.

You are just showing you are nothing but a lying blowhard.

> 
> The question is not whether or not HHH halts.

No, but DDD Halts if and only if HHH(DDD) halts

> 
> The question is does the finite string input to HHH mathematically
> map to behavior that halts when DDD is correctly emulated by HHH
> according to the semantics of the x86 language?
> 

No, as DDD needs to be a PROGRAM, which means it maps to behavior 
reguardless of what HHH's simulation says.

Now, DDD is dependent on the code of HHH, so we don't have a DDD until 
you choose have HHH is, and it also means that the input that shows DDD 
must include all of the code that DDD uses, including that of HHH , or 
you are just showing that you are a LIAR.

Your problem is you chose a game where you must go first, and second 
player wins, and you are trying to change things so you go second, but 
that breaks the rules.

You are just stuck with your LIES.