Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

Op 12.mei.2025 om 18:32 schreef olcott:

On 5/12/2025 6:46 AM, dbush wrote:

On 5/12/2025 12:11 AM, olcott wrote:

On 5/11/2025 10:54 PM, dbush wrote:

On 5/11/2025 11:50 PM, olcott wrote:

On 5/11/2025 10:39 PM, dbush wrote:

On 5/11/2025 11:36 PM, olcott wrote:

>
You try to get away with changing the subject
because you know that you are lying about one
or more steps of DDD emulated by HHH according
to the rules of the x86 language

>
It is you who are changing the subject away from DDD emulated by HHH (not one or more steps of DDD emulated by HHH) which you have admitted for the record is not correct:
>

>
I am only referring to the hypothetical infinite
set of pure x86 emulators specified below:
>

>
Each of which is processing a distinct, different input.
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Changing the input is not allowed.

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I am examining all the elements of a infinite set
such that each HHH/DDD pair has a specific fixed form.

>
And each one has a distinctly different input, and as such the halt status of one is unrelated to the halt status of another.
>
Changing the input is not allowed.

 In other words you are clueless about mathematical induction.
 It is true that when one step of DDD is emulated by HHH that
DDD does not halt.
 It is true that when N steps of DDD are emulated by HHH that
DDD does not halt.
 

But when enough steps of the *unchanged* DDD are emulated, then DDD does halt, as proven by HHH1. Only when you also change the input, then more steps are needed, but changing the input is not allowed.