Re: D(D) simulated by H cannot possibly reach past its own line 03

On 6/4/2024 3:19 AM, joes wrote:

Am Mon, 03 Jun 2024 13:09:30 -0500 schrieb olcott:

On 6/3/2024 9:17 AM, Mikko wrote:

On 2024-06-03 12:25:48 +0000, olcott said:

On 6/3/2024 2:39 AM, Mikko wrote:

On 2024-06-02 14:50:26 +0000, olcott said:

On 6/2/2024 4:50 AM, Fred. Zwarts wrote:

Op 03.mei.2024 om 15:40 schreef olcott:

 

Already addressed in another reply.

>
Which reply? The one where you said you made a mistake? Or typo?
>

I am not going to answer the same question from multiple people.

>
Your choice. But you can't keep multiple people from seeing your lack
of answer.

This is my canned reply that no one has attempted to refute because they
know it is irrefutable. When we are analyzing x86 code and someone
disagrees that is like disagreeing that 2 + 3 = 5.

If you regard it as irrefutable, you may ignore all attempts to do so.
No need to spam it.
 

That is why it is essential that I get agreement on the x86/C code
before moving on to Turing Machines. If they are going to lie about
arithmetic we cannot trust them with more complex math.
>
DD correctly emulated by any HH that can possibly exist DOES NOT HALT

Then H also doesn't halt, making it not a decider.
 

*Not at all as I prove right here*
Published hundreds of times here any other places,
now publishing it again right here:
On 6/4/2024 11:28 AM, olcott wrote:
[Proof that executed HH(DD,DD) and simulated HH(DD,DD)
  simulate DD correctly -- Mike Terry]
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer