Re: Every D(D) simulated by H presents non-halting behavior to H @@@

On 5/8/24 12:03 AM, olcott wrote:

On 5/7/2024 10:29 PM, olcott wrote:

On 5/7/2024 9:51 PM, Richard Damon wrote:

On 5/7/24 10:39 PM, olcott wrote:

On 5/7/2024 9:29 PM, Richard Damon wrote:

On 5/7/24 7:30 PM, olcott wrote:

On 5/7/2024 5:42 PM, Richard Damon wrote:

On 5/7/24 1:31 PM, olcott wrote:
>

Once someone has definitely proven to not be telling the truth
about any specific point it is correct to assume any other
assertions about this same point are also false until evidence
arises to the contrary.

>
Then I guess we can just go and ignore everything you have said.
>
PERIOD.

>
*Below I prove that you are not telling the truth about this point*
*Below I prove that you are not telling the truth about this point*
*Below I prove that you are not telling the truth about this point*
*Below I prove that you are not telling the truth about this point*
>
Message-ID: <v0ummt$2qov3$2@i2pn2.org>
*When you interpret*
On 5/1/2024 7:28 PM, Richard Damon wrote:
 > On 5/1/24 11:51 AM, olcott wrote:
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
as *D NEVER simulated by H*
>
you have shown a reckless disregard for the truth
that would win a defamation case.

>
Nope, It is clear you don't understand the logic of qualifiers.
>

>
*Prove it on this point*
Exactly how can ALWAYS: ∀x be construed as NEVER: ∄x

>
if there are no x.
>

>
00 int H(ptr x, ptr x)  // ptr is pointer to int function
01 int D(ptr x)
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 int main()
10 {
11   H(D,D);
12 }
>
The above template defines an infinite set of finite string H/D pairs where each D(D) that is simulated by H(D,D) also calls this same H(D,D).
>
I have one concrete fully operational instance of H/D pairs so
we know that more than zero of them exist.
>
I can adapt this one concrete instance to be the 7 shown below and
we can extrapolate the trend from there:
>
1st element of H/D pairs 1 step of D is simulated by H
2nd element of H/D pairs 2 steps of D are simulated by H
3rd element of H/D pairs 3 steps of D are simulated by H
>
4th element of H/D pairs 4 steps of D are simulated by H
this begins the first recursive simulation at line 01
>
5th element of H/D pairs 5 steps of D are simulated by
next step of the first recursive simulation at line 02
>
6th element of H/D pairs 6 steps of D are simulated by
last step of the first recursive simulation at line 03
>
7th element of H/D pairs 7 steps of D are simulated by H
this begins the second recursive simulation at line 01

 The one great thing that Mike did was confirm that HH can look
into all of the details of the internal state of its simulated
DD machine and still be a computable function. I suspected
that I might be, yet Mike confirmed that I am correct about this.
 That does not work in reverse though. A computable function is
is not allowed to look at its caller.
 

Right, and thus D can change itself to call the H deciding it, and thus needs to be defined to call just one H.
Also, the question for H isn't the behavior of its caller, but the behavior of the machine described by its input.
There is nothing in that definition that EXCLUDES that machine calling H (though, due to the structure of Turing Machines, that needs to be another copy of that machine, as the Linz proof does).
Your problem is you don't seem to understand the basic English words of the problem, maybe because you logic is based on lying and thus needs to avoid being to tied to actual definitions.