On 5/8/24 10:30 AM, wij wrote:

On Wed, 2024-05-08 at 08:21 -0500, olcott wrote:

On 5/8/2024 6:39 AM, Richard Damon wrote:

On 5/7/24 11:29 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$2qov3$[email protected]>
*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


But some is not all.

Thus, you demonstrate that you do not know how logic works, but think
that proof by example is a valid proof method for universal qualifiers.

The template specifies that D(D) is calling the same H(D,D)
that invokes it. All instances conform to the template.

I have one concrete instance as fully operational code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
line 555 u32 HH(ptr P, ptr I) its input in on
line 932 int DD(int (*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 specifies 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

8th element of H/D pairs 8 steps of D are simulated by H
next step of the second recursive simulation at line 02

9th element of H/D pairs 9 steps of D are simulated by H
this ends the second recursive simulation before line 03

The fact that I have shown how to build an H that does what you say no H >>> can do shows your "proof" is wrong, and thus your basic logic is incorrect. >>>
PROVEN.

You are solving POOP. Why do you changed the Halting Problem to POOP?

I am pointing out that olcott doesn't even solve his POOP as defined.

It was proven years ago that he wasn't solving Halting, which is why he switched to the poorly defined Termination Analyzer with D "correctly" simulated by H.

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