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

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$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
  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 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.