Re: Bad faith and dishonesty

On 5/30/2025 10:29 AM, Mike Terry wrote:

On 30/05/2025 10:19, vallor wrote:

On Thu, 29 May 2025 19:40:57 +0100, Richard Heathfield <rjh@cpax.org.uk>
wrote in <101a9np$gl7$1@dont-email.me>:
>

On 29/05/2025 19:14, olcott wrote:

On 5/29/2025 12:40 PM, Richard Heathfield wrote:

On 29/05/2025 16:49, olcott wrote:

On 5/28/2025 4:16 PM, Richard Heathfield wrote:

On 28/05/2025 22:05, dbush wrote:

On 5/28/2025 2:38 PM, olcott wrote:

My only aim is to show that the conventional halting problem proof
is wrong.

>
But why would you care whether or not the proof is wrong when
you've gone on record (multiple times) as stating that what the
proof proves is correct?

>
It would certainly earn him a place in history's footnotes, which
might well be considered sufficient motive. But he'd have to be able
to explain why he's right, which of course he can't.
>
<snip>
>
>

See my post: [Disagreeing with tautologies is always incorrect]

>
And it seems you still can't.
>
I have already read your article "Disagreeing with tautologies is
always incorrect"[1], which completely fails to explain your proof.

>
Maybe you have no idea what a tautology is.

>
Maybe you think that asserting something is true is sufficient to make
it true. It isn't.
>
>

Its the same thing as a self-evident truth.

>
Maybe you think that asserting something is self-evidently true is
sufficient to make it self-evidently true. It isn't.
>

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
      If simulating halt decider H correctly simulates its input D
      until H correctly determines that its simulated D would never
      stop running unless aborted then
>
It is a tautology that any input D to termination analyzer H that
*would never stop running unless aborted*
DOES SPECIFY NON-TERMINATING BEHAVIOR.

>
But in making that claim you assume that you correctly know the
termination behaviour of D.
>
I can easily sketch out a program that your HHH analyser would
impatiently abort as non-terminating, but which could conceivably stop
running this year, next year, sometime... or never.

>
Was wondering when someone would mention that...what does his HHH()
do with arbitrary programs?
>
$ cat ddd.c
#include <stdio.h>
>
void ddd(int r)
     {
     r--;
     if(r <= 0) return;
     fprintf(stderr,"calling ddd(%d)\n",r);
     ddd(r);
     fprintf(stderr,"returning, r=%d\n",r);
     return;
     }
>
>
int main(void)
{
>
ddd(50);
>
return 0;
}
>
I'd bet his HHH() would say this is non-terminating.
>

 It does what it does with DDD:  it simulates the program while monitoring the simulation progress, looking for what it considers to be signs of non-halting behaviour.  If it spots what it thinks is non- halting behaviour, it decides non-halting.  If the simulation halts, it decides halts. Otherwise it will continue simulating indefinitely.
 @Richard:  so you cannot make HHH decide non-halting simply by looping for a long long time, hoping HHH will get fed up!  That would just result in HHH simulating for a corresponding long long time. You need to feed it a program that halts, but matches one of his non-halting behaviour tests.  For example DDD.
 @vallor:  so with your example, ddd halts, and I don't believe any of HHH's tests would match, so HHH would simulate your ddd to completion and decide it halts.
 Mike.
 

When DDD is correctly emulated by HHH the first four
instructions of DDD are emulated. Then HHH(DDD) is
called from DDD then HHH emulates itself emulating DDD.
No matter how many times HHH emulates itself emulating
DDD the emulated DDD cannot possibly reach its "ret"
instruction final halt state. This proves that DDD emulated
by HHH is non-halting.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer