Re: DDD specifies recursive emulation to HHH and halting to HHH1

On 3/31/2025 6:19 AM, Richard Damon wrote:

On 3/30/25 11:16 PM, olcott wrote:

On 3/30/2025 9:40 PM, Richard Damon wrote:

On 3/30/25 10:17 PM, olcott wrote:

On 3/30/2025 7:35 PM, Richard Damon wrote:

On 3/30/25 5:56 PM, olcott wrote:

On 3/30/2025 4:05 PM, Richard Damon wrote:

On 3/30/25 4:32 PM, olcott wrote:

On 3/30/2025 1:52 PM, Richard Damon wrote:

On 3/30/25 2:27 PM, olcott wrote:

On 3/30/2025 3:12 AM, joes wrote:

Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:

On 3/29/2025 3:14 PM, dbush wrote:

On 3/29/2025 4:01 PM, olcott wrote:

>

We can know that when this adapted UTM simulates a finite number of
steps of its input that this finite number of steps were simulated
correctly.

And therefore does not do a correct UTM simulation that matches the
behavior of the direct execution as it is incomplete.

It is dishonest to expect non-terminating inputs to complete.

A complete simulation of a nonterminating input doesn't halt.
>

2) changing the input is not allowed

The input is unchanged. There never was any indication that the input
was in any way changed.

False, if the starting function calls UTM and UTM changes, you're
changing the input.

When UTM1 is a UTM that has been adapted to only simulate a finite
number of steps

So not an UTM.
>

and input D calls UTM1 then the behavior of D simulated
by UTM1 never reaches its final halt state.
When D is simulated by ordinary UTM2 that D does not call Then D reaches
its final halt state.

Doesn't matter if it calls it, but if the UTM halts.
>

Changing the input is not allowed.

I never changed the input. D always calls UTM1.
thus is the same input to UTM1 as it is to UTM2.

You changed UTM1, which is part of the input D.
>

>
UTM1 simulates D that calls UTM1
simulated D NEVER reaches final halt state
>
UTM2 simulates D that calls UTM1
simulated D ALWAYS reaches final halt state
>

>
Only because UTM1 isn't actually a UTM, but a LIE since it only does a partial simulation, not a complete as REQURIED by the definition of a UTM.
>

>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
DDD EMULATED BY HHH DOES SPECIFY THAT IT
CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
>
THAT IS WHAT IT SAYS AND ANYONE THAT DISAGREES
IS A DAMNED LIAR OR STUPID.
>

>
How is that DDD correctly emulated beyond the call HHH instruction by a program that is a pure function, and thus only looks at its input?
>

>
*THE ENTIRE SCOPE IS*
DDD EMULATED BY HHH DOES SPECIFY THAT IT
CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.

>
 From where? Remember, the Halting problem is SPECIFICALLY

>
OFF F-CKING TOPIC. WE ABOUT ONE F-CKING STEP OF MY PROOF.
WE HAVE BEEN TALKING ABOUT ONE F-CKING STEP OF MY PROOF
FOR THREE F-CKING YEARS.
>
DDD correctly emulated by HHH DOES NOT F-CKING HALT !!!
>
>

>
Your proof is just off topic ranting.
>
The problem is that DDD is NOT correctly emulated by HHH,

>
You are a damned liar when you try to get away
with implying that HHH does not emulate itself
emulating DDD in recursive emulation according
to the semantics of the x86 language.
>
>

 Of course it doesn't CORRECTLY emulate itself emulating DDD (and omitting that CORRECTLY is a key point to your fraud), as it stops part way, and CORRECT emulation that determines behavior doesn't stop until the end is reached.

It is ALWAYS CORRECT for any simulating termination
analyzer to stop simulating and reject any input
that would otherwise prevent its own termination.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer