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

On 3/28/2025 4:58 PM, Richard Damon wrote:

On 3/28/25 2:13 PM, olcott wrote:

On 3/28/2025 8:50 AM, Richard Damon wrote:

On 3/27/25 10:11 PM, olcott wrote:

On 3/27/2025 9:02 PM, Richard Damon wrote:

On 3/27/25 9:10 PM, olcott wrote:

On 3/27/2025 7:47 PM, Richard Damon wrote:

On 3/27/25 8:11 PM, olcott wrote:

On 3/27/2025 4:56 PM, joes wrote:

Am Thu, 27 Mar 2025 13:10:46 -0500 schrieb olcott:

On 3/27/2025 6:02 AM, Richard Damon wrote:

On 3/26/25 11:47 PM, olcott wrote:

On 3/26/2025 10:28 PM, Richard Damon wrote:

On 3/26/25 11:09 PM, olcott wrote:

On 3/26/2025 8:22 PM, Richard Damon wrote:

>

Non-Halting is that the machine won't reach its final staste even
if an unbounded number of steps are emulated. Since HHH doesn't do
that, it isn't showing non-halting.

DDD emulated by any HHH will never reach its final state in an
unbounded number of steps.

But DDD emulated by an actually correct emulator will,

If you were not intentionally persisting in a lie you would
acknowledge the dead obvious that DDD emulated by HHH according to the
semantics of the x86 language cannot possibly correctly reach its
final halt state.

>
_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]
>

Yes, HHH is not a correct simulator.
>

>
You say that it is not a correct simulator on the basis
of your ignorance of the x86 language that conclusively
proves that HHH does correctly simulate the first four
instructions of DDD and correctly simulates itself
simulating the first four instructions of DDD.
>

>
It isn't a correct simulator,

>
You know that you are lying about this or you would
show how DDD emulated by HHH would reach its final state
ACCORDING TO THE SEMANTICS OF THE X86 LANGUAGE.
>
>

>
It can't be, because your HHH doesn't meet your requirement.
>

>
You cannot show that because you know you are lying about that.
>

>
Sure we can, make a main that directly calls HHH and then DDD, then call HHH1(DDD)
>
That HHH will return 0, saying that DDD is non-halting, but the DDD wll return, showing that DDD is halting.
>
Look at the trace that HHH generates, and that HHH1 generates, HHH's will be a subset of the trace that HHH1 generates, showing that it is NOT proof that this program is non-halting as that exact same initial segment halts.
>
Your argument about changing HHH shows that it doesn't halt is just invalid, as then you either changed the input, or demonstrated that you input was a class error as it didn't contain the COMPLETE representation of the code of DDD.
>
Sorry, This is what you have been told for years, but you refuse to look at the truth, because you have been brainwashed by your lies.
>
Look

>
I can't understand how that confused mess addresses
the point of this thread:
>
It is a verified fact that the finite string of machine
code of DDD emulated by HHH according to the semantics of
the x86 language has different behavior than DDD emulated
by HHH1 according to the semantics of the x86 language.
>

 Where did you "verify" that LIE.
 You claim fails the simple test:
 What is the first instruction actually correctly emulated by the rules of the x86 language by HHH and HHH1 that had a different result.
 

When DDD emulated by HHH calls HHH(DDD) this call NEVER returns.
When DDD emulated by HHH1 calls HHH(DDD) this call returns.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer