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

Am Sat, 29 Mar 2025 10:28:15 -0500 schrieb olcott:

On 3/29/2025 4:56 AM, Mikko wrote:

On 2025-03-28 22:41:12 +0000, olcott said:

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:

 

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.

The x86 language or my supposed ignorance thereof doesn't prove shit.
HHH does not simulate the infinite stack of recursive simulations,
for obvious reasons.
 

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.

One cannot show something impossible.
 

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.

>
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.

Non sequitur.
 

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.

>
When DDD is correctly emulated the call HHH(DDD) returns.
>

When are you going to understand that disagreeing with the semantics of
the x86 language IS NOT ALLOWED?

Disagree with what semantics exactly? The call to HHH *must* return.