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On 6/18/2025 4:16 AM, Mikko wrote:Above is only required that HHH partially simulates the behaviourOn 2025-06-17 14:45:42 +0000, olcott said:*That is a given* for this thought experiment.
On 6/17/2025 3:42 AM, Mikko wrote:Can you prove that my HHH does simulate itself?On 2025-06-16 19:35:52 +0000, olcott said:void Infinite_Loop()
On 6/16/2025 5:36 AM, Mikko wrote:I have. I have shown that there is a simulating terminationOn 2025-06-15 13:49:51 +0000, olcott said:No one has ever even attempted to show the details
On 6/15/2025 3:24 AM, Fred. Zwarts wrote:No, they don't meet the second cireterion. HHH does not correctlyOp 14.jun.2025 om 15:38 schreef olcott:<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>On 6/14/2025 4:10 AM, Fred. Zwarts wrote:It seems I understand it better than you do. You seem to think that every recursion is a infinite recursion. As soon as you see a recursion, you think it has been proven that it is an infinite recursion, even if the code specifies an abort and halt.Op 13.jun.2025 om 17:53 schreef olcott:So you still don't understand what recursive simulation is?On 6/13/2025 5:51 AM, Mikko wrote:Indeed, HHH fails where other world-class simulators have no problem to simulate the program specified in the input.On 2025-06-12 15:30:05 +0000, olcott said:void DDD()
int DD()Strachey only informally presents the idea of the proof. Formalism
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
It is a verified fact that DD() *is* one of the forms
of the counter-example input as such an input would
be encoded in C. Christopher Strachey wrote his in CPL.
// rec routine P
// §L :if T[P] go to L
// Return §
// https://academic.oup.com/comjnl/article/7/4/313/354243
void Strachey_P()
{
L: if (HHH(Strachey_P)) goto L;
return;
}
https://academic.oup.com/comjnl/article- abstract/7/4/313/354243? redirectedFrom=fulltext
and details needed in a rigorous proof is not shown.
{
HHH(DDD);
return;
}
_DDD()
[00002192] 55 push ebp
[00002193] 8bec mov ebp,esp
[00002195] 6892210000 push 00002192
[0000219a] e833f4ffff call 000015d2 // call HHH
[0000219f] 83c404 add esp,+04
[000021a2] 5d pop ebp
[000021a3] c3 ret
Size in bytes:(0018) [000021a3]
Exactly how would DDD correctly emulated by HHH
reach its own "ret" instruction final halt state?
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
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
It is an easily verified fact that the input to HHH(DDD) and
the input to HHH(DD) meets the above self-evidently true criteria.
determine that its input would never stop running unless aborted.
Perhaps you may deceive with someting like equivocation someone to
believe it does but in reality it does not.
of how this is not correct:
void DDD()
{
HHH(DDD);
return;
}
When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
then this correctly simulated DDD never reaches its
simulated "return" statement final halt state.
analyser that has the name HHH and that simulates until it
finds either a call to HHH or termination. If it finds HHH
it continues simulation after the call. If it finds a return
from the input runction it returns 1. Your "any simulating
termination analyzer HHH" does not exclude my HHH. Therefore
your claim is false.
{
HERE: goto HERE;
return;
}
void Infinite_Recursion()
{
Infinite_Recursion();
}
void DDD()
{
HHH(DDD);
return;
}
When it is understood that HHH does simulate itself
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