Re: DDD INcorrectly emulated by HHH is INCorrectly rejected as non-halting V2

On 7/21/24 9:34 AM, olcott wrote:

On 7/21/2024 4:34 AM, Mikko wrote:

On 2024-07-20 13:11:03 +0000, olcott said:
>

On 7/20/2024 3:21 AM, Mikko wrote:

On 2024-07-19 14:08:24 +0000, olcott said:
>

When we use your incorrect reasoning we would conclude
that Infinite_Loop() is not an infinite loop because it
only repeats until aborted and is aborted.

>
You and your HHH can reason or at least conclude correctly about
Infinite_Loop but not about DDD. Possibly because it prefers to
say "no", which is correct about Infinte_loop but not about DDD.
>

>
*Because this is true I don't understand how you are not simply lying*
int main
{
   DDD();
}
>
Calls HHH(DDD) that must abort the emulation of its input
or {HHH, emulated DDD and executed DDD} never stop running.

>
You are the lying one.
>
If HHH(DDD) abrots its simulation and returns true it is correct as a
halt decider for DDD really halts.
>

 (b) We know that a decider is not allowed to report on the behavior
computation that itself is contained within. Deciders only take finite
string inputs. They do not take executing processes as inputs. Thus HHH
is not allowed to report on the behavior of this int main() { DDD(); }.

 From what?
I think this is another of your "Diagonalization" claims.
Yes, we can not ask in the form "The machine that calls you", but we can ask about that machine by giving its representation.

 Even the Linz proof makes this same mistake

Nope, not a mistake, just your failure to understand because you made yourself into an ignorant idiot.

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 Although embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly correctly report on
its own behavior because its input does the opposite of whatever
it reports embedded_H is only allowed to report on the behavior
that its input specifies.

Which is EXACTLY the behavior of H^ (H^) since that is the computation represented by it.

 Turing machines never take actual Turing machines as inputs.
They only take finite strings as inputs and an actual executing
Turing machine is not itself a finite string.

Right, they take representations of machines,

 <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

And you don't understand the meaning of "correct simulation" making you comments just ignorant pathological lies based on a reckless disregard of the truth.

      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>
 Since we ourselves can directly see that UTM based embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
must abort the simulation of its input otherwise this input would
never stop running we know that the criteria have been met.
 

A "UTM Based embedded_H" *CAN'T* abort as that is a violation of the property of it being a UTM. Your definition is based on a contradiction and thus is not correct.
Sorry, you are just proving yourself to be that ignorant pathological lying idiot that recklessly disregards the truth because he has gaslite and brainwashed himself into beleiving his own lies.
You are just killing and burying your ideas by piling on your lies.