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

On 3/29/25 11:25 AM, olcott wrote:

On 3/29/2025 4:27 AM, joes wrote:

Am Fri, 28 Mar 2025 14:38:35 -0500 schrieb olcott:

On 3/28/2025 2:20 PM, dbush wrote:

On 3/28/2025 3:15 PM, olcott wrote:

On 3/28/2025 4:33 AM, Fred. Zwarts wrote:

Op 28.mrt.2025 om 02:21 schreef olcott:

On 3/27/2025 8:09 PM, dbush wrote:

On 3/27/2025 9:07 PM, olcott wrote:

On 3/27/2025 7:38 PM, dbush wrote:

On 3/27/2025 8:34 PM, olcott wrote:

>

I did not say that no TM can ever report on behavior that matches
the behavior of a directly executing TM.

 

Why can't HHH do it? Explain what pathology is and what it does.
>

No TM can every directly see the behavior of the direct execution of
any other TM because no TM can take a directly executing TM as an
input.

 

Ridiculous strawman, nobody said that. Are you saying that nothing at
all can be computed about TMs?
>

 If HHH must report on the direct execution of DDD
then it must see the behavior of the direct
execution of DDD and this is always impossible
for every pair of TMs.
 

So we agree that the answer for:
'Is there an algorithm that can determine for all possible inputs
whether the input specifies a program that (according to the
semantics of the machine language) halts when directly executed?'
is 'no'. Correct?

>
In the same way: Is there an algorithm that correctly determines the
square root of a box of rocks?

 

Can you just say yes or no for once?
>

The inability to determine whether or not
this sentence: "What time is it?" is true
or false is not any instance of undecidability.
 The inability of any TM to report on the behavior
of the direct execution of any other TM is also
not any instance of undecidability.
 

In other words, you're saying that there's a TM/input where the
question of whether or not it halts when executed directly has no
correct yes or no answer.
Show it.
>

I proved it many times and because you are a Troll you ignored the proof
that by definition no TM can take an executing TM as its input, thus
cannot possibly report on something that it does not see.

 

Where is the proof that some TM has no definite halting status?
>

 No TM can ever report on the behavior of any directly
executed TM because no TM has any access to this behavior.

Sure it does. That is what a UTM is.

 

Failure to do so in your next message is your on-the-record admission
that the above question is valid.

>
When include ALL of the relevant details to the question it becomes:
What Boolean value can decider H correctly return when input D is able
to do the opposite of whatever value that H returns?

 

And the answer is none, ergo the assumption that an H exists is wrong.
>

 Likewise by the same reasoning we can prove that some
questions have no correct answer by allowing incorrect
questions.

yes, YOUR question is the incorrect one.
The Halting Problem asks a correct one: "Does the TM described by the input Halt (when run)?"
That has an answer, just not the one that your decider gives.

 

We can reject this question entirely when we discard its false
assumption. D is unable to do the opposite of whatever value that H
returns when H is a simulating halt decider.

 

Oh. That's a rather unorthodox resolution. How do you show that D
is impossible.
>

 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 The contradictory part is unreachable code to
DD correctly emulated by HHH.
 

But HHH doesn't correectly emulate DD, so you statement is just a LIE.
Note, to actually do this, you need to include in the "input" DD, the full code of the sub-program HHH, or you are just making a category error in your question.