Re: Functions computed by Turing Machines MUST apply finite string transformations to inputs --- MT

On 5/4/2025 11:28 PM, olcott wrote:

On 5/4/2025 10:12 PM, dbush wrote:

On 5/4/2025 11:04 PM, olcott wrote:

On 5/4/2025 9:58 PM, dbush wrote:

On 5/4/2025 9:23 PM, olcott wrote:

On 5/4/2025 8:04 PM, Ben Bacarisse wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:
...

As explained above, UTM(⟨Ĥ⟩ ⟨Ĥ⟩) simulates Ĥ run with input Ĥ (having the
same halting behaviour) and Ĥ run with input Ĥ HALTS.  So embedded_H does
not "gather enough information to deduce that UTM(⟨Ĥ⟩ ⟨Ĥ⟩) would never
halt".  THAT IS JUST A FANTASY THAT YOU HAVE.
>
UTM(⟨Ĥ⟩ ⟨Ĥ⟩) DOES halt, so embedded_H can't possibly gather information
that genuinely implies it DOESN'T halt.  The explanation is obvious:
embedded_H gathers information that *YOU* believe implies that UTM(⟨Ĥ⟩ ⟨Ĥ⟩)
would never halt, but *YOU ARE SIMPLY WRONG*.

>
He used to claim that false ("does not halt") was the correct answer,
/even though/ the computation in question halts!  Those were simpler
days.  Of course cranks will never admit to having been wrong about
anything other than a detail or two, so anyone who could be bothered
could try to get him to retract that old claim.
>

>
<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
>
     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>
>

>
And you *CONTINUE* to lie by implying that Sipser agrees with you when it's been repeated proven that he does not:
>
On Monday, March 6, 2023 at 2:41:27 PM UTC-5, Ben Bacarisse wrote:
 > I exchanged emails with him about this. He does not agree with anything
 > substantive that PO has written. I won't quote him, as I don't have
 > permission, but he was, let's say... forthright, in his reply to me.
>
This demonstrates a reckless disregard for the truth on your part.
>

>
Let the record show that you made no attempt to refute the above,

 I HAVE ALREADY ADDRESSED THIS
It is true that he did not take the time to understand
recursive emulation thus could not possibly see the
significance of my work without this.
 THAT DOES NOT MEAN THAT HE DID NOT AGREED
TO LET ME QUOTE HIS AGREEMENT WITH MY WORDS.

But you present that quote to imply that he agreed with what you meant, and it is proven above that he did not.
That you continue to do so shows your dishonesty.

 

constituting your admission that you are *intentionally* lying to push your agenda.
>

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
In other words embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct to
reject its input if
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Would not halt.
>

>
In other words, you change the input.
>
Changing the input is not allowed.

>
D *WOULD NEVER STOP RUNNING UNLESS*

>
Which means you're changing the input.
>

 *Quoted from below*
"D would never stop running unless aborted then"

Which is code for "changing the input".
Changing the input is not allowed.

 *D AND hypothetical D explained below*
"D" // above is the D
"stop running unless unless" // above is the hypothetical D.

So you're hypothetically changing the input.
Changing the input, hypothetically or otherwise, is not allowed.