Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/12/25 1:23 PM, olcott wrote:

On 5/12/2025 11:46 AM, Ben Bacarisse wrote:

wij <wyniijj5@gmail.com> writes:
>

On Sat, 2025-05-10 at 20:56 -0500, olcott wrote:

On 5/10/2025 8:44 PM, wij wrote:

...

Try to convert it to TM language to know you know nothing.

...

To refute the HP, you need to understand what it exactly means in TM.

>
I have known this for 22 years.

>
A working TM. Build it explicitly from transition function, then explain
your derivation. You know nothing.

>
He did, in 2018, claim to have exactly such a thing:
>
   "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
   specs does not exist. I now have a fully encoded pair of Turing
   Machines H / Ĥ proving them wrong."
>
   "I [...] encoded all of the exact TMD instructions of the Linz Turing
   machine H that correctly decides halting for its fully encoded input
   pair: (Ĥ, Ĥ)."
>
   "I provide the exact ⊢* wildcard states after the Linz H.q0 and after
   Ĥ.qx (Linz incorrectly uses q0 twice) showing exactly how the actual
   Linz H would correctly decide the actual Linz (Ĥ, Ĥ)."
>
Of course, no such "fully encoded Turing Machines" were ever produced.
He spent months rowing back this claim, eventually settling on the
notion that he was using "poetic licence" rather than admit he was wrong.
>

 <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>
 On 10/17/2022 10:23 AM, Ben Bacarisse wrote:
 > ...D(D) would not halt unless H stops the simulation.
 > H /can/ correctly determine this silly criterion (in this one case)...
 

Which means the correct simulation of the exact input D that was given to H, which WILL use the algorithm of that H, so if H ultimately return non-halting, as you claim is correct, then the correct emulation (which isn't done by H) will show that, and thus H is wrong.
Your problem is your "D" ends up not being the required program, and neither is your H.
Thus, your whole argument is based on category errors and lies.