Re: Peter Olcott here seems to consistently lie about this

On 8/1/24 10:09 PM, olcott wrote:

*This algorithm is used by all the simulating termination analyzers*
<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>

Which means that the decider H must correctly determine that if this exact input is given to a non-aborting simulator it will never halt.

 DDD is correctly emulated by HHH according to the x86
language semantics of DDD and HHH including when DDD
emulates itself emulating DDD

Nope, since it doesn't actually emulated the call instruction and follow the instructions in the HHH that the DDD calls.
And it thinks that its replacement of the call HHH with an unconditional

 *UNTIL*
 HHH correctly determines that never aborting this
emulation would cause DDD and HHH to endlessly repeat.

Nope, it determines that a DIFFERENT input based on a DIFFERENT HHH would never halt. You "logic" is based on programs not being fixed code, and thus you logic is based on lies.

 When I say everyone I mean:
Joes, Fred, Richard, Mike, Mikko, Andy, André...
 *Excluding only Ben Bacarisse*
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game.  PO really /has/
 > an H (it's trivial to do for this one case) that correctly
 > determines that P(P) *would* never stop running *unless*
 > aborted.
...
 > But H determines (correctly) that D would not halt if
 > it were not halted.  That much is a truism.
>

Nope. THIS D that calls TH(S H will halt, and its correct emulation reaches an end, but that isn't HHH's emulation, which is aborted and thus NOT CORRECT.
You logic is just based on rules your computation system doesn't support.
Since you intertwined the input and the decider, the decider if fixed and can't be vaired, and thus you can't argue about "replacing" it in place with something different, only giving this input to another decider at a diffetent address.
You claims that this isn't allowed just says you admit your system isn't actually Turing Complete and thus your proof is meaningless, as many non-turing complete systems can do halt deciding.