Re: Mike Terry Proves --- How the requirements that Professor Sipser agreed to are exactly met

On 5/20/25 10:08 PM, olcott wrote:

On 5/20/2025 8:06 PM, Mike Terry wrote:

On 20/05/2025 18:46, Fred. Zwarts wrote:

Op 20.mei.2025 om 16:37 schreef olcott:

On 5/20/2025 2:06 AM, Mikko wrote:

On 2025-05-20 04:20:54 +0000, olcott said:
>

<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
>
Do you understand that we are only evaluating whether
or not HHH/DDD meets this above criteria?

>
I do understand that the meaning of the behaviour is not mentioned
in the creteria and is therefore irrelevant, an obvious consequence
of which is that your "WRONG!" above is false.
>

>
*H correctly simulates its input D until*
specifies that HHH must simulate DDD according
to the meaning of the rules of the x86 language.
>
The meaning of every step of the behavior is
precisely specified by the x86 language.
>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
*H correctly simulates its input D*
00002172 00002173 00002175 0000217a
H correctly simulates itself simulating DDD
00002172 00002173 00002175 0000217a
>
*until H correctly determines that its simulated D*
*would never stop running unless aborted*

>
That is a wild guess of HHH, not a correct determination. When it sees the call to HHH and we know that HHH halts, we know that there is only a finite recursion, so the 'would never stop running' exists only in your dreams.
The input is a finite string that includes the code of Halt7.c, which specifies that the simulation will abort. So, HHH is wrong when it assumes that an abort is needed for this input to prevent  a never stop running.
Face the facts, not your dreams. Try a real argument, instead a repetition of your dream. Try to get out of rebuttal mode.
>

>
H sees DDD call the same function with the same
parameter and there are no conditional branch
instructions from the beginning of DDD to calling
HHH(DDD) again. This repeating pattern proves
non-termination.
>
>

>
HHH does not even see a full cycle, so it cannot know that there are no conditional branches in the cycle. You can view a full cycle in different ways:
1) from the first start of DDD up to the second start of DDD. The second beginning of DDD is reached after many steps of the simulation, which contains a lot of conditional branching instruction.
2) From the first start of HHH up to the second start of HHH. In this cycle there are also many conditional branch instructions within HHH.
So, it is misleading to say that there are no conditional branch instruction in the full cycle.
That a small part of the cycle does not have conditional branch instructions does not prove anything.
Face the facts. Stop repeating your dreams. Come out of rebuttal mode and try a serious honest dialogue.
>

>
Yes, that all correct.  There are loads of conditional branch instructions performed by HHH as part of DDD.  This makes a nonsense of the implementation of PO's "infinite recursion" test.
>
But there is a worse nonsense here:  even if there were indeed no conditional branches between the matching call statements in the simulation, THAT STILL WOULD NOT BE ENOUGH TO GUARANTEE INFINITE RECURSION!
>

 You are assuming details of HHH that are not included
in its specification. A DDD that is only simulated by
HHH *is* infinite recursion.

But such an HHH isn't the needed decider, so not the DDD that we are looking at when we have a decider HHH.
Sorry,

 void DDD()
{
   HHH(DDD);
   return;
}
 Any moron can see that DDD simulated by HHH cannot possibly halt.
It does not matter how many steps of DDD are simulated by HHH.
No DDD every reaches its own "return" statement final halt state.

And any HHH that just simulates its input isn't a decider, and thus fails.
Every DDD/HHH pair is different, so we need to look at the actual DDD that a given HHH is given. not some other one.

 *The same thing goes for Linz*
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

But that isn't the H^ of the problem, unless your H *IS* just UTM.
I guess you are just showing you like to lie.
All you are doing is proving your stupidity.

 

So his whole test idea is just nonsense, PLUS it has a nonsense implementation that ignores the bulk of simulated code for no justifiable reason.  Nonsense^2.
>
>
Mike.
>

 *Mike you are not paying close enough attention*
 <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
 All HHH have the same machine code
*Unless HHH aborts its simulated DDD, then*
a) Simulated DDD   NEVER STOPS RUNNING
b) Executed  DDD() NEVER STOPS RUNNING
c) Simulated HHH   NEVER STOPS RUNNING
d) Executed  HHH() (with all of its conditionals) NEVER STOPS RUNNING
      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>
 This PROVES that HHH is correct to abort the simulation of
its DDD and reject DDD as non-halting according to the above
criteria.