On Tuesday, June 28, 2022 at 10:47:01 AM UTC-4, olcott wrote:
> On 6/28/2022 9:22 AM, Dennis Bush wrote:
> > On Tuesday, June 28, 2022 at 10:14:33 AM UTC-4, olcott wrote:
> >> On 6/27/2022 5:03 PM, Dennis Bush wrote:
> >>> On Monday, June 27, 2022 at 5:58:55 PM UTC-4, olcott wrote:
> >>>> On 6/27/2022 4:46 PM, Dennis Bush wrote:
> >>>>> On Monday, June 27, 2022 at 5:38:02 PM UTC-4, olcott wrote:
> >>>>>> On 6/27/2022 4:14 PM, Dennis Bush wrote:
> >>>>>>> On Monday, June 27, 2022 at 5:11:30 PM UTC-4, olcott wrote:
> >>>>>>>> On 6/27/2022 4:02 PM, Dennis Bush wrote:
> >>>>>>>>> On Monday, June 27, 2022 at 4:52:17 PM UTC-4, olcott wrote:
> >>>>>>>>>> On 6/27/2022 2:45 PM, Dennis Bush wrote:
> >>>>>>>>>>> On Monday, June 27, 2022 at 2:13:44 PM UTC-4, Dennis Bush wrote:
> >>>>>>>>>>>> On Monday, June 27, 2022 at 2:11:45 PM UTC-4, olcott wrote:
> >>>>>>>>>>>>> On 6/27/2022 1:03 PM, Dennis Bush wrote:
> >>>>>>>>>>>>>> On Monday, June 27, 2022 at 1:20:39 PM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>> On 6/27/2022 12:03 PM, Dennis Bush wrote:
> >>>>>>>>>>>>>>>> On Monday, June 27, 2022 at 9:28:22 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>>> On 6/27/2022 7:58 AM, Dennis Bush wrote:
> >>>>>>>>>>>>>>>>>> On Monday, June 27, 2022 at 8:50:05 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>>>>> On 6/27/2022 7:40 AM, Dennis Bush wrote:
> >>>>>>>>>>>>>>>>>>>> On Monday, June 27, 2022 at 8:34:22 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>>>>>>> On 6/27/2022 7:26 AM, Dennis Bush wrote:
> >>>>>>>>>>>>>>>>>>>>>> On Monday, June 27, 2022 at 8:14:31 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>>>>>>>>> On 6/27/2022 6:57 AM, Dennis Bush wrote:
> >>>>>>>>>>>>>>>>>>>>>>>> On Monday, June 27, 2022 at 7:11:21 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>>>>>>>>>>> *This is the outline of the complete refutation*
> >>>>>>>>>>>>>>>>>>>>>>>>> *of the only rebuttal that anyone has left*
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> (a) The complete and correct x86 emulation of the input to H(P,P) by H
> >>>>>>>>>>>>>>>>>>>>>>>>> never reaches the "ret" instruction of P.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> It is known that H aborts its simulation and returns 0. That means that H *by definition* does not perform a complete and correct emulation. And since H (if it was constructed correctly) is a pure function of its inputs, it will ALWAYS return the same return for a given input.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> To state that H aborts *and* does a complete and correct simulation is simply nonsense.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> What *does* perform a complete and correct emulation of the input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> (b) The direct execution of P(P) does reach its "ret" instruction.
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> The above two are proven to be verified facts entirely on the basis of
> >>>>>>>>>>>>>>>>>>>>>>>>> the semantics of the x86 language.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> A nonsense statement can't be a "fact", verified or otherwise.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>>>>>>>>>>>>>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>>>>>>>>>>>>>>>>>>> specified by these inputs.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> Which for H(P,P) is BY DEFINITION is the behavior of P(P). The halting problem says that H MUST implement the following mapping (i.e. the halting function):
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> H(x,y)==1 if and only if x(y) halts, and
> >>>>>>>>>>>>>>>>>>>>>>>> H(x,y)==0 if and only if x(y) does not halt.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> H does not perform this mapping.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> P(P) is provably not the actual behavior of the actual input.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> It is BY DEFINITION. So if H comes up with something different that means it did something wrong, specifically it aborted a halting computation too soon.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> void P(u32 x)
> >>>>>>>>>>>>>>>>>>>>>>>>> {
> >>>>>>>>>>>>>>>>>>>>>>>>> if (H(x, x))
> >>>>>>>>>>>>>>>>>>>>>>>>> HERE: goto HERE;
> >>>>>>>>>>>>>>>>>>>>>>>>> return;
> >>>>>>>>>>>>>>>>>>>>>>>>> }
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> int main()
> >>>>>>>>>>>>>>>>>>>>>>>>> {
> >>>>>>>>>>>>>>>>>>>>>>>>> P(P);
> >>>>>>>>>>>>>>>>>>>>>>>>> }
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> _P()
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f0](01) 55 push ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f1](02) 8bec mov ebp,esp
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f3](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f6](01) 50 push eax
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f7](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fa](01) 51 push ecx
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fb](05) e820feffff call 00001020
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001200](03) 83c408 add esp,+08
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001203](02) 85c0 test eax,eax
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001205](02) 7402 jz 00001209
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001207](02) ebfe jmp 00001207
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001209](01) 5d pop ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [0000120a](01) c3 ret
> >>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000120a]
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> _main()
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001210](01) 55 push ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001211](02) 8bec mov ebp,esp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001213](05) 68f0110000 push 000011f0
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001218](05) e8d3ffffff call 000011f0
> >>>>>>>>>>>>>>>>>>>>>>>>> [0000121d](03) 83c404 add esp,+04
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001220](02) 33c0 xor eax,eax
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001222](01) 5d pop ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001223](01) c3 ret
> >>>>>>>>>>>>>>>>>>>>>>>>> Size in bytes:(0020) [00001223]
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> machine stack stack machine assembly
> >>>>>>>>>>>>>>>>>>>>>>>>> address address data code language
> >>>>>>>>>>>>>>>>>>>>>>>>> ======== ======== ======== ========= =============
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001210][00101fba][00000000] 55 push ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001211][00101fba][00000000] 8bec mov ebp,esp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001213][00101fb6][000011f0] 68f0110000 push 000011f0 // push P
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001218][00101fb2][0000121d] e8d3ffffff call 000011f0 // call P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f0][00101fae][00101fba] 55 push ebp // enter executed P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f1][00101fae][00101fba] 8bec mov ebp,esp
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f3][00101fae][00101fba] 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f6][00101faa][000011f0] 50 push eax // push P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f7][00101faa][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fa][00101fa6][000011f0] 51 push ecx // push P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fb][00101fa2][00001200] e820feffff call 00001020 // call H
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> Begin Simulation Execution Trace Stored at:21206e
> >>>>>>>>>>>>>>>>>>>>>>>>> Address_of_H:1020
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f0][0021205a][0021205e] 55 push ebp // enter emulated P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f1][0021205a][0021205e] 8bec mov ebp,esp
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f3][0021205a][0021205e] 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f6][00212056][000011f0] 50 push eax // push P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011f7][00212056][000011f0] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fa][00212052][000011f0] 51 push ecx // push P
> >>>>>>>>>>>>>>>>>>>>>>>>> [000011fb][0021204e][00001200] e820feffff call 00001020 // call emulated H
> >>>>>>>>>>>>>>>>>>>>>>>>> Infinitely Recursive Simulation Detected Simulation Stopped
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> H knows its own machine address and on this basis it can easily
> >>>>>>>>>>>>>>>>>>>>>>>>> examine its stored execution_trace of P (see above) to determine:
> >>>>>>>>>>>>>>>>>>>>>>>>> (a) P is calling H with the same arguments that H was called with.
> >>>>>>>>>>>>>>>>>>>>>>>>> (b) No instructions in P could possibly escape this otherwise infinitely
> >>>>>>>>>>>>>>>>>>>>>>>>> recursive emulation.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> FALSE. P contains these instructions:
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> if (Decide_Halting(&execution_trace, &decoded, code_end, &master_state,
> >>>>>>>>>>>>>>>>>>>>>>>> &slave_state, &slave_stack, Address_of_H, P, I))
> >>>>>>>>>>>>>>>>>>>>>>>> goto END_OF_CODE;
> >>>>>>>>>>>>>>>>>>>>>>>> return 0; // Does not halt
> >>>>>>>>>>>>>>>>>>>>>>>> END_OF_CODE:
> >>>>>>>>>>>>>>>>>>>>>>>> return 1; // Input has normally terminated
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> That *can* and *do* prevent infinite recursive emulation in a *correct* and *complete* emulation
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> (c) H aborts its emulation of P before its call to H is emulated.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> And in doing so it fails to see that the call to H, since H is a computation, will aborts *its* simulation of P and return 0 to the outer P being emulated, which would cause it to halt, making H(P,P)==0 wrong.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001200][00101fae][00101fba] 83c408 add esp,+08 // return to
> >>>>>>>>>>>>>>>>>>>>>>>>> executed P
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001203][00101fae][00101fba] 85c0 test eax,eax
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001205][00101fae][00101fba] 7402 jz 00001209
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001209][00101fb2][0000121d] 5d pop ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [0000120a][00101fb6][000011f0] c3 ret // return from
> >>>>>>>>>>>>>>>>>>>>>>>>> executed P
> >>>>>>>>>>>>>>>>>>>>>>>>> [0000121d][00101fba][00000000] 83c404 add esp,+04
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001220][00101fba][00000000] 33c0 xor eax,eax
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001222][00101fbe][00100000] 5d pop ebp
> >>>>>>>>>>>>>>>>>>>>>>>>> [00001223][00101fc2][00000000] c3 ret // ret from main
> >>>>>>>>>>>>>>>>>>>>>>>>> Number of Instructions Executed(878) / 67 = 13 pages
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> I've stated in detail why you're wrong. Now you need to explain in detail why I'm wrong.
> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>> Failure to do so, which includes simply restating your original point, will be taken as admission that what I've stated is correct.
> >>>>>>>>>>>>>>>>>>>>>>> H correctly predicts that its complete and correct x86 emulation of its
> >>>>>>>>>>>>>>>>>>>>>>> input never reaches the the "ret" instruction of P.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>> Repeating your original point.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>> This is the actual behavior of the input to H(P,P).
> >>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>> P(P) is not the actual behavior of the input to H(P,P).
> >>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>> Again, repeating your original point.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>> Never reaching the "ret" instruction means that P does not halt.
> >>>>>>>>>>>>>>>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>>>>>>>>>>>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>>>>>>>>>>>>>>>>> specified by these inputs.
> >>>>>>>>>>>>>>>>>>>>>>> Therefore H(P,P)==0 is correct.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>> And again, repeating your original point.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>> So no explanation as to why I'm wrong, just a repetition of your original points.
> >>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>> That means you accept that what I've said is true, and that H(P,P)==0 is wrong.
> >>>>>>>>>>>>>>>>>>>>> You rebuttal is based on rejecting the verified fact that a simulating
> >>>>>>>>>>>>>>>>>>>>> halt decider correctly predicts that its correctly simulated input will
> >>>>>>>>>>>>>>>>>>>>> never reach the final state "ret" instruction of this input.
> >>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>> FALSE. The correctly simulated input to H(P,P) is UTM(P,P) which halts, therefore H(P,P)==0 is wrong.
> >>>>>>>>>>>>>>>>>>> First of all I am not talking about UTM's.
> >>>>>>>>>>>>>>>>>>> I am talking about x86 emulation.
> >>>>>>>>>>>>>>>>>>> Try to encode what you mean, here is my guess:
> >>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>> void P(u32 x)
> >>>>>>>>>>>>>>>>>>> {
> >>>>>>>>>>>>>>>>>>> if (emulate_x86(x, x))
> >>>>>>>>>>>>>>>>>>> HERE: goto HERE;
> >>>>>>>>>>>>>>>>>>> return;
> >>>>>>>>>>>>>>>>>>> }
> >>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>> int main()
> >>>>>>>>>>>>>>>>>>> {
> >>>>>>>>>>>>>>>>>>> Output("Input_Halts = ", H((u32)P, (u32)P));
> >>>>>>>>>>>>>>>>>>> }
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>> P is a computation and therefore it can't be changed, nor can anything that it calls. I mean UTM(P,P). The simplest way to implement this is:
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>> void UTM(u32 x, u32 y)
> >>>>>>>>>>>>>>>>>> {
> >>>>>>>>>>>>>>>>>> x(y)
> >>>>>>>>>>>>>>>>>> }
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>> H cannot by definition perform a correct and complete emulation if it aborts, so the input must be passed to something that does, i.e. a UTM, for the source of truth.
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>> A simulating halt decider correctly simulates its input until it
> >>>>>>>>>>>>>>>>> correctly determines that this simulated input would never reach its
> >>>>>>>>>>>>>>>>> final state.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> But H isn't such a decider, because the simulated input does reach a final state when correctly simulated, i.e. by UTM(P,P), therefore H(P,P)==0 is wrong.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> Every simulating decider that correctly simulates its input until it
> >>>>>>>>>>>>>>> correctly determines that this simulated input would never reach its
> >>>>>>>>>>>>>>> final state correctly determines the halt status of this input.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> And H isn't such a decider, because it does not correctly determine that the input to H(P,P) would never halt as demonstrated by UTM(P,P) halting.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> The specification for H:
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> H(x,y)==0 if and only if x(y) does not halt, and
> >>>>>>>>>>>>>> H(x,y)==1 if and only if x(y) halts.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> H does not meet the specification.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>> The spec is provably incorrect in this case:
> >>>>>>>>>>>> A spec can't be incorrect. It just is. If it can't be satisfied, then that validates the halting problem proofs that say it can't be satisfied.
> >>>>>>>>>>>
> >>>>>>>>>>>
> >>>>>>>>>>> No response to this I see. So you agree that your H does not meet the above spec that the halting problem requires?
> >>>>>>>>>> You are just playing word games so I will use a different word:
> >>>>>>>>>> "criterion measure".
> >>>>>>>>>> A halt decider must compute the mapping from its inputs to an accept or
> >>>>>>>>>> reject state on the basis of the actual behavior that is actually
> >>>>>>>>>> specified by these inputs.
> >>>>>>>>>
> >>>>>>>>> As per this specification:
> >>>>>>>>>
> >>>>>>>>> H(x,y)==0 if and only if x(y) does not halt, and
> >>>>>>>>> H(x,y)==1 if and only if x(y) halts
> >>>>>>>>>
> >>>>>>>>> Which means the actual behavior of the actual input to H(P,P) is defined to be the behavior of P(P).
> >>>>>>>>>
> >>>>>>>>> Remember, the halting problem is ultimately about algorithms, not some simulator's simulation of it.
> >>>>>>>>>
> >>>>>>>>> This also means that if the result of H(P,P) doesn't match the behavior of P(P) then H is wrong by definition.
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>>> P(P) is provably not the actual behavior of the actual input,
> >>>>>>>>>
> >>>>>>>>> If you mean that H(P,P) can't simulate its input to a final state, then any simulator that aborts its input for any reason is necessarily correct, which is clearly nonsense.
> >>>>>>>> The formal mathematical semantics of the x86 language conclusively
> >>>>>>>> proves that P(P) is not the actual behavior of the actual input to H(P,P).
> >>>>>>>
> >>>>>>> UTM(P,P) halting proves otherwise, demonstrating that H aborted its simulation too soon.
> >>>>>> BUSTED!
> >>>>>> I GOT YOU NOW!
> >>>>>>
> >>>>>> How many recursive emulations does H need to wait before its emulated P
> >>>>>> to reaches its final instruction?
> >>>>>
> >>>>> Since H has a fixed algorithm (which we'll refer to as Ha, and the P that calls it we'll refer to as Pa to make that point clear) to to stop after N recursive emulations, it would need to simulate for N+1 emulations. But its fixed algorithm prevents it from simulating any more than N emulations.
> >>>>
> >>>> "H aborted its simulation too soon."
> >>>> Liar liar pants on fire.
> >>>
> >>> PO-speak for "I know you proved me wrong and I don't know what else to say"
>
> >> It is PO-speak for I just proved that you are lying:
> >> On 6/27/2022 4:14 PM, Dennis Bush wrote:
> >>> demonstrating that H aborted its simulation too soon.
>
> >> How many recursive emulations must H(P,P) emulate its input until its
> >> emulated input reaches the "ret" instruction of the emulated P?
> >
> > The fixed algorithm of H, which we'll refer to as Ha and the P that calls it we'll refer to as Pa, simulates for N recursive emulations. Therefore the input must be simulated for N+1 recursive emulations. The fixed algorthm Ha is unable to simulate for this many cycles and therefore gets the wrong answer.
> >
> *Because I boxed you into a corner* you are finally implicitly admitting
> that there is no integer value N number of cycles of recursive emulation
> that H(P,P) can emulate its input such that the emulated P reaches its
> final state "ret" instruction.

I already told you: N+1 cycles for Pa which is built on Ha.

Remember, the halting problem is about *algorithms*. Simply stated, the halt problem asks the question "does an algorithm exist that can determine if any arbitrary algorithm will halt given a particular input?"

What is being decided on in this case is the *algorithm* Pa. That algorithm includes the function P and the function H, which means neither can be changed without changing what is being decided on. And because Ha aborts, this means that the input must be passed to a *separate* function as a measure of the actual behavior of the input. So Ha(Pa,Pa) cannot simulate its input for enough cycles, but UTM(Pa,Pa) can simulate this same input to a final state, proving that Ha(Pa,Pa)==0 is wrong.

Now, if what you mean is that no potential simulating halt decider H can simulate the P built from it to completion, that would be correct. But the question isn't "can H simulate P to a final state". If that was the case, then Ha3(N,5)==0 would be correct because Ha3 is unable to simulate its input to a final state. For any Hi that simulates for X cycles and answers Hi(PiPi)==0, there is Hj that simulates for X+1 cycles and can run Hj(Pi,Pi)==1 to show that the former is wrong.

So ultimately, your claim is that the input to Ha(Pa,Pa) represents Pn(Pn) which, as you would put it, is "quite nuts" as the specification for any H:

H(x,y)==1 if and only if x(y) halts, and
H(x,y)==0 if and only if x(y) does not halt

Explicitly says otherwise.