On 5/20/2022 8:51 PM, Dennis Bush wrote:
> On Friday, May 20, 2022 at 9:42:11 PM UTC-4, olcott wrote:
>> On 5/20/2022 8:20 PM, Dennis Bush wrote:
>>> On Friday, May 20, 2022 at 8:56:09 PM UTC-4, olcott wrote:
>>>> On 5/20/2022 7:32 PM, Dennis Bush wrote:
>>>>> On Friday, May 20, 2022 at 8:15:31 PM UTC-4, olcott wrote:
>>>>>> On 5/20/2022 7:09 PM, Dennis Bush wrote:
>>>>>>> On Friday, May 20, 2022 at 7:37:15 PM UTC-4, olcott wrote:
>>>>>>>> On 5/20/2022 6:31 PM, Dennis Bush wrote:
>>>>>>>>> On Friday, May 20, 2022 at 7:25:15 PM UTC-4, olcott wrote:
>>>>>>>>>> On 5/20/2022 6:21 PM, Dennis Bush wrote:
>>>>>>>>>>> On Friday, May 20, 2022 at 7:13:37 PM UTC-4, olcott wrote:
>>>>>>>>>>>> On 5/20/2022 6:11 PM, Dennis Bush wrote:
>>>>>>>>>>>>> On Friday, May 20, 2022 at 7:02:24 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>> On 5/20/2022 5:53 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>> On Friday, May 20, 2022 at 6:51:42 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>> On 5/20/2022 5:44 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>> On Friday, May 20, 2022 at 6:33:29 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> You have known that the input to H(P,P) is simulated correctly proving
>>>>>>>>>>>>>>>>>>>> that H(P,P)==0 is correct for the whole six months
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> If H is intended to be a halt decider (even if only for the one case you
>>>>>>>>>>>>>>>>>>> claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> If a "correct simulation of the input to H(P,P)" is a simulation of the
>>>>>>>>>>>>>>>>>>> function call P(P), then H(P,P) == 0 is wrong because both P(P) and a
>>>>>>>>>>>>>>>>>>> correct simulation of P(P) halt.
>>>>>>>>>>>>>>>>>> So in other words even though there are no lines of code that are
>>>>>>>>>>>>>>>>>> simulated incorrectly you still claim that the simulation is incorrect.
>>>>>>>>>>>>>>>>>> THAT IS WHAT A DAMNED LYING BASTARD WOULD SAY.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So if we look at the execution trace of Ha3(N,5),
>>>>>>>>>>>>>>>> Yet another God damned lying bastard that uses the strawman error in an
>>>>>>>>>>>>>>>> attempt to get away with deception.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So if you believe that Ha3(N,5)==0 is either wrong or irrelevent you should have no problem explaining why without just saying "strawman".
>>>>>>>>>>>>>> I am talking about the correct simulation of the machine language of P
>>>>>>>>>>>>>> being correctly emulated by another 100% specific finite string of
>>>>>>>>>>>>>> machine language named H.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And by the criteria you use to show that H(P,P)==0 is correct,
>>>>>>>>>>>> More than that.
>>>>>>>>>>>> I prove that the criteria that I use is correct and I prove the
>>>>>>>>>>>> H(P,P)==0 is correct on the basis of that criteria.
>>>>>>>>>>>
>>>>>>>>>>> The *only* criteria for whether the result of H(P,P) is correct is whether P(P) halts, *by definition*.
>>>>>>>>>> Not for the specific machine language that I am referring to.
>>>>>>>>>> So OK you prove that you intend to stay in despicable liar mode.
>>>>>>>>>>
>>>>>>>>>> You proclaim by lying bastard dictatorial fiat that the input to H(P,P)
>>>>>>>>>> halts even though its correct execution trace proves otherwise.
>>>>>>>>>
>>>>>>>>> The input to H(P,P) represents P(P) by definition.
>>>>>>>> Not when we are talking about C functions and x86 machine-code Jackass.
>>>>>>>
>>>>>>> Then you're not talking about the halting problem, and H is computing something other that the halting function
>>>>>> None-the-less H(P,P)==0 is proven to be correct.
>>>>>
>>>>> Only for something that is not the halting problem.
>>>> So you agree that H(P,P)==0 is correct when H computes the halt status
>>>> of the machine language of P that is passed to H on the stack as a pair
>>>> of pointers to the literal string of P?
>>>
>>> If you're referring to *just* the function P, it doesn't have a halt status because it's not a complete program.
>>>
>> When we stipulate that the input to H(P,P) only halts when its correct
>> simulation reaches its machine address of [000009f0] otherwise it is
>> non-halting then H(P,P)==0 is the correct halt status of the input to
>> H(P,P).
>>
>> _P()
>> [000009d6](01) 55 push ebp
>> [000009d7](02) 8bec mov ebp,esp
>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>> [000009dc](01) 50 push eax // push P
>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>> [000009e0](01) 51 push ecx // push P
>> [000009e1](05) e840feffff call 00000826 // call H
>> [000009e6](03) 83c408 add esp,+08
>> [000009e9](02) 85c0 test eax,eax
>> [000009eb](02) 7402 jz 000009ef
>> [000009ed](02) ebfe jmp 000009ed
>> [000009ef](01) 5d pop ebp
>> [000009f0](01) c3 ret // Final state
>> Size in bytes:(0027) [000009f0]
>
> But H(P,P) does *not* perform a correct of its input. A correct simulation of this input is done Hb(P,P) which does in fact reach a final state. Therefore H(P,P)==0 is wrong.

OK then when the literal string of the machine code of H correctly
emulates the pair of literal strings of the machine code of P we can
verify that this execution trace is correct:

Begin Local Halt Decider Simulation Execution Trace Stored at:212352
....[00001352][0021233e][00212342] 55 push ebp // enter P
....[00001353][0021233e][00212342] 8bec mov ebp,esp
....[00001355][0021233e][00212342] 8b4508 mov eax,[ebp+08]
....[00001358][0021233a][00001352] 50 push eax // push P
....[00001359][0021233a][00001352] 8b4d08 mov ecx,[ebp+08]
....[0000135c][00212336][00001352] 51 push ecx // push P
....[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
....[00001352][0025cd66][0025cd6a] 55 push ebp // enter P
....[00001353][0025cd66][0025cd6a] 8bec mov ebp,esp
....[00001355][0025cd66][0025cd6a] 8b4508 mov eax,[ebp+08]
....[00001358][0025cd62][00001352] 50 push eax // push P
....[00001359][0025cd62][00001352] 8b4d08 mov ecx,[ebp+08]
....[0000135c][0025cd5e][00001352] 51 push ecx // push P
....[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

On the basis of this x86 source-code of this literal string:
558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx
[0000135d](05) e840feffff call 000011a2 // call H
[00001362](03) 83c408 add esp,+08
[00001365](02) 85c0 test eax,eax
[00001367](02) 7402 jz 0000136b
[00001369](02) ebfe jmp 00001369
[0000136b](01) 5d pop ebp
[0000136c](01) c3 ret
Size in bytes:(0027) [0000136c]

Specifies the trace that H provides.

>
> This of course is all academic because H is not computing the halting function:
>
> D(X,Y)==1 if and only if X(Y) halts, and
> D(X,Y)==0 if and only if X(Y) does not halt
>
> All you're doing is trying to explain how it comes up with the wrong answer.
>
> It doesn't matter whether or not this fits the description of what you think a decider should be. That's the problem description, period.
>
> If you state that no D can compute this function, then you're explicitly agreeing with the conclusion of the halting problem proofs.
>
>
>>> The machine language of P, as a *complete program*, includes the fixed machine language of H since P calls H, plus everything that H calls all the way down. And P(P) halts, therefore H(P,P)==0 is wrong.
>>>
>>> Your error is thinking that H doesn't matter, but it does.
>>>
>>>>>
>>>>>>
>>>>>> When the HP requires that a non-decider
>>>>>
>>>>> It doesn't matter what it's called. What matters is that it computes the halting function.
>>>>>
>>>>>> that is not a computation
>>>>>
>>>>> If it always gives the same result for the same input, it *is* a computation
>>>>>
>>>>>> is required to compute halting for a non-input
>>>>>
>>>>> Which is exactly represented by the input
>>>>>
>>>>>> it contradicts the axioms of computer science
>>>>>
>>>>> Which means what exactly? You're just making stuff up.
>>>>>
>>>>>> and therefore errs, thus making H(P,P)==0 relevant.
>>>>>
>>>>> So in other words you're saying that the conclusion of the halting problem proofs is correct.
>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>> The input to the C function H is a pair of pointers to the the literal
>>>>>>>> string of the machine code of P, passed on the stack.
>>>>>>>
>>>>>>> And the H that P calls is considered part of the program P so it's behavior must be considered. And since the fixed code of H *will* return 0, a correct simulation of the input, like that performed by Hb, does reach a final state.
>>>>>> The correctly simulated input to H(P,P) cannot possibly ever reach its
>>>>>> own final state therefore is correctly determined to be non-halting.
>>>>>
>>>>> The correctly simulated input to H(P,P) is performed by Hb(P,P) which does simulate it to a final state, therefore H(P,P)==0 is wrong.
>>>> --
>>>> Copyright 2022 Pete Olcott
>>>>
>>>> "Talent hits a target no one else can hit;
>>>> Genius hits a target no one else can see."
>>>> Arthur Schopenhauer
>>
>>
>> --
>> Copyright 2022 Pete Olcott
>>
>> "Talent hits a target no one else can hit;
>> Genius hits a target no one else can see."
>> Arthur Schopenhauer

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer