On Friday, May 20, 2022 at 9:59:27 PM UTC-4, olcott wrote:
> 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

Which is doesn't as demonstrated by Hb(P,P) reaching a final state and returning 1.

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

You left out the fixed code of H which simulates P and will eventually abort which Hb(P,P) is able to see but not H because H aborts too soon.

> ...[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

Plus the x86 source code of the fixed implementation of H and everything H calls all the way down.

>
> _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.

And none of this matters if H(P,P) isn't telling us if P(P) halts.

> >
> > 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.