On Friday, May 20, 2022 at 11:13:25 PM UTC-4, olcott wrote:
> On 5/20/2022 10:06 PM, Dennis Bush wrote:
> > On Friday, May 20, 2022 at 10:45:12 PM UTC-4, olcott wrote:
> >> On 5/20/2022 9:32 PM, Dennis Bush wrote:
> >>> On Friday, May 20, 2022 at 10:17:18 PM UTC-4, olcott wrote:
> >>>> On 5/20/2022 9:09 PM, Dennis Bush wrote:
> >>>>> 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.
> >>>> H is at machine address 00000826 every other variation is a dishonest
> >>>> dodge away from the actual question.
> >>>
> >>> And since the fixed H is programmed to abort
> >> This is not true and we are not yet discussing that aspect of it yet.
> >
> > It must be, otherwise you have nothing to decide on. Remember, P is a *complete* program, which means the H is called must be fully defined and fixed.
> >
> It is ridiculously stupid to say that a halt decider has a fixed behavior.

You have it backwards. It is ridiculously stupid to say that *any* computation, *including* a halt decider, does NOT have a fixed behavior. That's how a given input always gives the same output.

>
> given an input of the function domain it can return the corresponding
> output. https://en.wikipedia.org/wiki/Computable_function

And for the halting function, its domain is a turing machine M and its input w, and its output is: accept if M applied to w halts, and reject if M applied to w does not halt. And the question of the halting problem is: Is the halting function computable?

>
> In any case I am only talking about the simulation of the input to
> H(P,P). Are you freaking stupid of what?
> > Otherwise what you're saying is:
> >
> > Since H1(P1,P1)==0,
> > and H2(P2,P2)==0,
> > and H3(P3,P3)==0,
> > and H4(P4,P4)==0,
> > and H5(P5,P5)==0,
> > ....
> > and Hn(Pn,Pn) does not halt,
> > Then for any i, Hi(Pi,Pi) == 0 is correct
> >
> > These P's are all separate and distinct computations having nothing to do with each other.
> >
> This is the P that I am referring to knucklehead:
> 558bec8b4508508b4d0851e840feffff83c40885c07402ebfe5dc3

So since you aren't specifying H, then you do in fact mean:

Since H1(P1,P1)==0,
and H2(P2,P2)==0,
and H3(P3,P3)==0,
and H4(P4,P4)==0,
and H5(P5,P5)==0,
.....
and Hn(Pn,Pn) does not halt,
Then for any i, Hi(Pi,Pi) == 0 is correct

Which is invalid because each Pi is a separate and distinct computation, so one has nothing to do with another.