On 5/22/2022 12:36 PM, Richard Damon wrote:
> On 5/22/22 12:42 PM, olcott wrote:
>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>> On 5/22/22 12:31 PM, olcott wrote:
>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be for
>>>>>>>>>>>>>>>>>>>> one emulation and one nested emulation we can see
>>>>>>>>>>>>>>>>>>>> that the correctly emulated input to H(P,P) would
>>>>>>>>>>>>>>>>>>>> never reach its final state at machine address
>>>>>>>>>>>>>>>>>>>> [0000136c].
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution path
>>>>>>>>>>>>>>>>>>> of two independent execution units.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution trace
>>>>>>>>>>>>>>>>>> of one simulation and one nested simulation of the
>>>>>>>>>>>>>>>>>> input to H(P,P).
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the equivalent
>>>>>>>>>>>>>>>>> of a of a square circle.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is easy to
>>>>>>>>>>>>>>>> show when H simulates its input, yet another execution
>>>>>>>>>>>>>>>> trace of the input to H(P,P) that was invoked by P is
>>>>>>>>>>>>>>>> "like a square circle" can't possibly exist?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece of
>>>>>>>>>>>>>>> the first.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The fact you don't understand that says you just don't
>>>>>>>>>>>>>>> know how computers or programs actually work.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a UTM
>>>>>>>>>>>>>> that simulates a
>>>>>>>>>>>>>> TM description all of this is simply data on the first
>>>>>>>>>>>>>> UTM's tape and the only actual executable is the first UTM.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>> states that the second UTM is going through, but NOT the
>>>>>>>>>>>>> states that second UTM simulates in its own processing.
>>>>>>>>>>>>>
>>>>>>>>>>>>> That second UTM might produce its OWN trace of the states
>>>>>>>>>>>>> that it has simulated, but that is a SEPERATE trace, and
>>>>>>>>>>>>> NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a part
>>>>>>>>>>>> of its own data.
>>>>>>>>>>>
>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it isn't
>>>>>>>>>>> part of the trace generated by that UTM.
>>>>>>>>>>
>>>>>>>>>> The only actual executable is the outer UTM everything else is
>>>>>>>>>> a part of the same nested process.
>>>>>>>>>
>>>>>>>>> So the only actual valid trace is what that outer simulator
>>>>>>>>> actual simulated.
>>>>>>>>>
>>>>>>>>
>>>>>>>> There is a valid trace of every line of code that is emulated.
>>>>>>>> Operating system code has its trace tuned off. This only leaves
>>>>>>>> the user code such as P() and main(). Then we see the 14 lines
>>>>>>>> execution trace of the two level simulation of the input to H(P,P)
>>>>>>>
>>>>>>> No, because the second level emulation is NOT emulated by the top
>>>>>>> level emulator, its emulator is.
>>>>>>>
>>>>>>> Unless you are lying about what H does, you are just lying that
>>>>>>> the second level code is emulated by the same emulation process
>>>>>>> that the first is. (That may well be true, but it means you logic
>>>>>>> is still built on a lie).
>>>>>>>
>>>>>>
>>>>>> If you are too stupid to understand that H(P,P) derives the same
>>>>>> execution trace of its input every time it is called you are far
>>>>>> too stupid to evaluate my work.
>>>>>
>>>>> Ok, then why does the H(P,P) that P calls get stuck in an infinite
>>>>> recursion wneh the top level doesn't?
>>>>
>>>> It is a verified fact that the correct simulation of the input to
>>>> H(P,P) never reaches its final instruction thus conclusively proving
>>>> that it never halts.
>>>>
>>>
>>> The only machine that you have shown that does a correct simulation
>>> is the version that never aborts. That version fails to answer the
>>> question, so fails to be a halt decider.
>>>
>>> Any version of H that aborts, and returns a not-halting answer
>>> changes P into a Halting Compuation.
>>>
>>> The "pathological" use of H by P lets it change as you change H, so
>>> if H aborts, it is wrong because THAT P halts, if it doesn't, then it
>>> is wrong for not answering.
>>>
>>> You seem to miss this fact because you just don't understand the
>>> basics of how computations work. Part of your problem is you keep on
>>> trying to define H by rules that aren't an actual algorithm, so can't
>>> actually be written.
>>>
>>
>>
>>
>> It is an easily verifiable fact that the C function H does correctly
>> determine that the C function named P would never reach its last
>> instruction when correctly emulated by H.
>
> Don't just "Claim" it, so an ACTUAL verification, or you just show
> yourself to be a liar.
>
>>
>> Everyone disagreeing with verified facts is incorrect on the basis of
>> lack of technical competency or lack of honesty.
>
> You haven't verified ANY fact, you have made claims using FAKE data that
> don't even really support your claim.
>

On 5/22/22 5:06 PM, olcott wrote:
> On 5/22/2022 12:36 PM, Richard Damon wrote:
>> On 5/22/22 12:42 PM, olcott wrote:
>>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>>> On 5/22/22 12:31 PM, olcott wrote:
>>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be for
>>>>>>>>>>>>>>>>>>>>> one emulation and one nested emulation we can see
>>>>>>>>>>>>>>>>>>>>> that the correctly emulated input to H(P,P) would
>>>>>>>>>>>>>>>>>>>>> never reach its final state at machine address
>>>>>>>>>>>>>>>>>>>>> [0000136c].
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution path
>>>>>>>>>>>>>>>>>>>> of two independent execution units.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution
>>>>>>>>>>>>>>>>>>> trace of one simulation and one nested simulation of
>>>>>>>>>>>>>>>>>>> the input to H(P,P).
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the equivalent
>>>>>>>>>>>>>>>>>> of a of a square circle.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is easy to
>>>>>>>>>>>>>>>>> show when H simulates its input, yet another execution
>>>>>>>>>>>>>>>>> trace of the input to H(P,P) that was invoked by P is
>>>>>>>>>>>>>>>>> "like a square circle" can't possibly exist?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece of
>>>>>>>>>>>>>>>> the first.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The fact you don't understand that says you just don't
>>>>>>>>>>>>>>>> know how computers or programs actually work.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a UTM
>>>>>>>>>>>>>>> that simulates a
>>>>>>>>>>>>>>> TM description all of this is simply data on the first
>>>>>>>>>>>>>>> UTM's tape and the only actual executable is the first UTM.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>>> states that the second UTM is going through, but NOT the
>>>>>>>>>>>>>> states that second UTM simulates in its own processing.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> That second UTM might produce its OWN trace of the states
>>>>>>>>>>>>>> that it has simulated, but that is a SEPERATE trace, and
>>>>>>>>>>>>>> NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a part
>>>>>>>>>>>>> of its own data.
>>>>>>>>>>>>
>>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it isn't
>>>>>>>>>>>> part of the trace generated by that UTM.
>>>>>>>>>>>
>>>>>>>>>>> The only actual executable is the outer UTM everything else
>>>>>>>>>>> is a part of the same nested process.
>>>>>>>>>>
>>>>>>>>>> So the only actual valid trace is what that outer simulator
>>>>>>>>>> actual simulated.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> There is a valid trace of every line of code that is emulated.
>>>>>>>>> Operating system code has its trace tuned off. This only leaves
>>>>>>>>> the user code such as P() and main(). Then we see the 14 lines
>>>>>>>>> execution trace of the two level simulation of the input to H(P,P)
>>>>>>>>
>>>>>>>> No, because the second level emulation is NOT emulated by the
>>>>>>>> top level emulator, its emulator is.
>>>>>>>>
>>>>>>>> Unless you are lying about what H does, you are just lying that
>>>>>>>> the second level code is emulated by the same emulation process
>>>>>>>> that the first is. (That may well be true, but it means you
>>>>>>>> logic is still built on a lie).
>>>>>>>>
>>>>>>>
>>>>>>> If you are too stupid to understand that H(P,P) derives the same
>>>>>>> execution trace of its input every time it is called you are far
>>>>>>> too stupid to evaluate my work.
>>>>>>
>>>>>> Ok, then why does the H(P,P) that P calls get stuck in an infinite
>>>>>> recursion wneh the top level doesn't?
>>>>>
>>>>> It is a verified fact that the correct simulation of the input to
>>>>> H(P,P) never reaches its final instruction thus conclusively
>>>>> proving that it never halts.
>>>>>
>>>>
>>>> The only machine that you have shown that does a correct simulation
>>>> is the version that never aborts. That version fails to answer the
>>>> question, so fails to be a halt decider.
>>>>
>>>> Any version of H that aborts, and returns a not-halting answer
>>>> changes P into a Halting Compuation.
>>>>
>>>> The "pathological" use of H by P lets it change as you change H, so
>>>> if H aborts, it is wrong because THAT P halts, if it doesn't, then
>>>> it is wrong for not answering.
>>>>
>>>> You seem to miss this fact because you just don't understand the
>>>> basics of how computations work. Part of your problem is you keep on
>>>> trying to define H by rules that aren't an actual algorithm, so
>>>> can't actually be written.
>>>>
>>>
>>>
>>>
>>> It is an easily verifiable fact that the C function H does correctly
>>> determine that the C function named P would never reach its last
>>> instruction when correctly emulated by H.
>>
>> Don't just "Claim" it, so an ACTUAL verification, or you just show
>> yourself to be a liar.
>>
>>>
>>> Everyone disagreeing with verified facts is incorrect on the basis of
>>> lack of technical competency or lack of honesty.
>>
>> You haven't verified ANY fact, you have made claims using FAKE data
>> that don't even really support your claim.
>>
>
> Software engineering experts
> can reverse-engineer what the correct x86 emulation of the input to
> H(P,P) would be for one emulation and one nested emulation thus
> confirming that the provided execution trace is correct. They can do
> this entirely on the basis of the x86 source-code for P with no need to
> see the source-code or execution trace of H.

On 5/22/2022 4:34 PM, Richard Damon wrote:
>
> On 5/22/22 5:06 PM, olcott wrote:
>> On 5/22/2022 12:36 PM, Richard Damon wrote:
>>> On 5/22/22 12:42 PM, olcott wrote:
>>>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>>>> On 5/22/22 12:31 PM, olcott wrote:
>>>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be
>>>>>>>>>>>>>>>>>>>>>> for one emulation and one nested emulation we can
>>>>>>>>>>>>>>>>>>>>>> see that the correctly emulated input to H(P,P)
>>>>>>>>>>>>>>>>>>>>>> would never reach its final state at machine
>>>>>>>>>>>>>>>>>>>>>> address [0000136c].
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution
>>>>>>>>>>>>>>>>>>>>> path of two independent execution units.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution
>>>>>>>>>>>>>>>>>>>> trace of one simulation and one nested simulation of
>>>>>>>>>>>>>>>>>>>> the input to H(P,P).
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the
>>>>>>>>>>>>>>>>>>> equivalent of a of a square circle.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is easy
>>>>>>>>>>>>>>>>>> to show when H simulates its input, yet another
>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) that was
>>>>>>>>>>>>>>>>>> invoked by P is "like a square circle" can't possibly
>>>>>>>>>>>>>>>>>> exist?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece of
>>>>>>>>>>>>>>>>> the first.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> The fact you don't understand that says you just don't
>>>>>>>>>>>>>>>>> know how computers or programs actually work.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a UTM
>>>>>>>>>>>>>>>> that simulates a
>>>>>>>>>>>>>>>> TM description all of this is simply data on the first
>>>>>>>>>>>>>>>> UTM's tape and the only actual executable is the first UTM.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>>>> states that the second UTM is going through, but NOT the
>>>>>>>>>>>>>>> states that second UTM simulates in its own processing.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> That second UTM might produce its OWN trace of the states
>>>>>>>>>>>>>>> that it has simulated, but that is a SEPERATE trace, and
>>>>>>>>>>>>>>> NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a
>>>>>>>>>>>>>> part of its own data.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it isn't
>>>>>>>>>>>>> part of the trace generated by that UTM.
>>>>>>>>>>>>
>>>>>>>>>>>> The only actual executable is the outer UTM everything else
>>>>>>>>>>>> is a part of the same nested process.
>>>>>>>>>>>
>>>>>>>>>>> So the only actual valid trace is what that outer simulator
>>>>>>>>>>> actual simulated.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> There is a valid trace of every line of code that is emulated.
>>>>>>>>>> Operating system code has its trace tuned off. This only
>>>>>>>>>> leaves the user code such as P() and main(). Then we see the
>>>>>>>>>> 14 lines execution trace of the two level simulation of the
>>>>>>>>>> input to H(P,P)
>>>>>>>>>
>>>>>>>>> No, because the second level emulation is NOT emulated by the
>>>>>>>>> top level emulator, its emulator is.
>>>>>>>>>
>>>>>>>>> Unless you are lying about what H does, you are just lying that
>>>>>>>>> the second level code is emulated by the same emulation process
>>>>>>>>> that the first is. (That may well be true, but it means you
>>>>>>>>> logic is still built on a lie).
>>>>>>>>>
>>>>>>>>
>>>>>>>> If you are too stupid to understand that H(P,P) derives the same
>>>>>>>> execution trace of its input every time it is called you are far
>>>>>>>> too stupid to evaluate my work.
>>>>>>>
>>>>>>> Ok, then why does the H(P,P) that P calls get stuck in an
>>>>>>> infinite recursion wneh the top level doesn't?
>>>>>>
>>>>>> It is a verified fact that the correct simulation of the input to
>>>>>> H(P,P) never reaches its final instruction thus conclusively
>>>>>> proving that it never halts.
>>>>>>
>>>>>
>>>>> The only machine that you have shown that does a correct simulation
>>>>> is the version that never aborts. That version fails to answer the
>>>>> question, so fails to be a halt decider.
>>>>>
>>>>> Any version of H that aborts, and returns a not-halting answer
>>>>> changes P into a Halting Compuation.
>>>>>
>>>>> The "pathological" use of H by P lets it change as you change H, so
>>>>> if H aborts, it is wrong because THAT P halts, if it doesn't, then
>>>>> it is wrong for not answering.
>>>>>
>>>>> You seem to miss this fact because you just don't understand the
>>>>> basics of how computations work. Part of your problem is you keep
>>>>> on trying to define H by rules that aren't an actual algorithm, so
>>>>> can't actually be written.
>>>>>
>>>>
>>>>
>>>>
>>>> It is an easily verifiable fact that the C function H does correctly
>>>> determine that the C function named P would never reach its last
>>>> instruction when correctly emulated by H.
>>>
>>> Don't just "Claim" it, so an ACTUAL verification, or you just show
>>> yourself to be a liar.
>>>
>>>>
>>>> Everyone disagreeing with verified facts is incorrect on the basis
>>>> of lack of technical competency or lack of honesty.
>>>
>>> You haven't verified ANY fact, you have made claims using FAKE data
>>> that don't even really support your claim.
>>>
>>
>> Software engineering experts
>> can reverse-engineer what the correct x86 emulation of the input to
>> H(P,P) would be for one emulation and one nested emulation thus
>> confirming that the provided execution trace is correct. They can do
>> this entirely on the basis of the x86 source-code for P with no need
>> to see the source-code or execution trace of H.
>
> Interesting point, that you eed to talk about "reverse-engineering" this
> output implies that you don't actually have a program to generate it.
>

On 5/22/22 5:59 PM, olcott wrote:
> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>
>> On 5/22/22 5:06 PM, olcott wrote:
>>> On 5/22/2022 12:36 PM, Richard Damon wrote:
>>>> On 5/22/22 12:42 PM, olcott wrote:
>>>>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>>>>> On 5/22/22 12:31 PM, olcott wrote:
>>>>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be
>>>>>>>>>>>>>>>>>>>>>>> for one emulation and one nested emulation we can
>>>>>>>>>>>>>>>>>>>>>>> see that the correctly emulated input to H(P,P)
>>>>>>>>>>>>>>>>>>>>>>> would never reach its final state at machine
>>>>>>>>>>>>>>>>>>>>>>> address [0000136c].
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution
>>>>>>>>>>>>>>>>>>>>>> path of two independent execution units.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution
>>>>>>>>>>>>>>>>>>>>> trace of one simulation and one nested simulation
>>>>>>>>>>>>>>>>>>>>> of the input to H(P,P).
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the
>>>>>>>>>>>>>>>>>>>> equivalent of a of a square circle.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is easy
>>>>>>>>>>>>>>>>>>> to show when H simulates its input, yet another
>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) that was
>>>>>>>>>>>>>>>>>>> invoked by P is "like a square circle" can't possibly
>>>>>>>>>>>>>>>>>>> exist?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece
>>>>>>>>>>>>>>>>>> of the first.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> The fact you don't understand that says you just don't
>>>>>>>>>>>>>>>>>> know how computers or programs actually work.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a UTM
>>>>>>>>>>>>>>>>> that simulates a
>>>>>>>>>>>>>>>>> TM description all of this is simply data on the first
>>>>>>>>>>>>>>>>> UTM's tape and the only actual executable is the first
>>>>>>>>>>>>>>>>> UTM.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>>>>> states that the second UTM is going through, but NOT the
>>>>>>>>>>>>>>>> states that second UTM simulates in its own processing.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> That second UTM might produce its OWN trace of the
>>>>>>>>>>>>>>>> states that it has simulated, but that is a SEPERATE
>>>>>>>>>>>>>>>> trace, and NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a
>>>>>>>>>>>>>>> part of its own data.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it isn't
>>>>>>>>>>>>>> part of the trace generated by that UTM.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The only actual executable is the outer UTM everything else
>>>>>>>>>>>>> is a part of the same nested process.
>>>>>>>>>>>>
>>>>>>>>>>>> So the only actual valid trace is what that outer simulator
>>>>>>>>>>>> actual simulated.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> There is a valid trace of every line of code that is
>>>>>>>>>>> emulated. Operating system code has its trace tuned off. This
>>>>>>>>>>> only leaves the user code such as P() and main(). Then we see
>>>>>>>>>>> the 14 lines execution trace of the two level simulation of
>>>>>>>>>>> the input to H(P,P)
>>>>>>>>>>
>>>>>>>>>> No, because the second level emulation is NOT emulated by the
>>>>>>>>>> top level emulator, its emulator is.
>>>>>>>>>>
>>>>>>>>>> Unless you are lying about what H does, you are just lying
>>>>>>>>>> that the second level code is emulated by the same emulation
>>>>>>>>>> process that the first is. (That may well be true, but it
>>>>>>>>>> means you logic is still built on a lie).
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> If you are too stupid to understand that H(P,P) derives the
>>>>>>>>> same execution trace of its input every time it is called you
>>>>>>>>> are far too stupid to evaluate my work.
>>>>>>>>
>>>>>>>> Ok, then why does the H(P,P) that P calls get stuck in an
>>>>>>>> infinite recursion wneh the top level doesn't?
>>>>>>>
>>>>>>> It is a verified fact that the correct simulation of the input to
>>>>>>> H(P,P) never reaches its final instruction thus conclusively
>>>>>>> proving that it never halts.
>>>>>>>
>>>>>>
>>>>>> The only machine that you have shown that does a correct
>>>>>> simulation is the version that never aborts. That version fails to
>>>>>> answer the question, so fails to be a halt decider.
>>>>>>
>>>>>> Any version of H that aborts, and returns a not-halting answer
>>>>>> changes P into a Halting Compuation.
>>>>>>
>>>>>> The "pathological" use of H by P lets it change as you change H,
>>>>>> so if H aborts, it is wrong because THAT P halts, if it doesn't,
>>>>>> then it is wrong for not answering.
>>>>>>
>>>>>> You seem to miss this fact because you just don't understand the
>>>>>> basics of how computations work. Part of your problem is you keep
>>>>>> on trying to define H by rules that aren't an actual algorithm, so
>>>>>> can't actually be written.
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>> It is an easily verifiable fact that the C function H does
>>>>> correctly determine that the C function named P would never reach
>>>>> its last instruction when correctly emulated by H.
>>>>
>>>> Don't just "Claim" it, so an ACTUAL verification, or you just show
>>>> yourself to be a liar.
>>>>
>>>>>
>>>>> Everyone disagreeing with verified facts is incorrect on the basis
>>>>> of lack of technical competency or lack of honesty.
>>>>
>>>> You haven't verified ANY fact, you have made claims using FAKE data
>>>> that don't even really support your claim.
>>>>
>>>
>>> Software engineering experts
>>> can reverse-engineer what the correct x86 emulation of the input to
>>> H(P,P) would be for one emulation and one nested emulation thus
>>> confirming that the provided execution trace is correct. They can do
>>> this entirely on the basis of the x86 source-code for P with no need
>>> to see the source-code or execution trace of H.
>>
>> Interesting point, that you eed to talk about "reverse-engineering"
>> this output implies that you don't actually have a program to generate
>> it.
>>
>
> Not at all. This program took me a whole man-year.

On 5/22/2022 5:34 PM, Richard Damon wrote:
>
> On 5/22/22 5:59 PM, olcott wrote:
>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>>
>>> On 5/22/22 5:06 PM, olcott wrote:
>>>> On 5/22/2022 12:36 PM, Richard Damon wrote:
>>>>> On 5/22/22 12:42 PM, olcott wrote:
>>>>>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>>>>>> On 5/22/22 12:31 PM, olcott wrote:
>>>>>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be
>>>>>>>>>>>>>>>>>>>>>>>> for one emulation and one nested emulation we
>>>>>>>>>>>>>>>>>>>>>>>> can see that the correctly emulated input to
>>>>>>>>>>>>>>>>>>>>>>>> H(P,P) would never reach its final state at
>>>>>>>>>>>>>>>>>>>>>>>> machine address [0000136c].
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution
>>>>>>>>>>>>>>>>>>>>>>> path of two independent execution units.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution
>>>>>>>>>>>>>>>>>>>>>> trace of one simulation and one nested simulation
>>>>>>>>>>>>>>>>>>>>>> of the input to H(P,P).
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the
>>>>>>>>>>>>>>>>>>>>> equivalent of a of a square circle.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is easy
>>>>>>>>>>>>>>>>>>>> to show when H simulates its input, yet another
>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) that was
>>>>>>>>>>>>>>>>>>>> invoked by P is "like a square circle" can't
>>>>>>>>>>>>>>>>>>>> possibly exist?
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece
>>>>>>>>>>>>>>>>>>> of the first.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> The fact you don't understand that says you just
>>>>>>>>>>>>>>>>>>> don't know how computers or programs actually work.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a UTM
>>>>>>>>>>>>>>>>>> that simulates a
>>>>>>>>>>>>>>>>>> TM description all of this is simply data on the first
>>>>>>>>>>>>>>>>>> UTM's tape and the only actual executable is the first
>>>>>>>>>>>>>>>>>> UTM.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>>>>>> states that the second UTM is going through, but NOT
>>>>>>>>>>>>>>>>> the states that second UTM simulates in its own
>>>>>>>>>>>>>>>>> processing.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> That second UTM might produce its OWN trace of the
>>>>>>>>>>>>>>>>> states that it has simulated, but that is a SEPERATE
>>>>>>>>>>>>>>>>> trace, and NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a
>>>>>>>>>>>>>>>> part of its own data.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it isn't
>>>>>>>>>>>>>>> part of the trace generated by that UTM.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The only actual executable is the outer UTM everything
>>>>>>>>>>>>>> else is a part of the same nested process.
>>>>>>>>>>>>>
>>>>>>>>>>>>> So the only actual valid trace is what that outer simulator
>>>>>>>>>>>>> actual simulated.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> There is a valid trace of every line of code that is
>>>>>>>>>>>> emulated. Operating system code has its trace tuned off.
>>>>>>>>>>>> This only leaves the user code such as P() and main(). Then
>>>>>>>>>>>> we see the 14 lines execution trace of the two level
>>>>>>>>>>>> simulation of the input to H(P,P)
>>>>>>>>>>>
>>>>>>>>>>> No, because the second level emulation is NOT emulated by the
>>>>>>>>>>> top level emulator, its emulator is.
>>>>>>>>>>>
>>>>>>>>>>> Unless you are lying about what H does, you are just lying
>>>>>>>>>>> that the second level code is emulated by the same emulation
>>>>>>>>>>> process that the first is. (That may well be true, but it
>>>>>>>>>>> means you logic is still built on a lie).
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> If you are too stupid to understand that H(P,P) derives the
>>>>>>>>>> same execution trace of its input every time it is called you
>>>>>>>>>> are far too stupid to evaluate my work.
>>>>>>>>>
>>>>>>>>> Ok, then why does the H(P,P) that P calls get stuck in an
>>>>>>>>> infinite recursion wneh the top level doesn't?
>>>>>>>>
>>>>>>>> It is a verified fact that the correct simulation of the input
>>>>>>>> to H(P,P) never reaches its final instruction thus conclusively
>>>>>>>> proving that it never halts.
>>>>>>>>
>>>>>>>
>>>>>>> The only machine that you have shown that does a correct
>>>>>>> simulation is the version that never aborts. That version fails
>>>>>>> to answer the question, so fails to be a halt decider.
>>>>>>>
>>>>>>> Any version of H that aborts, and returns a not-halting answer
>>>>>>> changes P into a Halting Compuation.
>>>>>>>
>>>>>>> The "pathological" use of H by P lets it change as you change H,
>>>>>>> so if H aborts, it is wrong because THAT P halts, if it doesn't,
>>>>>>> then it is wrong for not answering.
>>>>>>>
>>>>>>> You seem to miss this fact because you just don't understand the
>>>>>>> basics of how computations work. Part of your problem is you keep
>>>>>>> on trying to define H by rules that aren't an actual algorithm,
>>>>>>> so can't actually be written.
>>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> It is an easily verifiable fact that the C function H does
>>>>>> correctly determine that the C function named P would never reach
>>>>>> its last instruction when correctly emulated by H.
>>>>>
>>>>> Don't just "Claim" it, so an ACTUAL verification, or you just show
>>>>> yourself to be a liar.
>>>>>
>>>>>>
>>>>>> Everyone disagreeing with verified facts is incorrect on the basis
>>>>>> of lack of technical competency or lack of honesty.
>>>>>
>>>>> You haven't verified ANY fact, you have made claims using FAKE data
>>>>> that don't even really support your claim.
>>>>>
>>>>
>>>> Software engineering experts
>>>> can reverse-engineer what the correct x86 emulation of the input to
>>>> H(P,P) would be for one emulation and one nested emulation thus
>>>> confirming that the provided execution trace is correct. They can do
>>>> this entirely on the basis of the x86 source-code for P with no need
>>>> to see the source-code or execution trace of H.
>>>
>>> Interesting point, that you eed to talk about "reverse-engineering"
>>> this output implies that you don't actually have a program to
>>> generate it.
>>>
>>
>> Not at all. This program took me a whole man-year.
>
> Then why do we need to "reverse-engineer" the trace?
>

On 5/22/22 6:50 PM, olcott wrote:
> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>
>> On 5/22/22 5:59 PM, olcott wrote:
>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>>>
>>>> On 5/22/22 5:06 PM, olcott wrote:
>>>>> On 5/22/2022 12:36 PM, Richard Damon wrote:
>>>>>> On 5/22/22 12:42 PM, olcott wrote:
>>>>>>> On 5/22/2022 11:40 AM, Richard Damon wrote:
>>>>>>>> On 5/22/22 12:31 PM, olcott wrote:
>>>>>>>>> On 5/22/2022 11:21 AM, Richard Damon wrote:
>>>>>>>>>> On 5/22/22 10:57 AM, olcott wrote:
>>>>>>>>>>> On 5/22/2022 6:06 AM, Richard Damon wrote:
>>>>>>>>>>>> On 5/22/22 1:02 AM, olcott wrote:
>>>>>>>>>>>>> On 5/21/2022 11:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 5/21/22 11:59 PM, olcott wrote:
>>>>>>>>>>>>>>> On 5/21/2022 10:54 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 5/21/22 11:36 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 5/21/2022 10:27 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>> On 5/21/22 10:48 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 5/21/2022 9:37 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>> On 5/21/22 10:28 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 5/21/2022 9:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> On 5/21/22 9:23 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 5/21/2022 8:05 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 5/21/22 3:38 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>> On 5/20/2022 5:25 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>> olcott <NoOne@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.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> When we correctly reverse-engineer what the
>>>>>>>>>>>>>>>>>>>>>>>>> execution trace of the input to H(P,P) would be
>>>>>>>>>>>>>>>>>>>>>>>>> for one emulation and one nested emulation we
>>>>>>>>>>>>>>>>>>>>>>>>> can see that the correctly emulated input to
>>>>>>>>>>>>>>>>>>>>>>>>> H(P,P) would never reach its final state at
>>>>>>>>>>>>>>>>>>>>>>>>> machine address [0000136c].
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> A nonsense trace, as it is mixing the execution
>>>>>>>>>>>>>>>>>>>>>>>> path of two independent execution units.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> In other words you acknowledge that you are
>>>>>>>>>>>>>>>>>>>>>>> technically incompetent to provide the execution
>>>>>>>>>>>>>>>>>>>>>>> trace of one simulation and one nested simulation
>>>>>>>>>>>>>>>>>>>>>>> of the input to H(P,P).
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> No, I am saying that you are asking for the
>>>>>>>>>>>>>>>>>>>>>> equivalent of a of a square circle.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> So an execution trace of the input to H(P,P) is
>>>>>>>>>>>>>>>>>>>>> easy to show when H simulates its input, yet
>>>>>>>>>>>>>>>>>>>>> another execution trace of the input to H(P,P) that
>>>>>>>>>>>>>>>>>>>>> was invoked by P is "like a square circle" can't
>>>>>>>>>>>>>>>>>>>>> possibly exist?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> The problem is that your second trace is NOT a piece
>>>>>>>>>>>>>>>>>>>> of the first.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> The fact you don't understand that says you just
>>>>>>>>>>>>>>>>>>>> don't know how computers or programs actually work.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> When a UTM simulates a TM description that calls a
>>>>>>>>>>>>>>>>>>> UTM that simulates a
>>>>>>>>>>>>>>>>>>> TM description all of this is simply data on the
>>>>>>>>>>>>>>>>>>> first UTM's tape and the only actual executable is
>>>>>>>>>>>>>>>>>>> the first UTM.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Yes, and a trace made by that outer UTM will show the
>>>>>>>>>>>>>>>>>> states that the second UTM is going through, but NOT
>>>>>>>>>>>>>>>>>> the states that second UTM simulates in its own
>>>>>>>>>>>>>>>>>> processing.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> That second UTM might produce its OWN trace of the
>>>>>>>>>>>>>>>>>> states that it has simulated, but that is a SEPERATE
>>>>>>>>>>>>>>>>>> trace, and NOT part of the trace from the OUTER UTM.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> And this trace is written to the outer UTM's tape as a
>>>>>>>>>>>>>>>>> part of its own data.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Yes, the DATA is there, ENCODED on the tape, but it
>>>>>>>>>>>>>>>> isn't part of the trace generated by that UTM.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The only actual executable is the outer UTM everything
>>>>>>>>>>>>>>> else is a part of the same nested process.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> So the only actual valid trace is what that outer
>>>>>>>>>>>>>> simulator actual simulated.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> There is a valid trace of every line of code that is
>>>>>>>>>>>>> emulated. Operating system code has its trace tuned off.
>>>>>>>>>>>>> This only leaves the user code such as P() and main(). Then
>>>>>>>>>>>>> we see the 14 lines execution trace of the two level
>>>>>>>>>>>>> simulation of the input to H(P,P)
>>>>>>>>>>>>
>>>>>>>>>>>> No, because the second level emulation is NOT emulated by
>>>>>>>>>>>> the top level emulator, its emulator is.
>>>>>>>>>>>>
>>>>>>>>>>>> Unless you are lying about what H does, you are just lying
>>>>>>>>>>>> that the second level code is emulated by the same emulation
>>>>>>>>>>>> process that the first is. (That may well be true, but it
>>>>>>>>>>>> means you logic is still built on a lie).
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> If you are too stupid to understand that H(P,P) derives the
>>>>>>>>>>> same execution trace of its input every time it is called you
>>>>>>>>>>> are far too stupid to evaluate my work.
>>>>>>>>>>
>>>>>>>>>> Ok, then why does the H(P,P) that P calls get stuck in an
>>>>>>>>>> infinite recursion wneh the top level doesn't?
>>>>>>>>>
>>>>>>>>> It is a verified fact that the correct simulation of the input
>>>>>>>>> to H(P,P) never reaches its final instruction thus conclusively
>>>>>>>>> proving that it never halts.
>>>>>>>>>
>>>>>>>>
>>>>>>>> The only machine that you have shown that does a correct
>>>>>>>> simulation is the version that never aborts. That version fails
>>>>>>>> to answer the question, so fails to be a halt decider.
>>>>>>>>
>>>>>>>> Any version of H that aborts, and returns a not-halting answer
>>>>>>>> changes P into a Halting Compuation.
>>>>>>>>
>>>>>>>> The "pathological" use of H by P lets it change as you change H,
>>>>>>>> so if H aborts, it is wrong because THAT P halts, if it doesn't,
>>>>>>>> then it is wrong for not answering.
>>>>>>>>
>>>>>>>> You seem to miss this fact because you just don't understand the
>>>>>>>> basics of how computations work. Part of your problem is you
>>>>>>>> keep on trying to define H by rules that aren't an actual
>>>>>>>> algorithm, so can't actually be written.
>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> It is an easily verifiable fact that the C function H does
>>>>>>> correctly determine that the C function named P would never reach
>>>>>>> its last instruction when correctly emulated by H.
>>>>>>
>>>>>> Don't just "Claim" it, so an ACTUAL verification, or you just show
>>>>>> yourself to be a liar.
>>>>>>
>>>>>>>
>>>>>>> Everyone disagreeing with verified facts is incorrect on the
>>>>>>> basis of lack of technical competency or lack of honesty.
>>>>>>
>>>>>> You haven't verified ANY fact, you have made claims using FAKE
>>>>>> data that don't even really support your claim.
>>>>>>
>>>>>
>>>>> Software engineering experts
>>>>> can reverse-engineer what the correct x86 emulation of the input to
>>>>> H(P,P) would be for one emulation and one nested emulation thus
>>>>> confirming that the provided execution trace is correct. They can
>>>>> do this entirely on the basis of the x86 source-code for P with no
>>>>> need to see the source-code or execution trace of H.
>>>>
>>>> Interesting point, that you eed to talk about "reverse-engineering"
>>>> this output implies that you don't actually have a program to
>>>> generate it.
>>>>
>>>
>>> Not at all. This program took me a whole man-year.
>>
>> Then why do we need to "reverse-engineer" the trace?
>>
>
> It is the only way that you can verify that this trace is correct:

On 5/22/2022 5:34 PM, Richard Damon wrote:
>
> On 5/22/22 5:59 PM, olcott wrote:
>> On 5/22/2022 4:34 PM, Richard Damon wrote:

>> So then you now agree that the nested invocation of H(P,P) would
>> derive the same execution trace of its input that the the outer one did?
>>
>
> Yes seem to have a confusion. Yes, the second simulator will produce and
> identical trace to the first one, AS A SEPERATE TRACE.

So the first invocation of H(P,P) derives this trace
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

And the second (nested) invocation of H(P,P) derives this trace
....[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

From this we can see the pattern that the C/x86 function named P will
never reach its last instruction, thus conclusively proving that when
H(P,P) rejects its input as non-halting it is correct.

>
> The issue is you can not validly "combine" them into a single trace,
> that is just a lie. The first simulator needs to show its trace of the
> instructions of the second simulator doing its simulation.
>

So when I show the trace of main() this trace is a lie? I show the trace
of every line of user code and none of these traces are a lie.

> I suppose if you want, you can omit that with a note of omitted detail,
> but the the trace goes to the call, says details omited, then the notice
> that the trace was aborted, but then you don't have any grounds to claim
> that it was correct (since there aren't any).
>
>> In order to contradict me you must contradict computer science.
>>
>>
>
> Nope, YOU do that enough.

--
Copyright 2022 Pete Olcott

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

On 5/22/22 7:16 PM, olcott wrote:
> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>
>> On 5/22/22 5:59 PM, olcott wrote:
>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>
>>> So then you now agree that the nested invocation of H(P,P) would
>>> derive the same execution trace of its input that the the outer one did?
>>>
>>
>> Yes seem to have a confusion. Yes, the second simulator will produce
>> and identical trace to the first one, AS A SEPERATE TRACE.
>
>
> So the first invocation of H(P,P) derives this trace
> 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
>
> And the second (nested) invocation of H(P,P) derives this trace
> ...[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
>
> From this we can see the pattern that the C/x86 function named P will
> never reach its last instruction, thus conclusively proving that when
> H(P,P) rejects its input as non-halting it is correct.
>

No, because the whole second trace is conditioned on the halt deciding
of H. So, if the top level H aborts at that point, then the trace become
incorrect by being incomplete (it doesn't show the FULL Halting behavior
of the input, which refers to the machine it represents, which did not
halt there.

When we put this input into an ACTUAL CORRECT emulator that doesn't
abort, it will show a third (nested) simulation followed by the
simulator that P called deciding to abort, and then P Halting.

>> The issue is you can not validly "combine" them into a single trace,
>> that is just a lie. The first simulator needs to show its trace of the
>> instructions of the second simulator doing its simulation.
>>
>
> So when I show the trace of main() this trace is a lie? I show the trace
> of every line of user code and none of these traces are a lie.

The phrase "User Code" is a lie, as H needs to be User Code, as that is
all Turing Machines have (at least as far as tracing goes).

I think the problem is that your H doesn't actually exist as a real
independent algorithm, but is enmeshed in your UTM x86, and thus isn't
actually a Computation.

>
>> I suppose if you want, you can omit that with a note of omitted
>> detail, but the the trace goes to the call, says details omited, then
>> the notice that the trace was aborted, but then you don't have any
>> grounds to claim that it was correct (since there aren't any).
>>
>>> In order to contradict me you must contradict computer science.
>>>
>>>
>>
>> Nope, YOU do that enough.
>
>

On 5/22/2022 6:52 PM, Richard Damon wrote:
> On 5/22/22 7:16 PM, olcott wrote:
>> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>>
>>> On 5/22/22 5:59 PM, olcott wrote:
>>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>
>>>> So then you now agree that the nested invocation of H(P,P) would
>>>> derive the same execution trace of its input that the the outer one
>>>> did?
>>>>
>>>
>>> Yes seem to have a confusion. Yes, the second simulator will produce
>>> and identical trace to the first one, AS A SEPERATE TRACE.
>>
>>
>> So the first invocation of H(P,P) derives this trace
>> 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
>>
>> And the second (nested) invocation of H(P,P) derives this trace
>> ...[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
>>
>>  From this we can see the pattern that the C/x86 function named P will
>> never reach its last instruction, thus conclusively proving that when
>> H(P,P) rejects its input as non-halting it is correct.
>>
>
> No, because the whole second trace is conditioned on the halt deciding
> of H. So, if the top level H aborts at that point, then the trace become
> incorrect by being incomplete (it doesn't show the FULL Halting behavior
> of the input, which refers to the machine it represents, which did not
> halt there.
The fact that I just proved that the simulated input to H(P,P) never
reaches its own final state conclusively proves that it is non-halting
regardless of whether or not it stops running.

--
Copyright 2022 Pete Olcott

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

On 5/22/22 7:59 PM, olcott wrote:
> On 5/22/2022 6:52 PM, Richard Damon wrote:
>> On 5/22/22 7:16 PM, olcott wrote:
>>> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>>>
>>>> On 5/22/22 5:59 PM, olcott wrote:
>>>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>>
>>>>> So then you now agree that the nested invocation of H(P,P) would
>>>>> derive the same execution trace of its input that the the outer one
>>>>> did?
>>>>>
>>>>
>>>> Yes seem to have a confusion. Yes, the second simulator will produce
>>>> and identical trace to the first one, AS A SEPERATE TRACE.
>>>
>>>
>>> So the first invocation of H(P,P) derives this trace
>>> 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
>>>
>>> And the second (nested) invocation of H(P,P) derives this trace
>>> ...[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
>>>
>>>  From this we can see the pattern that the C/x86 function named P
>>> will never reach its last instruction, thus conclusively proving that
>>> when H(P,P) rejects its input as non-halting it is correct.
>>>
>>
>> No, because the whole second trace is conditioned on the halt deciding
>> of H. So, if the top level H aborts at that point, then the trace
>> become incorrect by being incomplete (it doesn't show the FULL Halting
>> behavior of the input, which refers to the machine it represents,
>> which did not halt there.
> The fact that I just proved that the simulated input to H(P,P) never
> reaches its own final state conclusively proves that it is non-halting
> regardless of whether or not it stops running.
>

You proved that the INCOMPLETE simulation of the input by H doesn't
reach a final state.

NOT that the CORRECT simulation of the input by a CORRECT simulator doesn't.

You just are proving that you don't know the meaning of CORRECT. (or are
just lying).

On 5/22/2022 7:21 PM, Richard Damon wrote:
> On 5/22/22 7:59 PM, olcott wrote:
>> On 5/22/2022 6:52 PM, Richard Damon wrote:
>>> On 5/22/22 7:16 PM, olcott wrote:
>>>> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>>>>
>>>>> On 5/22/22 5:59 PM, olcott wrote:
>>>>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>>>
>>>>>> So then you now agree that the nested invocation of H(P,P) would
>>>>>> derive the same execution trace of its input that the the outer
>>>>>> one did?
>>>>>>
>>>>>
>>>>> Yes seem to have a confusion. Yes, the second simulator will
>>>>> produce and identical trace to the first one, AS A SEPERATE TRACE.
>>>>
>>>>
>>>> So the first invocation of H(P,P) derives this trace
>>>> 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
>>>>
>>>> And the second (nested) invocation of H(P,P) derives this trace
>>>> ...[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
>>>>
>>>>  From this we can see the pattern that the C/x86 function named P
>>>> will never reach its last instruction, thus conclusively proving
>>>> that when H(P,P) rejects its input as non-halting it is correct.
>>>>
>>>
>>> No, because the whole second trace is conditioned on the halt
>>> deciding of H. So, if the top level H aborts at that point, then the
>>> trace become incorrect by being incomplete (it doesn't show the FULL
>>> Halting behavior of the input, which refers to the machine it
>>> represents, which did not halt there.
>> The fact that I just proved that the simulated input to H(P,P) never
>> reaches its own final state conclusively proves that it is non-halting
>> regardless of whether or not it stops running.
>>
>
> You proved that the INCOMPLETE simulation of the input by H doesn't
> reach a final state.
>

So in other words you simply are not bright enough to understand that
the emulated P can't possibly ever reach its own last instruction.
I honestly can't believe that you are that stupid.

> NOT that the CORRECT simulation of the input by a CORRECT simulator
> doesn't.
>
> You just are proving that you don't know the meaning of CORRECT. (or are
> just lying).

--
Copyright 2022 Pete Olcott

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

On 5/22/22 9:01 PM, olcott wrote:
> On 5/22/2022 7:21 PM, Richard Damon wrote:
>> On 5/22/22 7:59 PM, olcott wrote:
>>> On 5/22/2022 6:52 PM, Richard Damon wrote:
>>>> On 5/22/22 7:16 PM, olcott wrote:
>>>>> On 5/22/2022 5:34 PM, Richard Damon wrote:
>>>>>>
>>>>>> On 5/22/22 5:59 PM, olcott wrote:
>>>>>>> On 5/22/2022 4:34 PM, Richard Damon wrote:
>>>>>
>>>>>>> So then you now agree that the nested invocation of H(P,P) would
>>>>>>> derive the same execution trace of its input that the the outer
>>>>>>> one did?
>>>>>>>
>>>>>>
>>>>>> Yes seem to have a confusion. Yes, the second simulator will
>>>>>> produce and identical trace to the first one, AS A SEPERATE TRACE.
>>>>>
>>>>>
>>>>> So the first invocation of H(P,P) derives this trace
>>>>> 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
>>>>>
>>>>> And the second (nested) invocation of H(P,P) derives this trace
>>>>> ...[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
>>>>>
>>>>>  From this we can see the pattern that the C/x86 function named P
>>>>> will never reach its last instruction, thus conclusively proving
>>>>> that when H(P,P) rejects its input as non-halting it is correct.
>>>>>
>>>>
>>>> No, because the whole second trace is conditioned on the halt
>>>> deciding of H. So, if the top level H aborts at that point, then the
>>>> trace become incorrect by being incomplete (it doesn't show the FULL
>>>> Halting behavior of the input, which refers to the machine it
>>>> represents, which did not halt there.
>>> The fact that I just proved that the simulated input to H(P,P) never
>>> reaches its own final state conclusively proves that it is
>>> non-halting regardless of whether or not it stops running.
>>>
>>
>> You proved that the INCOMPLETE simulation of the input by H doesn't
>> reach a final state.
>>
>
> So in other words you simply are not bright enough to understand that
> the emulated P can't possibly ever reach its own last instruction.
> I honestly can't believe that you are that stupid.

H can not emulate the input P to the final state, yes.

But that isn't the definition of Halting.

Halting is defined by the ACTUAL MACHINE P reaching the final state, or
a CORRECT (unaborted) simulation reaching the final state.

P DOES do this if H(P,P) returns 0, so that can't be the correct answer.

YOu are just confirming your ignorance of the topic, as you keep
thinking that an aborted simulation not reaching the final state shows
the machine is non-halting.

>
>> NOT that the CORRECT simulation of the input by a CORRECT simulator
>> doesn't.
>>
>> You just are proving that you don't know the meaning of CORRECT. (or
>> are just lying).
>
>

Peter Olcott wrote:
> ... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.

On 5/23/22 7:55 AM, Python wrote:
> Peter Olcott wrote:
>> ... it is non-halting regardless of whether or not it stops running.
>
> This is quite a serious cognitive dissonance to be able to
> write down such a sentence and believe it.
>
> Cancer is not the worse illness you have, Peter.
>
>

As near as I can figure it, Peter thinks that because H stops simulating
its P input, that P never halts, because that copy never gets to the
final state that a fully run P will get to.

Yes, he confuses all sorts of things, forcing distinctions between
things that are supposed to be identical and merges things that are
supposed to be distinct.

It seems he as confinced himself of his own false ideas, and twists
reality as much as needed to try to get it to conform to what he thinks
he needs to make it fit.

He can't afford to look at rebuttals, because they show him more areas
that he needs to twist to make things conform and he is running out of
things to distort to make it happen.

This is the typical result when you takes as a core foundation something
that isn't true.

On 5/23/2022 6:55 AM, Python wrote:
> Peter Olcott wrote:
>> ... it is non-halting regardless of whether or not it stops running.
>
> This is quite a serious cognitive dissonance to be able to
> write down such a sentence and believe it.
>
> Cancer is not the worse illness you have, Peter.
>
>

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company.

--
Copyright 2022 Pete Olcott

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

On 5/23/2022 7:22 AM, Richard Damon wrote:
> On 5/23/22 7:55 AM, Python wrote:
>> Peter Olcott wrote:
>>> ... it is non-halting regardless of whether or not it stops running.
>>
>> This is quite a serious cognitive dissonance to be able to
>> write down such a sentence and believe it.
>>
>> Cancer is not the worse illness you have, Peter.
>>
>>
>
> As near as I can figure it, Peter thinks that because H stops simulating
> its P input, that P never halts, because that copy never gets to the
> final state that a fully run P will get to.
>

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
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

In other words you are simply too freaking stupid to understand that the
correctly simulated P never reaches its last instruction whether or not
this simulation is ever aborted.

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[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]

--
Copyright 2022 Pete Olcott

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

Peter Olcott, demented kook, wrote:
> On 5/23/2022 6:55 AM, Python wrote:
>> Peter Olcott wrote:
>>> ... it is non-halting regardless of whether or not it stops running.
>>
>> This is quite a serious cognitive dissonance to be able to
>> write down such a sentence and believe it.
>>
>> Cancer is not the worse illness you have, Peter.
>>
>>
>
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234)
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company.

This is not even remotely related to your absurd claims, come on Peter!

On 5/23/2022 8:58 AM, Python wrote:
> Peter Olcott, demented kook, wrote:
>> On 5/23/2022 6:55 AM, Python wrote:
>>> Peter Olcott wrote:
>>>> ... it is non-halting regardless of whether or not it stops running.
>>>
>>> This is quite a serious cognitive dissonance to be able to
>>> write down such a sentence and believe it.
>>>
>>> Cancer is not the worse illness you have, Peter.
>>>
>>>
>>
>> computation that halts … the Turing machine will halt whenever it
>> enters a final state. (Linz:1990:234)
>>
>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>> Lexington/Toronto: D. C. Heath and Company.
>
> This is not even remotely related to your absurd claims, come on Peter!
>
>

Halting DOES NOT MEAN STOPS RUNNING,
Halting means reaches its final state.

--
Copyright 2022 Pete Olcott

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

Peter Olcott, old fart, wrote:
> On 5/23/2022 8:58 AM, Python wrote:
>> Peter Olcott, demented kook, wrote:
>>> On 5/23/2022 6:55 AM, Python wrote:
>>>> Peter Olcott wrote:
>>>>> ... it is non-halting regardless of whether or not it stops running.
>>>>
>>>> This is quite a serious cognitive dissonance to be able to
>>>> write down such a sentence and believe it.
>>>>
>>>> Cancer is not the worse illness you have, Peter.
>>>>
>>>>
>>>
>>> computation that halts … the Turing machine will halt whenever it
>>> enters a final state. (Linz:1990:234)
>>>
>>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>>> Lexington/Toronto: D. C. Heath and Company.
>>
>> This is not even remotely related to your absurd claims, come on Peter!
>>
>>
>
> Halting DOES NOT MEAN STOPS RUNNING,
> Halting means reaches its final state.
>

*facepalm*

On 5/23/2022 9:43 AM, Python wrote:
> Peter Olcott, old fart, wrote:
>> On 5/23/2022 8:58 AM, Python wrote:
>>> Peter Olcott, demented kook, wrote:
>>>> On 5/23/2022 6:55 AM, Python wrote:
>>>>> Peter Olcott wrote:
>>>>>> ... it is non-halting regardless of whether or not it stops running.
>>>>>
>>>>> This is quite a serious cognitive dissonance to be able to
>>>>> write down such a sentence and believe it.
>>>>>
>>>>> Cancer is not the worse illness you have, Peter.
>>>>>
>>>>>
>>>>
>>>> computation that halts … the Turing machine will halt whenever it
>>>> enters a final state. (Linz:1990:234)
>>>>
>>>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>>>> Lexington/Toronto: D. C. Heath and Company.
>>>
>>> This is not even remotely related to your absurd claims, come on Peter!
>>>
>>>
>>
>> Halting DOES NOT MEAN STOPS RUNNING,
>> Halting means reaches its final state.
>>
>
> *facepalm*

I know that you are ignorant about this, yet at least you are not a
liar. The definition of halting does mean reaches final state, this is a
term of the art of computer science thus overrides and supersedes its
common meaning.

--
Copyright 2022 Pete Olcott

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

On 5/23/2022 5:55 AM, Python wrote:
> Peter Olcott wrote:
>> ... it is non-halting regardless of whether or not it stops running.
>
> This is quite a serious cognitive dissonance to be able to
> write down such a sentence and believe it.
>
> Cancer is not the worse illness you have, Peter.
Insightful observation. I have been curious, since he is now a
pathological liar, whether he lies about cancer too. I wouldn't be
surprised. As to cognitive disorders, I believe he his learning
disabilities compounded by the attention span of a roach. I don't
believe that any thing said here will cure any of his observed issues.

As a test, I speculate he will respond to this message, if at all, with
either many identical lines of capitol letters or yet another copy of
his faked up trace. Gentlemen, Place Your Bets!
--
Jeff Barnett

On 5/23/2022 1:39 PM, Jeff Barnett wrote:
> On 5/23/2022 5:55 AM, Python wrote:
>> Peter Olcott wrote:
>>> ... it is non-halting regardless of whether or not it stops running.
>>
>> This is quite a serious cognitive dissonance to be able to
>> write down such a sentence and believe it.
>>
>> Cancer is not the worse illness you have, Peter.
> Insightful observation. I have been curious, since he is now a
> pathological liar, whether he lies about cancer too. I wouldn't be
> surprised. As to cognitive disorders, I believe he his learning
> disabilities compounded by the attention span of a roach. I don't
> believe that any thing said here will cure any of his observed issues.
>
> As a test, I speculate he will respond to this message, if at all, with
> either many identical lines of capitol letters or yet another copy of
> his faked up trace. Gentlemen, Place Your Bets!

_P()
[00001352](01) 55 push ebp
[00001353](02) 8bec mov ebp,esp
[00001355](03) 8b4508 mov eax,[ebp+08]
[00001358](01) 50 push eax // push P
[00001359](03) 8b4d08 mov ecx,[ebp+08]
[0000135c](01) 51 push ecx // push P
[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]

People with sufficient technical competence that are not God damned
liars can easily verify that the following trace provided for the x86
emulated input to H(P,P) for one emulation and one nested emulation is
correct entirely on the basis of the above x86 source code for P.

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

--
Copyright 2022 Pete Olcott

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

On 5/23/22 9:33 AM, olcott wrote:
> On 5/23/2022 7:22 AM, Richard Damon wrote:
>> On 5/23/22 7:55 AM, Python wrote:
>>> Peter Olcott wrote:
>>>> ... it is non-halting regardless of whether or not it stops running.
>>>
>>> This is quite a serious cognitive dissonance to be able to
>>> write down such a sentence and believe it.
>>>
>>> Cancer is not the worse illness you have, Peter.
>>>
>>>
>>
>> As near as I can figure it, Peter thinks that because H stops
>> simulating its P input, that P never halts, because that copy never
>> gets to the final state that a fully run P will get to.
>>
>
> 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
> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
> In other words you are simply too freaking stupid to understand that the
> correctly simulated P never reaches its last instruction whether or not
> this simulation is ever aborted.

No, you confuse the concept of a CORRECT simulation with what H deos.

H generates an ABORTED simulation. An ABORTED SIMULATION are NEVER
correct, because the machine they are simulating continue to run after
that point.

You just don't understand that the behavior of a PROGRAM is not
dependent on the behavior of an individual execution which isn't allowed
to complete.

P(P) Halts, the CORRECT simulation of the input to H(P,P), i.e the
equivalent to UTM(P,P) does too.

THe fact that H can't do that just shows that it can't do it, and it is
just plain wrong if it claims to be a Halt Decider.

If it gives up the claim to being a Halt Decider, then it could possibly
redefine what the "behavior of its input" is, but then is can't be used
to be a counter example for the Halting Problem.

Your failure to understand just proves that you do not understand the
very basics of computations, and apparently a substantial part of your
connection with reality.

>
> _P()
> [00001352](01)  55              push ebp
> [00001353](02)  8bec            mov ebp,esp
> [00001355](03)  8b4508          mov eax,[ebp+08]
> [00001358](01)  50              push eax      // push P
> [00001359](03)  8b4d08          mov ecx,[ebp+08]
> [0000135c](01)  51              push ecx      // push P
> [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]
>
>

On 5/23/22 9:29 AM, olcott wrote:
> On 5/23/2022 6:55 AM, Python wrote:
>> Peter Olcott wrote:
>>> ... it is non-halting regardless of whether or not it stops running.
>>
>> This is quite a serious cognitive dissonance to be able to
>> write down such a sentence and believe it.
>>
>> Cancer is not the worse illness you have, Peter.
>>
>>
>
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234)
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company.
>
>

Right, and nothing you have done shows that the TURING MACHINE didn't
halt when the Turing Machine that represent your decider said it wouldnt't

All you have done with all your arguments is prove that you just don't
understand what that is saying.

A partial simulation is NOT a Turing Machine, or shows what a Turing
Machine would do.

To make your arguments, you FIRST need to define the EXACT algorithm
your H uses, and once that is fixed, it becomes clear that for whatever
algorithm you give it, it is wrong about H^/P not halting.

All your proof over "All H" does is prove that NO H, built by your
template can ever prove that this H^/P will Halt, which does NOT prove
that it is non-Halting.

You logic is FULL of inconsistencies like the above that have been
pointed out to you. Your inability to see them shows a lack of logical
processing in your brain.

On 5/23/22 10:33 AM, olcott wrote:
> On 5/23/2022 8:58 AM, Python wrote:
>> Peter Olcott, demented kook, wrote:
>>> On 5/23/2022 6:55 AM, Python wrote:
>>>> Peter Olcott wrote:
>>>>> ... it is non-halting regardless of whether or not it stops running.
>>>>
>>>> This is quite a serious cognitive dissonance to be able to
>>>> write down such a sentence and believe it.
>>>>
>>>> Cancer is not the worse illness you have, Peter.
>>>>
>>>>
>>>
>>> computation that halts … the Turing machine will halt whenever it
>>> enters a final state. (Linz:1990:234)
>>>
>>> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
>>> Lexington/Toronto: D. C. Heath and Company.
>>
>> This is not even remotely related to your absurd claims, come on Peter!
>>
>>
>
> Halting DOES NOT MEAN STOPS RUNNING,
> Halting means reaches its final state.
>
And the fact that H stopped running P doesn't mean that it is Non-Halting.