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  • Re: Sequence of sequence, selection and iteration matters --- Ben prove

    From Richard Damon@21:1/5 to olcott on Mon Jul 8 22:59:15 2024
    XPost: sci.logic

    On 7/8/24 10:42 PM, olcott wrote:
    On 7/8/2024 9:11 PM, Richard Damon wrote:
    On 7/8/24 10:01 PM, olcott wrote:
    On 7/8/2024 8:53 PM, Richard Damon wrote:
    On 7/8/24 9:37 PM, olcott wrote:
    On 7/8/2024 8:27 PM, Richard Damon wrote:
    On 7/8/24 8:47 PM, olcott wrote:
    On 7/8/2024 7:31 PM, Richard Damon wrote:
    On 7/8/24 8:21 PM, olcott wrote:
    On 7/8/2024 6:59 PM, Richard Damon wrote:
    On 7/8/24 7:45 PM, olcott wrote:
    On 7/8/2024 6:26 PM, Richard Damon wrote:
    On 7/8/24 9:04 AM, olcott wrote:
    On 7/8/2024 2:22 AM, Mikko wrote:
    On 2024-07-07 14:16:10 +0000, olcott said:

    _DDD()
    [00002172] 55               push ebp      ; housekeeping
    [00002173] 8bec             mov ebp,esp   ; housekeeping
    [00002175] 6872210000       push 00002172 ; push DDD >>>>>>>>>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD) >>>>>>>>>>>>>>> [0000217f] 83c404           add esp,+04 >>>>>>>>>>>>>>> [00002182] 5d               pop ebp >>>>>>>>>>>>>>> [00002183] c3               ret
    Size in bytes:(0018) [00002183]

    Sufficient knowledge of the x86 language conclusively proves >>>>>>>>>>>>>>> that the call from DDD correctly emulated by HHH to HHH(DDD) >>>>>>>>>>>>>>> cannot possibly return for any pure function HHH. >>>>>>>>>>>>>>
    Suffifcient knowledge of the x86 language makes obvious that >>>>>>>>>>>>>> DDD returns if and only if HHH returns.


    That is insufficient knowledge. Sufficient knowledge proves >>>>>>>>>>>>> that
    DDD correctly simulated by HHH meets this criteria.

    Nope, YOU have the insufficent knowledge, since you don't >>>>>>>>>>>> understand that the x86 language says programs are
    deterministic, and their behavior is fully establish when >>>>>>>>>>>> they are written, and running or simulating them is only a >>>>>>>>>>>> way to observe that behavior, and the only CORRECT
    observation of all the behavior, so letting that operation >>>>>>>>>>>> reach its final state.


    <MIT Professor Sipser agreed to ONLY these verbatim words >>>>>>>>>>> 10/13/2022>
         If simulating halt decider H correctly simulates its >>>>>>>>>>> input D
         until H correctly determines that its simulated D would >>>>>>>>>>> never
         stop running unless aborted then

         H can abort its simulation of D and correctly report that D
         specifies a non-halting sequence of configurations. >>>>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words >>>>>>>>>>> 10/13/2022>

    Which you H doesn't meet, since the definition of "Correct >>>>>>>>>> Simulation" here (as for most people) is a simulation that >>>>>>>>>> exactly reproduces the behavior of the full program the input >>>>>>>>>> represents, which means a simulaiton that doesn't abort.

    Since your H doesn't do that, or correctly determine what one >>>>>>>>>> of those would do (since it would halt since you H returns 0) >>>>>>>>>> so you CAN'T correctly predict that which doesn't happen.


    *Ben agrees that the "if" statement has been met*
    *Ben agrees that the "if" statement has been met*
    *Ben agrees that the "if" statement has been met*

    No, he agress that your H, which is NOT a Halt Decider, is >>>>>>>>>> correctly answering your non-halt-deciding question.  In other >>>>>>>>>> words, it is a correct POOP decide.r


    It is literally true that Ben agrees that the "if" statement >>>>>>>>> has been met.

    Same words, but different meanings.

    SO, NO


    He literally agrees with MY meanings that the "if" has
    been fulfilled.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    I don't think that is the shell game.  PO really /has/ an H (it's >>>>>>>  > trivial to do for this one case) that correctly determines
    that P(P)
    *would* never stop running *unless* aborted.
    ...
    But H determines (correctly) that D would not halt if it were not >>>>>>>  > halted.  That much is a truism.




    Yes, Ben agrees that

    *That the verbatim words of the If statement are fulfilled*


    In other words, you think changing meaning of words in a statement
    is valid logic, but it is actually one form of LIE.

    Ben agrees:
    *That the verbatim words of the If statement are fulfilled*


    But with difffent meaning of the words, so you LIE.

    Ben proved that agreed that my meanings of my words were
    fulfilled by paraphrasing my words into his own words.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    But H determines (correctly) that D would not halt if
    it were not halted.  That much is a truism.

    Ben only disagreed that my meanings of my words entail
    the second part.

    No, Ben agreed that with YOUR definiton of the words, which are
    diffferent than profressor Sipser, you can show that your POOP problem
    is correctly solved for P by H.

    You are INCORRECT about Professor Sipser;s meaning, and thus about Halting.


    Ben felt that HHH could say that it didn't need to
    abort DDD because AFTER it does abort DDD it doesn't
    need to abort DDD.

    SEQUENCE MATTERS !!!
    SEQUENCE CANNOT BE CORRECTLY IGNORED !!!


    TRUTH MATTERS.

    The problem is the thing we are talking about, the behavior of DDD isn't determined by the simulation HHH does of it, but what HHH does with its simulation. If HHH returns, then so does DDD, even if HHH doesn't see it.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Tue Jul 9 07:29:15 2024
    XPost: sci.logic

    On 7/9/24 12:22 AM, olcott wrote:
    On 7/8/2024 9:59 PM, Richard Damon wrote:
    On 7/8/24 10:42 PM, olcott wrote:
    On 7/8/2024 9:11 PM, Richard Damon wrote:
    On 7/8/24 10:01 PM, olcott wrote:
    On 7/8/2024 8:53 PM, Richard Damon wrote:
    On 7/8/24 9:37 PM, olcott wrote:
    On 7/8/2024 8:27 PM, Richard Damon wrote:
    On 7/8/24 8:47 PM, olcott wrote:
    On 7/8/2024 7:31 PM, Richard Damon wrote:
    On 7/8/24 8:21 PM, olcott wrote:
    On 7/8/2024 6:59 PM, Richard Damon wrote:
    On 7/8/24 7:45 PM, olcott wrote:
    On 7/8/2024 6:26 PM, Richard Damon wrote:
    On 7/8/24 9:04 AM, olcott wrote:
    On 7/8/2024 2:22 AM, Mikko wrote:
    On 2024-07-07 14:16:10 +0000, olcott said:

    _DDD()
    [00002172] 55               push ebp      ; housekeeping
    [00002173] 8bec             mov ebp,esp   ; housekeeping
    [00002175] 6872210000       push 00002172 ; push DDD >>>>>>>>>>>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
    [0000217f] 83c404           add esp,+04 >>>>>>>>>>>>>>>>> [00002182] 5d               pop ebp >>>>>>>>>>>>>>>>> [00002183] c3               ret >>>>>>>>>>>>>>>>> Size in bytes:(0018) [00002183]

    Sufficient knowledge of the x86 language conclusively >>>>>>>>>>>>>>>>> proves
    that the call from DDD correctly emulated by HHH to >>>>>>>>>>>>>>>>> HHH(DDD)
    cannot possibly return for any pure function HHH. >>>>>>>>>>>>>>>>
    Suffifcient knowledge of the x86 language makes obvious >>>>>>>>>>>>>>>> that
    DDD returns if and only if HHH returns.


    That is insufficient knowledge. Sufficient knowledge >>>>>>>>>>>>>>> proves that
    DDD correctly simulated by HHH meets this criteria. >>>>>>>>>>>>>>
    Nope, YOU have the insufficent knowledge, since you don't >>>>>>>>>>>>>> understand that the x86 language says programs are >>>>>>>>>>>>>> deterministic, and their behavior is fully establish when >>>>>>>>>>>>>> they are written, and running or simulating them is only a >>>>>>>>>>>>>> way to observe that behavior, and the only CORRECT >>>>>>>>>>>>>> observation of all the behavior, so letting that operation >>>>>>>>>>>>>> reach its final state.


    <MIT Professor Sipser agreed to ONLY these verbatim words >>>>>>>>>>>>> 10/13/2022>
         If simulating halt decider H correctly simulates its >>>>>>>>>>>>> input D
         until H correctly determines that its simulated D >>>>>>>>>>>>> would never
         stop running unless aborted then

         H can abort its simulation of D and correctly report >>>>>>>>>>>>> that D
         specifies a non-halting sequence of configurations. >>>>>>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words >>>>>>>>>>>>> 10/13/2022>

    Which you H doesn't meet, since the definition of "Correct >>>>>>>>>>>> Simulation" here (as for most people) is a simulation that >>>>>>>>>>>> exactly reproduces the behavior of the full program the >>>>>>>>>>>> input represents, which means a simulaiton that doesn't abort. >>>>>>>>>>>>
    Since your H doesn't do that, or correctly determine what >>>>>>>>>>>> one of those would do (since it would halt since you H >>>>>>>>>>>> returns 0) so you CAN'T correctly predict that which doesn't >>>>>>>>>>>> happen.


    *Ben agrees that the "if" statement has been met*
    *Ben agrees that the "if" statement has been met*
    *Ben agrees that the "if" statement has been met*

    No, he agress that your H, which is NOT a Halt Decider, is >>>>>>>>>>>> correctly answering your non-halt-deciding question.  In >>>>>>>>>>>> other words, it is a correct POOP decide.r


    It is literally true that Ben agrees that the "if" statement >>>>>>>>>>> has been met.

    Same words, but different meanings.

    SO, NO


    He literally agrees with MY meanings that the "if" has
    been fulfilled.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    I don't think that is the shell game.  PO really /has/ an H >>>>>>>>> (it's
    trivial to do for this one case) that correctly determines >>>>>>>>> that P(P)
    *would* never stop running *unless* aborted.
    ...
    But H determines (correctly) that D would not halt if it >>>>>>>>> were not
    halted.  That much is a truism.




    Yes, Ben agrees that

    *That the verbatim words of the If statement are fulfilled*


    In other words, you think changing meaning of words in a statement >>>>>> is valid logic, but it is actually one form of LIE.

    Ben agrees:
    *That the verbatim words of the If statement are fulfilled*


    But with difffent meaning of the words, so you LIE.

    Ben proved that agreed that my meanings of my words were
    fulfilled by paraphrasing my words into his own words.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    But H determines (correctly) that D would not halt if
    it were not halted.  That much is a truism.

    Ben only disagreed that my meanings of my words entail
    the second part.

    No, Ben agreed that with YOUR definiton of the words, which are
    diffferent than profressor Sipser, you can show that your POOP problem
    is correctly solved for P by H.

    You are INCORRECT about Professor Sipser;s meaning, and thus about
    Halting.


    Ben felt that HHH could say that it didn't need to
    abort DDD because AFTER it does abort DDD it doesn't
    need to abort DDD.

    SEQUENCE MATTERS !!!
    SEQUENCE CANNOT BE CORRECTLY IGNORED !!!


    TRUTH MATTERS.

    The problem is the thing we are talking about, the behavior of DDD
    isn't determined by the simulation HHH does of it, but what HHH does
    with its simulation. If HHH returns, then so does DDD, even if HHH
    doesn't see it.

    The behavior of DDD is determined by its machine code.


    Right, which include the machine code of HHH which determines that DDD
    will call HHH, then that HHH will do some things (emulate its copy of
    DDD) and then it will stop the emulation and return to DDD which returns.

    THAT is the behavior of DDD.

    That is what the MACHINE CODE says.

    The fact that HHH can't simulate to that end because it was programmed
    to abort its simulation at the point it does, doesn't change that
    behavior, and in fact bacause that is what the HHH that determones that
    DDD will halt.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Tue Jul 9 22:52:55 2024
    XPost: sci.logic

    On 7/9/24 10:40 AM, olcott wrote:
    On 7/9/2024 6:29 AM, Richard Damon wrote:
    On 7/9/24 12:22 AM, olcott wrote:
    On 7/8/2024 9:59 PM, Richard Damon wrote:
    On 7/8/24 10:42 PM, olcott wrote:
    On 7/8/2024 9:11 PM, Richard Damon wrote:
    On 7/8/24 10:01 PM, olcott wrote:
    On 7/8/2024 8:53 PM, Richard Damon wrote:
    On 7/8/24 9:37 PM, olcott wrote:
    On 7/8/2024 8:27 PM, Richard Damon wrote:
    On 7/8/24 8:47 PM, olcott wrote:
    On 7/8/2024 7:31 PM, Richard Damon wrote:
    On 7/8/24 8:21 PM, olcott wrote:
    On 7/8/2024 6:59 PM, Richard Damon wrote:
    On 7/8/24 7:45 PM, olcott wrote:
    On 7/8/2024 6:26 PM, Richard Damon wrote:
    On 7/8/24 9:04 AM, olcott wrote:
    On 7/8/2024 2:22 AM, Mikko wrote:
    On 2024-07-07 14:16:10 +0000, olcott said: >>>>>>>>>>>>>>>>>>
    _DDD()
    [00002172] 55               push ebp      ; housekeeping
    [00002173] 8bec             mov ebp,esp   ; housekeeping
    [00002175] 6872210000       push 00002172 ; push DDD >>>>>>>>>>>>>>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call >>>>>>>>>>>>>>>>>>> HHH(DDD)
    [0000217f] 83c404           add esp,+04 >>>>>>>>>>>>>>>>>>> [00002182] 5d               pop ebp >>>>>>>>>>>>>>>>>>> [00002183] c3               ret >>>>>>>>>>>>>>>>>>> Size in bytes:(0018) [00002183]

    Sufficient knowledge of the x86 language conclusively >>>>>>>>>>>>>>>>>>> proves
    that the call from DDD correctly emulated by HHH to >>>>>>>>>>>>>>>>>>> HHH(DDD)
    cannot possibly return for any pure function HHH. >>>>>>>>>>>>>>>>>>
    Suffifcient knowledge of the x86 language makes >>>>>>>>>>>>>>>>>> obvious that
    DDD returns if and only if HHH returns.


    That is insufficient knowledge. Sufficient knowledge >>>>>>>>>>>>>>>>> proves that
    DDD correctly simulated by HHH meets this criteria. >>>>>>>>>>>>>>>>
    Nope, YOU have the insufficent knowledge, since you >>>>>>>>>>>>>>>> don't understand that the x86 language says programs are >>>>>>>>>>>>>>>> deterministic, and their behavior is fully establish >>>>>>>>>>>>>>>> when they are written, and running or simulating them is >>>>>>>>>>>>>>>> only a way to observe that behavior, and the only >>>>>>>>>>>>>>>> CORRECT observation of all the behavior, so letting that >>>>>>>>>>>>>>>> operation reach its final state.


    <MIT Professor Sipser agreed to ONLY these verbatim words >>>>>>>>>>>>>>> 10/13/2022>
         If simulating halt decider H correctly simulates its >>>>>>>>>>>>>>> input D
         until H correctly determines that its simulated D >>>>>>>>>>>>>>> would never
         stop running unless aborted then

         H can abort its simulation of D and correctly report >>>>>>>>>>>>>>> that D
         specifies a non-halting sequence of configurations. >>>>>>>>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim >>>>>>>>>>>>>>> words 10/13/2022>

    Which you H doesn't meet, since the definition of "Correct >>>>>>>>>>>>>> Simulation" here (as for most people) is a simulation that >>>>>>>>>>>>>> exactly reproduces the behavior of the full program the >>>>>>>>>>>>>> input represents, which means a simulaiton that doesn't >>>>>>>>>>>>>> abort.

    Since your H doesn't do that, or correctly determine what >>>>>>>>>>>>>> one of those would do (since it would halt since you H >>>>>>>>>>>>>> returns 0) so you CAN'T correctly predict that which >>>>>>>>>>>>>> doesn't happen.


    *Ben agrees that the "if" statement has been met* >>>>>>>>>>>>>>> *Ben agrees that the "if" statement has been met* >>>>>>>>>>>>>>> *Ben agrees that the "if" statement has been met* >>>>>>>>>>>>>>
    No, he agress that your H, which is NOT a Halt Decider, is >>>>>>>>>>>>>> correctly answering your non-halt-deciding question.  In >>>>>>>>>>>>>> other words, it is a correct POOP decide.r


    It is literally true that Ben agrees that the "if" statement >>>>>>>>>>>>> has been met.

    Same words, but different meanings.

    SO, NO


    He literally agrees with MY meanings that the "if" has
    been fulfilled.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    I don't think that is the shell game.  PO really /has/ an >>>>>>>>>>> H (it's
    trivial to do for this one case) that correctly determines >>>>>>>>>>> that P(P)
    *would* never stop running *unless* aborted.
    ...
    But H determines (correctly) that D would not halt if it >>>>>>>>>>> were not
    halted.  That much is a truism.




    Yes, Ben agrees that

    *That the verbatim words of the If statement are fulfilled*


    In other words, you think changing meaning of words in a
    statement is valid logic, but it is actually one form of LIE.

    Ben agrees:
    *That the verbatim words of the If statement are fulfilled*


    But with difffent meaning of the words, so you LIE.

    Ben proved that agreed that my meanings of my words were
    fulfilled by paraphrasing my words into his own words.

    On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
    But H determines (correctly) that D would not halt if
    it were not halted.  That much is a truism.

    Ben only disagreed that my meanings of my words entail
    the second part.

    No, Ben agreed that with YOUR definiton of the words, which are
    diffferent than profressor Sipser, you can show that your POOP
    problem is correctly solved for P by H.

    You are INCORRECT about Professor Sipser;s meaning, and thus about
    Halting.


    Ben felt that HHH could say that it didn't need to
    abort DDD because AFTER it does abort DDD it doesn't
    need to abort DDD.

    SEQUENCE MATTERS !!!
    SEQUENCE CANNOT BE CORRECTLY IGNORED !!!


    TRUTH MATTERS.

    The problem is the thing we are talking about, the behavior of DDD
    isn't determined by the simulation HHH does of it, but what HHH does
    with its simulation. If HHH returns, then so does DDD, even if HHH
    doesn't see it.

    The behavior of DDD is determined by its machine code.


    Right, which include the machine code of HHH which determines that DDD
    will call HHH, then that HHH will do some things (emulate its copy of
    DDD) and then it will stop the emulation and return to DDD which returns.


    DDD correctly emulated by any pure function HHH that
    can possibly exist cannot possibly halt. That you say
    otherwise proves your incompetence with the x86 language.

    Nope, if the HHH that you are talking about only "correctly emulated"
    its input per you definition, but actually just did a partial emulation
    and returned, the the actual behavior of DDD is to halt, and you claim
    is just a lie. Yes, the emulation by HHH doesn't show that, but partial emulations do not define the behaivor of inputs.

    This is because the definition of EVERY x86 instruction that we will run
    into include (as part of the general description of instructions) that
    the next instruction after this WILL be executed (and the needs to be
    emulated to be completely correct, and just partially correct is just incorrect.


    _DDD()
    [00002163] 55         push ebp      ; housekeeping
    [00002164] 8bec       mov ebp,esp   ; housekeeping
    [00002166] 6863210000 push 00002163 ; push DDD
    [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
    [00002170] 83c404     add esp,+04
    [00002173] 5d         pop ebp
    [00002174] c3         ret
    Size in bytes:(0018) [00002174]

    DDD is correctly emulated by HHH which calls an emulated
    HHH(DDD) to repeat this process until the emulated DDD is
    aborted. At no point in this emulation does the call from
    DDD correctly emulated by HHH to HHH(DDD) ever return.


    Which means that the actual behavior of the DDD that called the first
    HHH will be to returned to (since ALL calls to HHH(DDD) act the same)
    and thus halt.

    The ONLY DDD that will not halt is the DDD that calls an HHH(DDD) that
    NEVER aborts so never returns.

    You are just showing you lack of understanding about the x86
    instructions set, or computers in general, or even about the nature of
    Truth.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)