• PL Olcott
    626
    Here's where we are up to: can you explain how you reject diagonalisation for Gödel but not for Cantor? Or do you reject Cantor's argument, too?Banno

    I am just saying that when diagonalisation is evaluating {epistemological antinomies}
    it always makes sure to ignore all of the details. It never even notices that they are
    {epistemological antinomies}. Diagonalisation only looks for a 1 or a 0 on the
    diagonal and makes sure to ignore absolutely everything else.
  • Banno
    25.2k
    I can't work out what that means.

    So {epistemological antinomies} (why the curly brackets?) are, for example, the liar. Where is there an example of the Liar being used in a diagonalization? What might that look like? OR do you mean something else?
  • PL Olcott
    626
    So an {epistemological antinomies} (why the curly brackets?) is, for example, the liar. Where is there an example of the Liar being used in a diagonalization? What might that look like? OR do you mean something else?Banno

    We need to go back to more basic things.
    Do you understand why this question has no correct answer?
    The question is: >>>Is this sentence true: "This sentence is not true." ???<<<

    If that is too difficult then do you understand why this is not true or false?
    "This sentence is not true."
  • Banno
    25.2k
    Do you understand why this question has no correct answer?
    The question is: >>>Is this sentence true: "This sentence is not true." ???<<<
    PL Olcott

    But that's not right - you've been given several correct answers.
  • PL Olcott
    626
    But that's not right - you've been given several correct answers.Banno

    In other words you do not understand that it is an incorrect question.
    We go back one more step.

    Do you understand why this is not true or false?
    "This sentence is not true."
  • Banno
    25.2k
    In other words you do not understand that it is an incorrect question.PL Olcott

    Well, no. I’m pointing out that you only have a problem here if you restrict yourself to yes/no with no revision.

    Go on. But be brief. You seem to be repeating yourself yet again.
  • PL Olcott
    626
    Well, no. I’m pointing out that you only have a problem here if you restrict yourself to yes/no with no revision.Banno

    So in other words you think that a Turing machine halt decider might
    reply: "I don't know let me think about it?"

    The thought experiment must stipulate that any answer besides
    [yes, no] is a wrong answer to preserve the mathematical mapping
    from Carol's question to a Turing machine halt decider.

    The question was written by a PhD computer science professor to
    make the computer science of the halting problem easier to understand.
  • PL Olcott
    626
    PL Olcott. ok. Next.Banno

    Ah so you are totally convinced that Carol's question
    posed to Carol is a self-contradictory thus incorrect question?
  • Banno
    25.2k
    I think you are wasting my time.
  • PL Olcott
    626
    ↪PL Olcott I think you are wasting my time.Banno

    Although this forum has by far the very best people of any forum
    that I have ever participated in I still have to put checks and balances
    into the dialogue to prevent what always otherwise turns out to be
    infinite digression.

    If you agree we can go on to the next point if you don't agree
    then you lack the mandatory prerequisites required for the next point.
  • PL Olcott
    626
    PL Olcott. ok. Next.
    — Banno
    PL Olcott

    Self Referential Undecidability Construed as Incorrect Questions
    https://philpapers.org/archive/OLCSRU.pdf

    Has been reviewed by Professor Hehner and clarifications have been
    made corresponding to his review.
  • Banno
    25.2k
    You've moved back from Gödel to the halting problem. Ok.

    So check this out:
    The Halting Problem is:

    INPUT: A string P and a string I. We will think of P as a program.

    OUTPUT: 1, if P halts on I, and 0 if P goes into an infinite loop on I.

    Theorem (Turing circa 1940): There is no program to solve the Halting Problem.

    Proof: Assume to reach a contradiction that there exists a program Halt(P, I) that solves the halting problem, Halt(P, I) returns True if and only P halts on I. The given this program for the Halting Problem, we could construct the following string/code Z:

    Program (String x)
    If Halt(x, x) then
    Loop Forever
    Else Halt.
    End.

    Consider what happens when the program Z is run with input Z
    Case 1: Program Z halts on input Z. Hence, by the correctness of the Halt program, Halt returns true on input Z, Z. Hence, program Z loops forever on input Z. Contradiction.

    Case 1: Program Z loops forever on input Z. Hence, by the correctness of the Halt program, Halt returns false on input Z, Z. Hence, program Z halts on input Z. Contradiction.

    End Proof.
    Prof Kirk Pruhs

    This is a reductio argument:
    • Assume there is a program Halt
    • Show that Halt leads to a contradiction
    • Conclude that there can be no program such as Halt

    First and most obvious question is where in this the thing you called the "isomorphism from Carol's question to the halting problem proof counter- example template" is located. It's not there. But we can add it: "Will Program Z loop forever if fed itself as input?"

    Will Program Z loop forever if fed itself as input? The argument shows that we can't have an answer to that question. But that's exactly the point that shows that a program such as Halt cannot be written.

    So sure, "the inability of a halt decider to correctly provide the halt status of an input that does the opposite of whatever halt status is provided does not place any actual limit on computation." But the impossibility of writing the program Halt does.

    The argument is not that Z is impossible, but that H is.
  • PL Olcott
    626


    By skimming the paper to contrive some excuse for rebuttal you missed this:

    The bottom line of all of the above reasoning is that it is agreed that the halt status of some inputs to some halt deciders cannot possibly be correctly determined when the halt decider is required to report on the behavior of the direct execution of this input.

    The brand new insight by the PhD computer science professor and myself (since 2004) is that the inability of a halt decider to correctly provide the halt status of an input that does the opposite of whatever halt status is provided does not place any actual limit on computation.

    It is generally the case that the inability to do the logically impossible never places any actual limits on anyone of anything. That no CAD system can possibly correctly draw a square circle places no actual limits on computation.
  • PL Olcott
    626
    First and most obvious question is where in this the thing you called the "isomorphism from Carol's question to the halting problem proof counter- example template" is located. It's not there. But we can add it: "Will Program Z loop forever if fed itself as input?"Banno

    That Carol's question contradicts every yes/no answer that Carol can provide <is>
    isomorphic to input D to decider H that does that opposite of whatever Boolean value that H returns.
  • PL Olcott
    626
    So sure, "the inability of a halt decider to correctly provide the halt status of an input that does the opposite of whatever halt status is provided does not place any actual limit on computation." But the impossibility of writing the program Halt does.Banno

    It is equally a logical impossible for any CAD system to correctly draw a square circle.
    The inability to do the logically impossible never places any actual limits on anyone of anything.
  • Banno
    25.2k
    Have a think on it again. You have shown that Z is problematic. Sure, it is. That's what shows that H is impossible.
  • PL Olcott
    626
    Have a think on it again. You have shown that Z is problematic. Sure, it is. That's what shows that H is impossible.Banno

    I have had a think in this for thousands of hours since 2004 when someone
    else directly presented me with nearly the exact same question.

    It is equally a logical impossible for any CAD system to correctly draw a square circle. The inability to do the logically impossible never places any actual limits on anyone of anything.PL Olcott
  • Banno
    25.2k
    Your last few replies do not seem to be addressed to my point.

    Your question occurs with Z, not with H. Z is problematic, but Z is also a consequence of H, hence H is problematic.
  • PL Olcott
    626
    Your question occurs in Z, not in H. Z is problematic, but Z is also a consequence of H, hence H is problematic.Banno

    The whole halting problem proof depends on some input D that does the
    opposite of whatever Boolean value that H returns. Changing the names
    does not change this. When you change the names I ignore them.

    This is the whole point of my and Hehner's proof
    That it is a logical impossibility for H to return a value corresponding
    to the behavior of the direct execution of D(D) does not in any way
    limit computation because the inability to do the logically impossible
    is never any actual limit to anyone or anything.
  • Banno
    25.2k


    "Will Program Z loop forever if fed itself as input?"Banno

    IS this the equivalent of Carol's question? If not, what is?
  • PL Olcott
    626
    "Will Program Z loop forever if fed itself as input?"
    — Banno
    Banno

    I don't know and I don't care. Changing the subject is no form of rebuttal.

    // The following is written in C
    //
    01 typedef int (*ptr)(); // pointer to int function
    02 int H(ptr x, ptr y) // uses x86 emulator to simulate its input
    03
    04 int D(ptr x)
    05 {
    06   int Halt_Status = H(x, x);
    07   if (Halt_Status)
    08     HERE: goto HERE;
    09   return Halt_Status;
    10 }
    
  • Banno
    25.2k
    Your claim is that some equivalent of Carol's question occurs in the halting program proof. It's not unreasonable to ask you to show where it occurs.
  • PL Olcott
    626
    ↪PL Olcott Your claim is that some equivalent of Carol's question occurs in the halting program proof. It's not unreasonable to ask you to show where it occurs.Banno

    Already answered and you simply ignored.

    That Carol's question contradicts every yes/no answer that Carol can provide <is> isomorphic to input D to decider H that does that opposite of whatever Boolean value that H returns.PL Olcott
  • Banno
    25.2k
    This is pointless.
    That Carol's question contradicts every yes/no answer that Carol can provide <is> isomorphic to input D to decider H that does that opposite of whatever Boolean value that H returns.PL Olcott

    SO, Z?

    It shouldn't be this hard. I'm just checking that I've understood your point.
  • PL Olcott
    626
    SO, Z?

    It shouldn't be this hard. I'm just checking that I've understood your point.
    Banno

    The best that I can tell is that Z is an incorrect sloppy mess that has no actual
    name and no return value. Do you not know C?

    On the other hand D and H have been fully operational code (for two years now) of the x86utm operating system that I created.
  • Banno
    25.2k
    All I'm asking is where Carol's question occurs.
    Sure, show me in C.
  • PL Olcott
    626
    ↪PL Olcott All I'm asking is where Carol's question occurs.
    Sure, show me in C.
    Banno

    Isomorphic means <same kind of thing>.
    Carol's question for Carol and input D
    to decider H are the exact <same kind of thing>.

    In both cases their question contradicts every answer.
  • Banno
    25.2k
    I give up.

    My conclusion is that you're unable to present your thesis in a manner that is sufficiently clear to be evaluated.
  • PL Olcott
    626
    My conclusion is that you're unable to present your thesis in a manner that is sufficiently clear to be evaluated.Banno

    If that was true then professor Hehner would not have totally agreed with me today.

    I am thinking the the problem is that I assumed you had more technical knowledge
    than you do. Tell me exactly how much you know about computer programming
    and I can change my words to fit your knowledge level.

    I thought the my original version of a halt decider that simply
    tries to say what another program will do when this other program
    does the opposite of whatever it says is as clear as I can get.


    That is the <same kind of thing> as Carol's question for Carol.
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