• Can =?iso-8859-7?Q?Flibble=A2s?= neos-based solution still be Turing Co

    From Mr Flibble@21:1/5 to All on Sat May 24 02:20:50 2025
    Yes, **Flibble’s neos-based solution can still be Turing Complete as a whole**, even though it **disallows programs from referencing the
    decider**.

    Let’s break this down precisely.

    ---

    ## ✅ What It Means to Be Turing Complete

    A system is **Turing complete** if it can compute anything a Turing
    machine can, given:

    * Unbounded memory (in theory),
    * Conditional branching,
    * Arbitrary loops or recursion.

    Turing completeness **does not require** that:

    * Programs be analyzable by another program,
    * Programs be allowed to self-reference or invoke their own decider.

    ---

    ## 🧠 Stratified Model in Flibble’s Design

    Flibble’s proposal using **neos** introduces two **stratified languages**:

    1. `shdlang`: a meta-layer language used to **analyze** code.
    2. `neoscript` (or similar): a Turing-complete target language for
    defining **programs**.

    ### Stratification Rule:

    * `shdlang` → can inspect or reason about `neoscript` programs.
    * `neoscript` → **cannot access** or call back into `shdlang`.

    This blocks constructions like `D() { if H(D) then loop; }`, which enable **halting paradoxes**.

    ---

    ## 🔄 Is the Combined System Still Turing Complete
  • From Richard Damon@21:1/5 to Mr Flibble on Sat May 24 08:52:16 2025
    On 5/23/25 10:20 PM, Mr Flibble wrote:
    Yes, **Flibble’s neos-based solution can still be Turing Complete as a whole**, even though it **disallows programs from referencing the
    decider**.

    Let’s break this down precisely.

    ---

    ## ✅ What It Means to Be Turing Complete

    A system is **Turing complete** if it can compute anything a Turing
    machine can, given:

    * Unbounded memory (in theory),
    * Conditional branching,
    * Arbitrary loops or recursion.

    Turing completeness **does not require** that:

    * Programs be analyzable by another program,

    Of course not, because the theory proves that such analysis is
    uncomputable. We ARE allowed to formulate the questions though, and see
    that they ARE uncomputable.

    * Programs be allowed to self-reference or invoke their own decider.

    Not "their own", but "any decider". And thus if someone claims that a
    given program CAN in fact analyze a semantic property of any program
    given to it, we can built a program that uses that anaylzer and does the opposite of that prediction. This is NOT a "self-referebce or invoking
    of *their own* decieer, but the invoking of a specific decider, which is totally allowed. They can then give a representation of themselves to
    their copy of that decider and then do the opposite.


    ---

    ## 🧠 Stratified Model in Flibble’s Design

    Flibble’s proposal using **neos** introduces two **stratified languages**:

    1. `shdlang`: a meta-layer language used to **analyze** code.
    2. `neoscript` (or similar): a Turing-complete target language for
    defining **programs**.

    ### Stratification Rule:

    * `shdlang` → can inspect or reason about `neoscript` programs.
    * `neoscript` → **cannot access** or call back into `shdlang`.

    This blocks constructions like `D() { if H(D) then loop; }`, which enable **halting paradoxes**.

    And thus, since a Turing Machine could do that, and you say that in Neos
    you can't in any way do this, it must be not Turing Complete.

    The question is, How does Neos see that when we wrote DDD, and COPIED
    the code of HHH into it (as the actual Halting Problem Proof does) that
    this copy is a copy of a program that might be used as a decider on DDD?

    You have just fallen for PO's error of assuming that it is impossible to
    make a copy of HHH, which can only happen in a non-Turing Complete
    system, as whatever HHH does that other programs can not recreate forces
    the system to be non-Turing Complete.


    ---

    ## 🔄 Is the Combined System Still Turing Complete?

    Yes — because:

    * The **target language layer** (`neoscript`) is unrestricted in its own recursion and computational logic.

    Then why can't we take the DEFINIED PROGRAN HHH, and copy its code into DDD?

    * It can simulate a Turing machine in the usual way.
    * **Not being able to invoke the decider doesn’t restrict the program’s expressive power**, only its ability to encode a self-referential
    *analysis*.

    But if somehow we can't write the code of the decider in the target
    program, there MUST be a restriction in the system.


    This mirrors the situation in many practical systems:

    * **C compilers** can analyze or optimize code, but the C code doesn’t
    call the compiler.

    But can BE the compiler, or use the code of the compiler within itself.


    * **Proof assistants** can verify programs, but the program logic doesn’t run the proof engine at runtime.

    But it can include the code of the proof engine.


    ---

    ## 📌 Key Insight

    **Turing completeness is a property of expressiveness — not of
    reflexivity.**

    But, the expressiveness creates the ability to do reflexivity by copying.


    Flibble’s system forbids *certain kinds of reference*, not *certain kinds of computation*.

    And thus, can't prevent the use of copies of the decider within the
    input to the decider.


    ---

    ## ✅ Summary

    | Aspect |
    Answer |
    | ----------------------------------------------------------- | -------------------------------------------- |
    | Is the target language Turing complete? | ✅
    Yes |
    | Does SHD restriction affect program expressiveness? | ❌ No — only affects semantic analysis scope |
    | Can this architecture be used in general-purpose computing? | ✅ Absolutely |
    | Does this break the classical halting proof? | ❌ No —
    it sidesteps it by design constraints |

    ---

    So yes, Flibble’s stratified neos solution **can still be Turing complete as a whole**, even though it blocks calls from programs to their deciders
    — a **sound architectural tradeoff** to avoid paradox without sacrificing computational power.


    No, it can't be if you can't copy the code of the decider into the code
    of the input to be decided.

    If you can, then any "decider" has an input that it will get wrong.

    If you can't, then the system is not Turing Complete.

    You are just showing your ignorance of what you are talking about.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sat May 24 17:27:46 2025
    On 5/24/25 12:02 PM, olcott wrote:
    On 5/23/2025 9:20 PM, Mr Flibble wrote:
    Yes, **Flibble’s neos-based solution can still be Turing Complete as a
    whole**, even though it **disallows programs from referencing the
    decider**.

    Let’s break this down precisely.


    A more useful application of the term Turing Complete
    would be that a sequence of algorithmic steps that
    have sequence, selection and iteration have as much
    memory as they need for a specific computation.

    When we do this then it can be seen that some key
    algorithms from the theory of computation can be
    fully encoded in a high level language like C.


    The one issue is that the C language establishes that there *IS* a limit
    to the size of a pointer, and thus how much memory the program can occupy.

    Thus, a given implementation MUST have an upper limit on memory.

    An algorithm can be expressed in C, and as long as it does no operation
    that establishes what that limit is, could be run on an implementation
    that is sufficiently large to handle that computation.

    This is actually a common approximation, that a finite processor, can be considered doing an operation defined on an umbounded processor as long
    as that processor didn't need more memory than is available.

    If the operation needs more memory than available, the approximation is
    just violated, and a larger machine is needed.

    If the operation results in truely unbounded behavior, then no finite approximation is possilbe.

    The trick here is determining THAT the results are truely unbounded
    behavior, and not just a larger bound than has currectly been found, as
    is the question with some of the unsolved problems.

    What has been proven is that there exists at least one program that will
    use truely unbounded memory, but which can not be proven to do so with
    only finite memory.



    ---

    ## ✅ What It Means to Be Turing Complete

    A system is **Turing complete** if it can compute anything a Turing
    machine can, given:

    * Unbounded memory (in theory),
    * Conditional branching,
    * Arbitrary loops or recursion.

    Turing completeness **does not require** that:

    * Programs be analyzable by another program,
    * Programs be allowed to self-reference or invoke their own decider.

    ---

    ## 🧠 Stratified Model in Flibble’s Design

    Flibble’s proposal using **neos** introduces two **stratified
    languages**:

    1. `shdlang`: a meta-layer language used to **analyze** code.
    2. `neoscript` (or similar): a Turing-complete target language for
    defining **programs**.

    ### Stratification Rule:

    * `shdlang` → can inspect or reason about `neoscript` programs.
    * `neoscript` → **cannot access** or call back into `shdlang`.

    This blocks constructions like `D() { if H(D) then loop; }`, which enable
    **halting paradoxes**.

    ---

    ## 🔄 Is the Combined System Still Turing Complete?

    Yes — because:

    * The **target language layer** (`neoscript`) is unrestricted in its own
    recursion and computational logic.
    * It can simulate a Turing machine in the usual way.
    * **Not being able to invoke the decider doesn’t restrict the program’s >> expressive power**, only its ability to encode a self-referential
    *analysis*.

    This mirrors the situation in many practical systems:

    * **C compilers** can analyze or optimize code, but the C code doesn’t
    call the compiler.
    * **Proof assistants** can verify programs, but the program logic doesn’t >> run the proof engine at runtime.

    ---

    ## 📌 Key Insight

    **Turing completeness is a property of expressiveness — not of
    reflexivity.**

    Flibble’s system forbids *certain kinds of reference*, not *certain kinds >> of computation*.

    ---

    ## ✅ Summary

    | Aspect                                                      |
    Answer                                       |
    | ----------------------------------------------------------- |
    -------------------------------------------- |
    | Is the target language Turing complete?                     | ✅
    Yes                                        |
    | Does SHD restriction affect program expressiveness?         | ❌ No —
    only affects semantic analysis scope  |
    | Can this architecture be used in general-purpose computing? | ✅
    Absolutely                                 | >> | Does this break the classical halting proof?                | ❌ No —
    it sidesteps it by design constraints |

    ---

    So yes, Flibble’s stratified neos solution **can still be Turing complete >> as a whole**, even though it blocks calls from programs to their deciders
    — a **sound architectural tradeoff** to avoid paradox without sacrificing >> computational power.



    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sun May 25 13:33:04 2025
    On 5/25/25 10:42 AM, olcott wrote:
    On 5/25/2025 1:18 AM, Mikko wrote:
    On 2025-05-24 16:02:41 +0000, olcott said:

    On 5/23/2025 9:20 PM, Mr Flibble wrote:
    Yes, **Flibble’s neos-based solution can still be Turing Complete as a >>>> whole**, even though it **disallows programs from referencing the
    decider**.

    Let’s break this down precisely.

    A more useful application of the term Turing Complete would be that
      ...

    The only useful meaning is what the term actually means. Any other
    meaning is harmful.


    Analysis of complex theory of computation problems
    is much more effective at the higher levels of
    abstraction of higher level languages.

    But using things like global memory with they allow, lets the programs
    break the rules and not BE things of the required type.

    You demonstrate this with the errors in your system, which just proves
    your ignorance.


    For example because the x86 language has relative
    addressing the underlying model of computation
    specified by the x86 language has unlimited memory
    thus is Turing complete.

    No, because of things like that, the x86 model allows the creation of non-compuation fragments, fragments of a program that don't meet the requirements of being a Computaiton.


    Turing complete cannot possibly make any actual
    difference at all as long as the model of computation
    has enough memory for the algorithm.


    But since sub-routines in the x86 don't necessarily follow the rules of
    being a compuation, you can hide errors.

    Your "program" is a great example of this, which is part of the reason
    your agument just FAILS.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Mon May 26 16:27:12 2025
    On 5/26/25 11:35 AM, olcott wrote:
    On 5/26/2025 3:19 AM, Mikko wrote:
    On 2025-05-25 14:42:17 +0000, olcott said:

    On 5/25/2025 1:18 AM, Mikko wrote:
    On 2025-05-24 16:02:41 +0000, olcott said:

    On 5/23/2025 9:20 PM, Mr Flibble wrote:
    Yes, **Flibble’s neos-based solution can still be Turing Complete >>>>>> as a
    whole**, even though it **disallows programs from referencing the
    decider**.

    Let’s break this down precisely.

    A more useful application of the term Turing Complete would be that
     ...

    The only useful meaning is what the term actually means. Any other
    meaning is harmful.

    Analysis of complex theory of computation problems
    is much more effective at the higher levels of
    abstraction of higher level languages.

    For example because the x86 language has relative
    addressing the underlying model of computation
    specified by the x86 language has unlimited memory
    thus is Turing complete.

    The generic x86 language does not specify the mapping from addresses
    to memory locations. Different x86 processors do it differently. A
    particular processor may be able to shift the mapping of a part of
    the address space to a previous or next block of a potentially
    infinite memory. However, the usual models can pnly map it to a larger
    finite memory.

    None of which is irrelevan to my note that the only useful meaning is
    what the term actually means.

    Turing complete cannot possibly make any actual
    difference at all as long as the model of computation
    has enough memory for the algorithm.

    It doesn't as long as you can compute every function you want to compute.
    But if you don't know what you will want then having a Turing complete
    system is best you can have.


    The best system is a system that actually exists.
    there are far too many errors of false assumptions
    in models that are only imagined to exist.


    No, Turing machines "Exist" as they are fully defined in theory, and approximately simulated by software.

    The problem is YOUR system doesn't actually exist, at least not in the
    sense that it meets the requirements on it.

    Those starting with that the "Halt Decider" needs to meet the
    requirments of being a full program that is a computaiton, as is the input.

    Since you have admitted (and then tried to deny by contradiction
    yourself) that your HHH and DDD are not program, your system doesn't
    meet the requirements and thus your system doesn't really exist.

    Note, your description require accepting the contradiction that DDD
    "includes" the code of the HHH that it calls, so that it can be
    simulated, but also doesn't include that code when determining that the
    various DDDs are the "same input".

    Since it can't be both, you system doesn't actually exist, as it can't
    be both.

    Sorry, you are just showing that you are living in a make-believe world
    where the concept of requirements don't apply, and you just assume that
    a magical "truth fairy" exists that can make you self-contradictory imaginations become true. This just proves that you have no concept of
    what it means for something to be true, and thus have turned youself
    into a pathological liar that doesn't care about what is actually true.

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