Recursive definitions are fine, infinitely recursive definitions (such
as The Halting Problem) are INVALID.

RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.

/Flibble

Mr Flibble <flibble@reddwarf.jmc> writes:

> Recursive definitions are fine, infinitely recursive definitions (such
> as The Halting Problem) are INVALID.

(1) The halting problem is not an infinitely recursive definition.

(2) Infinitely recursive definitions are often fine. For example, the
list of Fibonacci numbers:

fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

or the list of factorials:

facts = prod 1 1 where prod p n = p : prod (p*n) (n+1)

--
Ben.

On 5/1/2022 9:39 AM, Ben wrote:
> Mr Flibble <flibble@reddwarf.jmc> writes:
>
>> Recursive definitions are fine, infinitely recursive definitions (such
>> as The Halting Problem) are INVALID.
>
> (1) The halting problem is not an infinitely recursive definition.
>
> (2) Infinitely recursive definitions are often fine. For example, the
> list of Fibonacci numbers:
>

This type is never fine: // Adapted from Clocksin & Mellish
foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))

Because Gödel says
14 Every epistemological antinomy can likewise be used for a similar
undecidability proof

G ↔ ¬Provable(F, G) is an epistemological antinomy therefore it is
necessarily sufficiently equivalent to his G.

Likewise with this one: LP ↔ ¬True(LP)
It can be evaluated as semantically incorrect without the need to define
True(). No matter how True() is defined LP ↔ ¬True(LP) is semantically
incorrect because it specifies this:

LP ↔ ¬True(¬True(¬True(¬True(¬True(¬True(¬True(¬True(LP))))))))
and Prolog can detect and reject this with unify_with_occurs_check.

> fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
>
> or the list of factorials:
>
> facts = prod 1 1 where prod p n = p : prod (p*n) (n+1)
>

--
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 Sun, 01 May 2022 15:39:35 +0100
Ben <ben.usenet@bsb.me.uk> wrote:

> Mr Flibble <flibble@reddwarf.jmc> writes:
>
> > Recursive definitions are fine, infinitely recursive definitions
> > (such as The Halting Problem) are INVALID.
>
> (1) The halting problem is not an infinitely recursive definition.
>
> (2) Infinitely recursive definitions are often fine. For example, the
> list of Fibonacci numbers:
>
> fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

FOR FUCKS SAKE, HOW STUPID ARE YOU? THAT IS A RECURSIVE DEFINITION NOT
AN *INFINITELY* RECURSIVE DEFINITION AS IT TERMINATES.

/Flibble

On 5/1/22 8:37 AM, Mr Flibble wrote:
> Recursive definitions are fine, infinitely recursive definitions (such
> as The Halting Problem) are INVALID.
>
> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>
> /Flibble
>

And the Halting Problem isn't recursive at all, so it can't be infinity
recursive. NOTHING in the definition refers to itself.

The counter example is, in a way, "recursive", but that recursion can
only become infinite if the proposed Halt Decider turns out to fail to
be a Halt Decider.

Otherwise, it is no more infinitely recursive then the recursive
definition of factorial.

On 5/1/22 12:05 PM, polcott wrote:
> On 5/1/2022 9:39 AM, Ben wrote:
>> Mr Flibble <flibble@reddwarf.jmc> writes:
>>
>>> Recursive definitions are fine, infinitely recursive definitions (such
>>> as The Halting Problem) are INVALID.
>>
>> (1) The halting problem is not an infinitely recursive definition.
>>
>> (2) Infinitely recursive definitions are often fine.  For example, the
>> list of Fibonacci numbers:
>>
>
> This type is never fine: // Adapted from Clocksin & Mellish
> foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(foo(...))))))))))))

Depends on the definition of foo.

If foo is the function "fact" I have presented, then the unwinding
becomes exactly that to a too dumb naive expansion.

Thus "never" is incorrect.

>
> Because Gödel says
> 14 Every epistemological antinomy can likewise be used for a similar
> undecidability proof
>
> G ↔ ¬Provable(F, G) is an epistemological antinomy therefore it is
> necessarily sufficiently equivalent to his G.
>
> Likewise with this one: LP ↔ ¬True(LP)
> It can be evaluated as semantically incorrect without the need to define
> True(). No matter how True() is defined LP ↔ ¬True(LP) is semantically
> incorrect because it specifies this:
>
> LP ↔ ¬True(¬True(¬True(¬True(¬True(¬True(¬True(¬True(LP))))))))
> and Prolog can detect and reject this with unify_with_occurs_check.

Because Prolog is too limited to be able to fully handle this sort of
recursion. Prolog apparently can't handle the recursive definition of
fact(n), which just proves that its rejection of something as
"infinitely recursive" is NOT "Proof" that it is invalid.

>
>
>>    fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
>>
>> or the list of factorials:
>>
>>     facts = prod 1 1 where prod p n = p : prod (p*n) (n+1)
>>
>
>

On 5/1/2022 12:08 PM, Richard Damon wrote:
> On 5/1/22 8:37 AM, Mr Flibble wrote:
>> Recursive definitions are fine, infinitely recursive definitions (such
>> as The Halting Problem) are INVALID.
>>
>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>
>> /Flibble
>>
>
> And the Halting Problem isn't recursive at all, so it can't be infinity
> recursive. NOTHING in the definition refers to itself.
>

In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and an
input, whether the program will finish running, or continue to run forever.

For any program H that might determine if programs halt, a
"pathological" program P, called with some input, can pass its own
source and its input to H and then specifically do the opposite of what
H predicts P will do.

void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
return;
}

https://en.wikipedia.org/wiki/Halting_problem

When H(P,P) is invoked H refers to itself embedded within P.
I have spent decades on the generic notion of pathological self-reference.

> The counter example is, in a way, "recursive", but that recursion can
> only become infinite if the proposed Halt Decider turns out to fail to
> be a Halt Decider.
>
> Otherwise, it is no more infinitely recursive then the recursive
> definition of factorial.

--
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 2022-05-01 11:08, Richard Damon wrote:
> On 5/1/22 8:37 AM, Mr Flibble wrote:
>> Recursive definitions are fine, infinitely recursive definitions (such
>> as The Halting Problem) are INVALID.
>>
>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>
>> /Flibble
>>
>
> And the Halting Problem isn't recursive at all, so it can't be infinity
> recursive. NOTHING in the definition refers to itself.
>
> The counter example is, in a way, "recursive", but that recursion can
> only become infinite if the proposed Halt Decider turns out to fail to
> be a Halt Decider.

Actually, even the counterexample is not recursive. It only becomes
recursive (or recursive-like) if one assumes that H bases its answer on
the simulation of its input. And the proof certainly does not require
that to be the case.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/1/22 1:23 PM, olcott wrote:
> On 5/1/2022 12:08 PM, Richard Damon wrote:
>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>> Recursive definitions are fine, infinitely recursive definitions (such
>>> as The Halting Problem) are INVALID.
>>>
>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>
>>> /Flibble
>>>
>>
>> And the Halting Problem isn't recursive at all, so it can't be
>> infinity recursive. NOTHING in the definition refers to itself.
>>
>
> In computability theory, the halting problem is the problem of
> determining, from a description of an arbitrary computer program and an
> input, whether the program will finish running, or continue to run forever.

And that ends the definition of the Halting Problem, which is NOT recursive.

>
> For any program H that might determine if programs halt, a
> "pathological" program P, called with some input, can pass its own
> source and its input to H and then specifically do the opposite of what
> H predicts P will do.
>
> void P(u32 x)
> {
>   if (H(x, x))
>     HERE: goto HERE;
>   return;
> }
>
> https://en.wikipedia.org/wiki/Halting_problem
>
> When H(P,P) is invoked H refers to itself embedded within P.
> I have spent decades on the generic notion of pathological self-reference.

And yes, the counter example is recursive, but NOT infinitely so if H
meets the requirements of being a decider.

If H doesn't meet the requirements of being a decider, and thus ALWAYS
answering, and doesn't answer for H(P,P), then it isn't a counter
example, as if fails to meet the requirements.

This is just like fact(n) is NOT infinitely recursive, for any finite n.

>
>> The counter example is, in a way, "recursive", but that recursion can
>> only become infinite if the proposed Halt Decider turns out to fail to
>> be a Halt Decider.
>>
>> Otherwise, it is no more infinitely recursive then the recursive
>> definition of factorial.
>
>

On 5/1/22 1:30 PM, André G. Isaak wrote:
> On 2022-05-01 11:08, Richard Damon wrote:
>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>> Recursive definitions are fine, infinitely recursive definitions (such
>>> as The Halting Problem) are INVALID.
>>>
>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>
>>> /Flibble
>>>
>>
>> And the Halting Problem isn't recursive at all, so it can't be
>> infinity recursive. NOTHING in the definition refers to itself.
>>
>> The counter example is, in a way, "recursive", but that recursion can
>> only become infinite if the proposed Halt Decider turns out to fail to
>> be a Halt Decider.
>
> Actually, even the counterexample is not recursive. It only becomes
> recursive (or recursive-like) if one assumes that H bases its answer on
> the simulation of its input. And the proof certainly does not require
> that to be the case.
>
> André
>

True, which is why I quoted "recursive", it is merely referential. And
even with simulation it sort of isn't recursive, since we are supposed
to be using different copies of the function, so the recursion is only
on a "meta" level that knows the input is based on the decider, but not
at the actual physical computation layer, since we are just invoking a
long series of machines with the same descriptions, since Turing
Machines are limited in the ways they can handle recursion, needing to
use their tape to implement it.

On 5/1/2022 12:30 PM, André G. Isaak wrote:
> On 2022-05-01 11:08, Richard Damon wrote:
>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>> Recursive definitions are fine, infinitely recursive definitions (such
>>> as The Halting Problem) are INVALID.
>>>
>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>
>>> /Flibble
>>>
>>
>> And the Halting Problem isn't recursive at all, so it can't be
>> infinity recursive. NOTHING in the definition refers to itself.
>>
>> The counter example is, in a way, "recursive", but that recursion can
>> only become infinite if the proposed Halt Decider turns out to fail to
>> be a Halt Decider.
>
> Actually, even the counterexample is not recursive. It only becomes
> recursive (or recursive-like) if one assumes that H bases its answer on
> the simulation of its input. And the proof certainly does not require
> that to be the case.
>
> André
>

Since the definition of H is wide open and can be anything at all that
meets the spec, if any of these definitions make the "impossible" input
decidable then this refutes the proofs.

--
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 2022-05-01 12:33, olcott wrote:
> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>> On 2022-05-01 11:08, Richard Damon wrote:
>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>> Recursive definitions are fine, infinitely recursive definitions (such
>>>> as The Halting Problem) are INVALID.
>>>>
>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>
>>>> /Flibble
>>>>
>>>
>>> And the Halting Problem isn't recursive at all, so it can't be
>>> infinity recursive. NOTHING in the definition refers to itself.
>>>
>>> The counter example is, in a way, "recursive", but that recursion can
>>> only become infinite if the proposed Halt Decider turns out to fail
>>> to be a Halt Decider.
>>
>> Actually, even the counterexample is not recursive. It only becomes
>> recursive (or recursive-like) if one assumes that H bases its answer
>> on the simulation of its input. And the proof certainly does not
>> require that to be the case.
>>
>> André
>>
>
> Since the definition of H is wide open and can be anything at all that
> meets the spec, if any of these definitions make the "impossible" input
> decidable then this refutes the proofs.

But the spec is as follows:

Hq0 <M> w ⊢* Hqy iff M applied to w halts
⊢* Hqn otherwise

Yours fails to meet this spec.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/1/2022 1:43 PM, André G. Isaak wrote:
> On 2022-05-01 12:33, olcott wrote:
>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>> Recursive definitions are fine, infinitely recursive definitions (such
>>>>> as The Halting Problem) are INVALID.
>>>>>
>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>
>>>> The counter example is, in a way, "recursive", but that recursion
>>>> can only become infinite if the proposed Halt Decider turns out to
>>>> fail to be a Halt Decider.
>>>
>>> Actually, even the counterexample is not recursive. It only becomes
>>> recursive (or recursive-like) if one assumes that H bases its answer
>>> on the simulation of its input. And the proof certainly does not
>>> require that to be the case.
>>>
>>> André
>>>
>>
>> Since the definition of H is wide open and can be anything at all that
>> meets the spec, if any of these definitions make the "impossible"
>> input decidable then this refutes the proofs.
>
> But the spec is as follows:
>
> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>           ⊢* Hqn otherwise
>
> Yours fails to meet this spec.
>
> André
>

You keep insisting that H must be a mind reader and base it decision on
something other than its input parameters when you already know that all
deciders compute the mapping from their inputs to their own final state.

Thus you gleefully contradict facts that you accept as true. Anyone that
contradicts facts that they know are true is a liar by definition.

--
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 5/1/22 2:33 PM, olcott wrote:
> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>> On 2022-05-01 11:08, Richard Damon wrote:
>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>> Recursive definitions are fine, infinitely recursive definitions (such
>>>> as The Halting Problem) are INVALID.
>>>>
>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>
>>>> /Flibble
>>>>
>>>
>>> And the Halting Problem isn't recursive at all, so it can't be
>>> infinity recursive. NOTHING in the definition refers to itself.
>>>
>>> The counter example is, in a way, "recursive", but that recursion can
>>> only become infinite if the proposed Halt Decider turns out to fail
>>> to be a Halt Decider.
>>
>> Actually, even the counterexample is not recursive. It only becomes
>> recursive (or recursive-like) if one assumes that H bases its answer
>> on the simulation of its input. And the proof certainly does not
>> require that to be the case.
>>
>> André
>>
>
> Since the definition of H is wide open and can be anything at all that
> meets the spec, if any of these definitions make the "impossible" input
> decidable then this refutes the proofs.
>

But it does need to meet the requirements as stated and can't say that
because the way I want to solve the problem leads to an "infinite
recursion" which is impossible, I get the change the requirements.

H needs to return the answer for what the computation H^ applied to the
representation of H^ will do, when that H^ is based on the H and calls
it in the way to ask H what it thinks H^ applied to the representation
of H^ does and then does the opposite.

If you can't ask H about the computation H^ applied to the
representaiton of H^, then H just fails to meet the requirements.

Since you say <H^> <H^> isn't the way to ask it, you need to define how
you do ask it, and then build your H^ on that method and show you get
the correct answer.

Since H^ asks the exact same question, and then does the opposite, and
all copies of a computation give the same answer for the same input, H
is stuck in always being wrong of failing to meet some requriement.

Until you can show that you can build a Turing Machine that gives a
different answer when asked as a sole machine compared to what it does
as an Turing Machine embedded in another Turing Machine even when given
the exact same tape in the two cases, you "proof" doesn't work.

On 2022-05-01 13:05, olcott wrote:
> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>> On 2022-05-01 12:33, olcott wrote:
>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>> (such
>>>>>> as The Halting Problem) are INVALID.
>>>>>>
>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>
>>>>>> /Flibble
>>>>>>
>>>>>
>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>
>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>> can only become infinite if the proposed Halt Decider turns out to
>>>>> fail to be a Halt Decider.
>>>>
>>>> Actually, even the counterexample is not recursive. It only becomes
>>>> recursive (or recursive-like) if one assumes that H bases its answer
>>>> on the simulation of its input. And the proof certainly does not
>>>> require that to be the case.
>>>>
>>>> André
>>>>
>>>
>>> Since the definition of H is wide open and can be anything at all
>>> that meets the spec, if any of these definitions make the
>>> "impossible" input decidable then this refutes the proofs.
>>
>> But the spec is as follows:
>>
>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>            ⊢* Hqn otherwise
>>
>> Yours fails to meet this spec.
>>
>> André
>>
>
> You keep insisting that H must be a mind reader and base it decision on
> something other than its input parameters when you already know that all
> deciders compute the mapping from their inputs to their own final state.

Even if your view that TM's can only answer about their inputs were
true, the spec is what the spec is. Not every spec can be met.

I can specify a Turing Machine as follows (where w is some string)

Mq0 w ⊢* Mqy iff the moon is full
⊢* Mqn otherwise

It's not possible to write such a TM, but the fact that the spec can't
be met doesn't entitle you to change it to something else that can be
met. You have to simply assert that the spec cannot be met.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On Sunday, 1 May 2022 at 20:05:17 UTC+1, olcott wrote:
> On 5/1/2022 1:43 PM, André G. Isaak wrote:
> > On 2022-05-01 12:33, olcott wrote:
> >> On 5/1/2022 12:30 PM, André G. Isaak wrote:
> >>> On 2022-05-01 11:08, Richard Damon wrote:
> >>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
> >>>>> Recursive definitions are fine, infinitely recursive definitions (such
> >>>>> as The Halting Problem) are INVALID.
> >>>>>
> >>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
> >>>>>
> >>>>> /Flibble
> >>>>>
> >>>>
> >>>> And the Halting Problem isn't recursive at all, so it can't be
> >>>> infinity recursive. NOTHING in the definition refers to itself.
> >>>>
> >>>> The counter example is, in a way, "recursive", but that recursion
> >>>> can only become infinite if the proposed Halt Decider turns out to
> >>>> fail to be a Halt Decider.
> >>>
> >>> Actually, even the counterexample is not recursive. It only becomes
> >>> recursive (or recursive-like) if one assumes that H bases its answer
> >>> on the simulation of its input. And the proof certainly does not
> >>> require that to be the case.
> >>>
> >>> André
> >>>
> >>
> >> Since the definition of H is wide open and can be anything at all that
> >> meets the spec, if any of these definitions make the "impossible"
> >> input decidable then this refutes the proofs.
> >
> > But the spec is as follows:
> >
> > Hq0 <M> w ⊢* Hqy iff M applied to w halts
> > ⊢* Hqn otherwise
> >
> > Yours fails to meet this spec.
> >
> > André
> >
> You keep insisting that H must be a mind reader and base it decision on
> something other than its input parameters when you already know that all
> deciders compute the mapping from their inputs to their own final state.
>
> Thus you gleefully contradict facts that you accept as true. Anyone that
> contradicts facts that they know are true is a liar by definition.
>
You must understand that if P(P) haltd ansd H(P,P) reports "non-halting",
anyone is going to say that you don't have a counterexample to Linz.

Now I appreciate that you have fairly consistently said that whilst
P(P) halts, "the input to H is non-halting". But I don't think anyone has
a handle on that. It just seems to be nonsense, but it must mean something.
The best I can guess is that the simulating Halt decider H reports
P(P) as "non-halting" and you have decided that H is correct. So its
simulation and halt decision is by definition not wrong.

I

On 5/1/2022 2:19 PM, André G. Isaak wrote:
> On 2022-05-01 13:05, olcott wrote:
>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>> On 2022-05-01 12:33, olcott wrote:
>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>>> (such
>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>
>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>
>>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>>> can only become infinite if the proposed Halt Decider turns out to
>>>>>> fail to be a Halt Decider.
>>>>>
>>>>> Actually, even the counterexample is not recursive. It only becomes
>>>>> recursive (or recursive-like) if one assumes that H bases its
>>>>> answer on the simulation of its input. And the proof certainly does
>>>>> not require that to be the case.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> Since the definition of H is wide open and can be anything at all
>>>> that meets the spec, if any of these definitions make the
>>>> "impossible" input decidable then this refutes the proofs.
>>>
>>> But the spec is as follows:
>>>
>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>            ⊢* Hqn otherwise
>>>
>>> Yours fails to meet this spec.
>>>
>>> André
>>>
>>
>> You keep insisting that H must be a mind reader and base it decision
>> on something other than its input parameters when you already know
>> that all deciders compute the mapping from their inputs to their own
>> final state.
>
> Even if your view that TM's can only answer about their inputs were
> true, the spec is what the spec is. Not every spec can be met.
>

I am saying that deciders are defined to compute the mapping from their
input parameters to their own accept or reject state, thus any
"specification" that contradicts this is fundamentally incorrect.

--
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 5/1/22 3:05 PM, olcott wrote:
> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>> On 2022-05-01 12:33, olcott wrote:
>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>> (such
>>>>>> as The Halting Problem) are INVALID.
>>>>>>
>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>
>>>>>> /Flibble
>>>>>>
>>>>>
>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>
>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>> can only become infinite if the proposed Halt Decider turns out to
>>>>> fail to be a Halt Decider.
>>>>
>>>> Actually, even the counterexample is not recursive. It only becomes
>>>> recursive (or recursive-like) if one assumes that H bases its answer
>>>> on the simulation of its input. And the proof certainly does not
>>>> require that to be the case.
>>>>
>>>> André
>>>>
>>>
>>> Since the definition of H is wide open and can be anything at all
>>> that meets the spec, if any of these definitions make the
>>> "impossible" input decidable then this refutes the proofs.
>>
>> But the spec is as follows:
>>
>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>            ⊢* Hqn otherwise
>>
>> Yours fails to meet this spec.
>>
>> André
>>
>
> You keep insisting that H must be a mind reader and base it decision on
> something other than its input parameters when you already know that all
> deciders compute the mapping from their inputs to their own final state.
>
> Thus you gleefully contradict facts that you accept as true. Anyone that
> contradicts facts that they know are true is a liar by definition.
>

We don't demand that it be a mind reader, only that it gives the right
answer.

If the input paramateres don't represent what is needed to give the
answer, the the inputs were formed incorrectly and you need to define
better what you need done to the input to let you give the right answer.

As has been asked, how DO you ask about the comutation H^ applied to the
representation of H^ (where the representation needed by H^ is the same
repesentation needed for H to decide this question).

If we can't ask the question, then H just plain fails to be the needed
decider.

Note, that the "traditional" method of converting a machine description
to a finite string for input to another Turing Machine is to define some
UTM that performs the computation.

Thus we get the equivalent definiton, that

H x y -> Hqy iff UTM x y Halts and -> Hqn iff UTM x y will never Halt.

Note, UTM is an ACTUAL UTM, and thus it will NEVER abort its simulation
but continue to the end or run forever.

H doesn't need to actually use that UTM, but could be based on it, but
if so, and it modifies that UTM to abort, then the test is still done on
the UNMODIFIED UTM that will never abort.

And yes, this does mean that H might be though of as needing to "mind
read" something it can do, but that just means the problem is shown to
be impossible.

On 5/1/2022 2:21 PM, Malcolm McLean wrote:
> On Sunday, 1 May 2022 at 20:05:17 UTC+1, olcott wrote:
>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>> On 2022-05-01 12:33, olcott wrote:
>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>> Recursive definitions are fine, infinitely recursive definitions (such
>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>
>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>
>>>>>>> /Flibble
>>>>>>>
>>>>>>
>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>
>>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>>> can only become infinite if the proposed Halt Decider turns out to
>>>>>> fail to be a Halt Decider.
>>>>>
>>>>> Actually, even the counterexample is not recursive. It only becomes
>>>>> recursive (or recursive-like) if one assumes that H bases its answer
>>>>> on the simulation of its input. And the proof certainly does not
>>>>> require that to be the case.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> Since the definition of H is wide open and can be anything at all that
>>>> meets the spec, if any of these definitions make the "impossible"
>>>> input decidable then this refutes the proofs.
>>>
>>> But the spec is as follows:
>>>
>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>> ⊢* Hqn otherwise
>>>
>>> Yours fails to meet this spec.
>>>
>>> André
>>>
>> You keep insisting that H must be a mind reader and base it decision on
>> something other than its input parameters when you already know that all
>> deciders compute the mapping from their inputs to their own final state.
>>
>> Thus you gleefully contradict facts that you accept as true. Anyone that
>> contradicts facts that they know are true is a liar by definition.
>>
> You must understand that if P(P) haltd ansd H(P,P) reports "non-halting",
> anyone is going to say that you don't have a counterexample to Linz.
>
> Now I appreciate that you have fairly consistently said that whilst
> P(P) halts, "the input to H is non-halting". But I don't think anyone has
> a handle on that. It just seems to be nonsense, but it must mean something.
> The best I can guess is that the simulating Halt decider H reports
> P(P) as "non-halting" and you have decided that H is correct. So its
> simulation and halt decision is by definition not wrong.

I am saying that deciders are defined to compute the mapping from their
input parameters to their own accept or reject state, thus any
"specification" that contradicts this is fundamentally incorrect.

H(P,P) does compute the mapping from its input parameters to its own
final reject state correctly.

--
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 2022-05-01 13:24, olcott wrote:
> On 5/1/2022 2:19 PM, André G. Isaak wrote:
>> On 2022-05-01 13:05, olcott wrote:
>>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>>> On 2022-05-01 12:33, olcott wrote:
>>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>>>> (such
>>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>>
>>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>>
>>>>>>>> /Flibble
>>>>>>>>
>>>>>>>
>>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>>
>>>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>>>> can only become infinite if the proposed Halt Decider turns out
>>>>>>> to fail to be a Halt Decider.
>>>>>>
>>>>>> Actually, even the counterexample is not recursive. It only
>>>>>> becomes recursive (or recursive-like) if one assumes that H bases
>>>>>> its answer on the simulation of its input. And the proof certainly
>>>>>> does not require that to be the case.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> Since the definition of H is wide open and can be anything at all
>>>>> that meets the spec, if any of these definitions make the
>>>>> "impossible" input decidable then this refutes the proofs.
>>>>
>>>> But the spec is as follows:
>>>>
>>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>>            ⊢* Hqn otherwise
>>>>
>>>> Yours fails to meet this spec.
>>>>
>>>> André
>>>>
>>>
>>> You keep insisting that H must be a mind reader and base it decision
>>> on something other than its input parameters when you already know
>>> that all deciders compute the mapping from their inputs to their own
>>> final state.
>>
>> Even if your view that TM's can only answer about their inputs were
>> true, the spec is what the spec is. Not every spec can be met.
>>
>
> I am saying that deciders are defined to compute the mapping from their
> input parameters to their own accept or reject state, thus any
> "specification" that contradicts this is fundamentally incorrect.

Specifications can't be 'correct' or 'incorrect'. They can be doable or
non-doable.

Since you claim to be a software engineer, consider the following:

Someone hires you to write some program. They tell you what the program
must do and what they ask for is simply not possible for a computer
program to do (for example, they want you to solve an NP-hard problem in
linear time).

Would you:

(a) Tell them that the specification is simply not doable

or

(b) Invest thousands of man-hours in writing a program that does
something not exactly like the spec on the assumption that they will pay
you for this after you explain to them that their specification was
incorrect and that therefore this must be what they actually wanted?

The spec is what you WANT the Turing Machine to do. It may or may not be
possible, but if it isn't possible that doesn't change what it was you
wanted it to do.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/1/2022 2:35 PM, André G. Isaak wrote:
> On 2022-05-01 13:24, olcott wrote:
>> On 5/1/2022 2:19 PM, André G. Isaak wrote:
>>> On 2022-05-01 13:05, olcott wrote:
>>>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>>>> On 2022-05-01 12:33, olcott wrote:
>>>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>>>> Recursive definitions are fine, infinitely recursive
>>>>>>>>> definitions (such
>>>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>>>
>>>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>>>
>>>>>>>>> /Flibble
>>>>>>>>>
>>>>>>>>
>>>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>>>
>>>>>>>> The counter example is, in a way, "recursive", but that
>>>>>>>> recursion can only become infinite if the proposed Halt Decider
>>>>>>>> turns out to fail to be a Halt Decider.
>>>>>>>
>>>>>>> Actually, even the counterexample is not recursive. It only
>>>>>>> becomes recursive (or recursive-like) if one assumes that H bases
>>>>>>> its answer on the simulation of its input. And the proof
>>>>>>> certainly does not require that to be the case.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> Since the definition of H is wide open and can be anything at all
>>>>>> that meets the spec, if any of these definitions make the
>>>>>> "impossible" input decidable then this refutes the proofs.
>>>>>
>>>>> But the spec is as follows:
>>>>>
>>>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>>>            ⊢* Hqn otherwise
>>>>>
>>>>> Yours fails to meet this spec.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> You keep insisting that H must be a mind reader and base it decision
>>>> on something other than its input parameters when you already know
>>>> that all deciders compute the mapping from their inputs to their own
>>>> final state.
>>>
>>> Even if your view that TM's can only answer about their inputs were
>>> true, the spec is what the spec is. Not every spec can be met.
>>>
>>
>> I am saying that deciders are defined to compute the mapping from
>> their input parameters to their own accept or reject state, thus any
>> "specification" that contradicts this is fundamentally incorrect.
>
> Specifications can't be 'correct' or 'incorrect'. They can be doable or
> non-doable.
>

If a spec says that cats are dogs or a decider computes the mapping from
non-inputs its contradict the definition of cat, dog, decider, thus
disagrees with established facts.

Anything that disagrees with established facts is always necessarily
incorrect.

--
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 5/1/22 3:24 PM, olcott wrote:
> On 5/1/2022 2:19 PM, André G. Isaak wrote:
>> On 2022-05-01 13:05, olcott wrote:
>>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>>> On 2022-05-01 12:33, olcott wrote:
>>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>>>> (such
>>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>>
>>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>>
>>>>>>>> /Flibble
>>>>>>>>
>>>>>>>
>>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>>
>>>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>>>> can only become infinite if the proposed Halt Decider turns out
>>>>>>> to fail to be a Halt Decider.
>>>>>>
>>>>>> Actually, even the counterexample is not recursive. It only
>>>>>> becomes recursive (or recursive-like) if one assumes that H bases
>>>>>> its answer on the simulation of its input. And the proof certainly
>>>>>> does not require that to be the case.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> Since the definition of H is wide open and can be anything at all
>>>>> that meets the spec, if any of these definitions make the
>>>>> "impossible" input decidable then this refutes the proofs.
>>>>
>>>> But the spec is as follows:
>>>>
>>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>>            ⊢* Hqn otherwise
>>>>
>>>> Yours fails to meet this spec.
>>>>
>>>> André
>>>>
>>>
>>> You keep insisting that H must be a mind reader and base it decision
>>> on something other than its input parameters when you already know
>>> that all deciders compute the mapping from their inputs to their own
>>> final state.
>>
>> Even if your view that TM's can only answer about their inputs were
>> true, the spec is what the spec is. Not every spec can be met.
>>
>
> I am saying that deciders are defined to compute the mapping from their
> input parameters to their own accept or reject state, thus any
> "specification" that contradicts this is fundamentally incorrect.
>
>

But an X Decider is defined to say that the mapping it computes must
match the property X.

Thus a Halt Decider must match the Halting Property which is defined
based on the compuation the input repesents.

If you are saying we can't define a representation to allow that, that
is sufficient to say that there can not be any Halt Deciders (defined as
ones that get all inputs correctly).

You don't get to escape that requirement.

Your H is, by your claim a "Decider", but since the input doesn't
actually match the required representation for a Halting Decider, it
isn't one, but just your Poop Decider.

On 2022-05-01 13:43, olcott wrote:

>> Specifications can't be 'correct' or 'incorrect'. They can be doable
>> or non-doable.
>>
>
> If a spec says that cats are dogs or a decider computes the mapping from
> non-inputs its contradict the definition of cat, dog, decider, thus
> disagrees with established facts.
>
> Anything that disagrees with established facts is always necessarily
> incorrect.

And, as usual, you snipped all of the material which came afterwards
which addressed your misconception. Why not actually answer the
hypothetical I asked?

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/1/22 3:43 PM, olcott wrote:
> On 5/1/2022 2:35 PM, André G. Isaak wrote:
>> On 2022-05-01 13:24, olcott wrote:
>>> On 5/1/2022 2:19 PM, André G. Isaak wrote:
>>>> On 2022-05-01 13:05, olcott wrote:
>>>>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>>>>> On 2022-05-01 12:33, olcott wrote:
>>>>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>>>>> Recursive definitions are fine, infinitely recursive
>>>>>>>>>> definitions (such
>>>>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>>>>
>>>>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>>>>
>>>>>>>>>> /Flibble
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>>>>
>>>>>>>>> The counter example is, in a way, "recursive", but that
>>>>>>>>> recursion can only become infinite if the proposed Halt Decider
>>>>>>>>> turns out to fail to be a Halt Decider.
>>>>>>>>
>>>>>>>> Actually, even the counterexample is not recursive. It only
>>>>>>>> becomes recursive (or recursive-like) if one assumes that H
>>>>>>>> bases its answer on the simulation of its input. And the proof
>>>>>>>> certainly does not require that to be the case.
>>>>>>>>
>>>>>>>> André
>>>>>>>>
>>>>>>>
>>>>>>> Since the definition of H is wide open and can be anything at all
>>>>>>> that meets the spec, if any of these definitions make the
>>>>>>> "impossible" input decidable then this refutes the proofs.
>>>>>>
>>>>>> But the spec is as follows:
>>>>>>
>>>>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>>>>            ⊢* Hqn otherwise
>>>>>>
>>>>>> Yours fails to meet this spec.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> You keep insisting that H must be a mind reader and base it
>>>>> decision on something other than its input parameters when you
>>>>> already know that all deciders compute the mapping from their
>>>>> inputs to their own final state.
>>>>
>>>> Even if your view that TM's can only answer about their inputs were
>>>> true, the spec is what the spec is. Not every spec can be met.
>>>>
>>>
>>> I am saying that deciders are defined to compute the mapping from
>>> their input parameters to their own accept or reject state, thus any
>>> "specification" that contradicts this is fundamentally incorrect.
>>
>> Specifications can't be 'correct' or 'incorrect'. They can be doable
>> or non-doable.
>>
>
> If a spec says that cats are dogs or a decider computes the mapping from
> non-inputs its contradict the definition of cat, dog, decider, thus
> disagrees with established facts.
>
> Anything that disagrees with established facts is always necessarily
> incorrect.
>
>

No, if someone contracts you to deliver a Cat that is a Dog, then you
need to deliver them a Cat that is a Dog, or just refuse the contract.
If you deliver a Cat that isn't a Dog, or a Dog that isn't a Cat, you
haven't met the terms of the contract. Saying the contract is
"illogical" doesn't mean that you get to change it. You need to either
accept it or reject it.

If Halting requires deciding on something that can't be given as an
input, then that just proves that you can't decide Halting with a Turing
Machine (or equivalent computation), thus PROVING the theory, not
refuting it.

The Problem is the Problem. The claim is that the Problem can't be
solved under the rules. If you say the only way to solve the problem is
to change the rules, that is AGREEING with the claim.

On 5/1/22 3:27 PM, olcott wrote:
> On 5/1/2022 2:21 PM, Malcolm McLean wrote:
>> On Sunday, 1 May 2022 at 20:05:17 UTC+1, olcott wrote:
>>> On 5/1/2022 1:43 PM, André G. Isaak wrote:
>>>> On 2022-05-01 12:33, olcott wrote:
>>>>> On 5/1/2022 12:30 PM, André G. Isaak wrote:
>>>>>> On 2022-05-01 11:08, Richard Damon wrote:
>>>>>>> On 5/1/22 8:37 AM, Mr Flibble wrote:
>>>>>>>> Recursive definitions are fine, infinitely recursive definitions
>>>>>>>> (such
>>>>>>>> as The Halting Problem) are INVALID.
>>>>>>>>
>>>>>>>> RECURSIVE and INFINITELY RECURSIVE are TWO DIFFERENT THINGS.
>>>>>>>>
>>>>>>>> /Flibble
>>>>>>>>
>>>>>>>
>>>>>>> And the Halting Problem isn't recursive at all, so it can't be
>>>>>>> infinity recursive. NOTHING in the definition refers to itself.
>>>>>>>
>>>>>>> The counter example is, in a way, "recursive", but that recursion
>>>>>>> can only become infinite if the proposed Halt Decider turns out to
>>>>>>> fail to be a Halt Decider.
>>>>>>
>>>>>> Actually, even the counterexample is not recursive. It only becomes
>>>>>> recursive (or recursive-like) if one assumes that H bases its answer
>>>>>> on the simulation of its input. And the proof certainly does not
>>>>>> require that to be the case.
>>>>>>
>>>>>> André
>>>>>>
>>>>>
>>>>> Since the definition of H is wide open and can be anything at all that
>>>>> meets the spec, if any of these definitions make the "impossible"
>>>>> input decidable then this refutes the proofs.
>>>>
>>>> But the spec is as follows:
>>>>
>>>> Hq0 <M> w ⊢* Hqy iff M applied to w halts
>>>>            ⊢* Hqn otherwise
>>>>
>>>> Yours fails to meet this spec.
>>>>
>>>> André
>>>>
>>> You keep insisting that H must be a mind reader and base it decision on
>>> something other than its input parameters when you already know that all
>>> deciders compute the mapping from their inputs to their own final state.
>>>
>>> Thus you gleefully contradict facts that you accept as true. Anyone that
>>> contradicts facts that they know are true is a liar by definition.
>>>
>> You must understand that if P(P) haltd ansd H(P,P) reports "non-halting",
>> anyone is going to say that you don't have a counterexample to Linz.
>>
>> Now I appreciate that you have fairly consistently said that whilst
>> P(P) halts, "the input to H  is non-halting". But I don't think anyone
>> has
>> a handle on that. It just seems to be nonsense, but it must mean
>> something.
>> The best I can guess is that the simulating Halt decider H reports
>> P(P) as "non-halting" and you have decided that H is correct. So its
>> simulation and halt decision is by definition not wrong.
>
> I am saying that deciders are defined to compute the mapping from their
> input parameters to their own accept or reject state, thus any
> "specification" that contradicts this is fundamentally incorrect.
>
> H(P,P) does compute the mapping from its input parameters to its own
> final reject state correctly.
>
>

It computes "A" Mapping, not "THE" Mapping, as THE mapping is defined by
the computation that the input represents, which you say it can't be
reposisible for, which just says it has rejected being a Halting
Decider, and is just some poop Decider.