Re: PROOF that numbers are countable

• Previous message: |-|erc: "Re: PROOF that emulators are impossible"
• In reply to: Will Twentyman: "Re: PROOF that numbers are countable"
• Next in thread: Will Twentyman: "Re: PROOF that numbers are countable"
• Reply: Will Twentyman: "Re: PROOF that numbers are countable"
• Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]

Date: Tue, 04 May 2004 23:54:37 GMT

"Will Twentyman" <wtwentyman@read.my.sig> wrote in > >
> >>>>>and turing machines themselves can construct *every function* and have
> >>>>>them countable, without subfunctions. These systems are as powerful
> >>>>>as all other computing systems, including your specific subfunction reliant
> >>>>>paradox succeptible system that is the only system that your proofs
> >>>>>produce their EVIDENTLY ABSTRACT conjectures.
> >>>>>
> >>>>>Define Function xyz (par1, par2) {
> >>>>> *body*
> >>>>>}
> >>>>>
> >>>>>is a construct by humans for humans, its not necessary. If you introduce
> >>>>>it don't enforce its idiosyncracies on valid countable complete function sets.
> >>>>
> >>>>It is not necessary, but it can be recoded into a register machine or
> >>>>turing machine program. It is therefore equivalent to a portion of a
> >>>>turing machine.
> >>>>
> >>>
> >>>
> >>>That would require F to be a subfunction emulated within UTM.
> >>>G(j) = F(j,j) mod 2 + 1
> >>>has to compile to a godel number within the system used in UTM.
> >>>
> >>>Right?
> >>>
> >>>
> >>>If UTM is simple, it will assign a godel number. Assume it allows
> >>>function decomposition to determine the godel number of a function.
> >>>If X compiles, then X mod 2 compiles
> >>>If X compiles, then X + 1 compiles
> >>>
> >>>Therefore if F(x,y) was a subfunction, F(j,j) mod 2 + 1 would compile to a godel number,
> >>>a godel number would be found that couldn't exist as a valid function in the
> >>>system, its definition would differ to its value in F.
> >>
> >>What do you mean by "a valid function in the system"? F having a godel
> >>number has nothing to do with whether F halts.
> >>
> >>
> >>>Sounds self defeating, the assumption is that a UTM exists that has a clever
> >>>godel scheme, not to find a particular impossible system.
> >>
> >>What do you mean by an "impossible system"?
> >>
> >>I get the sense you have several unstated assumptions about what is
> >>going on. For example, I suspect you want a function to have a Godel
> >>number if and only if it is guaranteed to halt. The problem, of course,
> >>is how do you know if a particular function, when presented, will
> >>actually halt?
> >
> > That's your problem, you're the one claiming such an enumeration of halting
> > functions is impossible.
>
> Based on the standard definition, it is.

What ? come off it you've all be PROVING it to me for 2 months,
now your proof is SQUAT you're relying on a definition.. get over it.

>
> > We merely define a system of computation that does not utilise function names.
> > This system contains all possible functions but it does not contain functions
> > like diagonal functions, because they cannot be defined at all.
>
> Here you are changing the definition of what counts as a function. More
> importantly, you are changing the definition of what is a Turing
> Program. Turing Programs do not have names. A C program can be
> translated into assembly, which does not have names for functions, does
> not have functions, and does not even have variables.

rubbish, assembler just uses the numbers of the address like basic, goto 100

>
> You appear to be doing your mathematics based on your programming
> language of choice, rather than looking at the definitions.

No YOU ARE reliant on a particular paradigm of functions to establish a contradiction.
I have several choices of consistent enumerable function representations, you have
ONE - declaring names for functions. This has been my argument for weeks strange
you would turn it around. I'm just pointing out your diag proof doesn't work
in *this* paradigm, I only need one inconsistency to prove it wrong.

>
> If you want to analyze whether your special functions are guaranteed to
> halt, feel free. Personally, I'm far from convinced of that.

thats because your textbook is wrong so you evidently aren't thinking for yourself.

>
> > Anyone here have a copy of Shadows Of The Mind? There's a model of
> > functions using only a single function twice.
> >
> > (Twice suc) 1 = 2
> >
> > Different rearrangements make different functions, twice() is used as the number 1,
> > twice(twice()) is the number 2.
> >
> > Addition is some strange looking high level function like so :
> >
> > (twice((twice(twice))twice))
> >
> > And you apply it to numbers in the correct form
> >
> > (twice((twice(twice))twice)) (twice(twice)) (twice)
> >
> > = (twice(twice(twice)))
> >
> > i.e. 2 + 1 = 3.
> >
> > It does mulitplication, fibonacci, its as powerful as TMs.
> > I'm unsure, but it might be a construct in which all functions halt, depends
> > if it does recursion or not.
>
> If it is as powerful as TMs, then there are constructs that do not halt.

proof?

>
> > If you have a copy of the book or know of the article post the functions up.
> >
> > Notice it does not use function names, you can't define
> > ADD(x, y) = (twice((twice(twice))twice)) x y
> >
> > and use ADD in further functions, that would not longer be a pure function syntax,
> > allowing that level of construction does not increase the class of functions
> > it can compute, it just changes the type of computation and the function twice
> > is no longer necessary.
> >
> >
> >
> >
> > 1 0 1 r 2
> > 1 1 1 l 3
> > .
> > 2 0 0 l 1
> > 2 1 0 r 2
> > .
> > 3 0 1 l 1
> > 3 1 1 l 99
> > .
> >
> > Here's a TM that counts in binary. You can basically write the digits into
> > its godel number, 011 10 110 11 001 01 101 10 011 01 111 00
> >
> > Several billion large. There is no encoding of F(j,j) in this system, you may
> > be able to access a form of F(j,j) within the turing machine but you will
> > have to manually derive all the values from primitive operations. None of
> > the values of F(j,k) depend on any values from F(x, y) where x =/= j.
> >
> > All the functions are independant.
>
> Herc, either you are dealing with all TMs, or a subset of them. Either
> way is fine. If you want to announce that you have a subset of the TMs
> that are guaranteed to halt, and do not have a diagonalization that is
> within the subset, so be it. You will have to do some work, but that's ok.
>
> The problem right now, as I see it, is that you make a claim that
> appears to refer to ALL TM's, then specify that you are only working
> with a special selection of them. You don't always prove that your
> claim applies to your special selection, and rarely show that your
> special selection has anything to do with the behavior of ALL TM's.
>
> Perhaps you would have more luck introducing a new notation, such as a
> HM is a Turing Machine with the following restrictions: [list
> restrictions]. Then you can work out the properties of HM's and see if
> you can use HM's to show anything interesting. You could also see which
> of your results for HM's correspond to work on TM's.
>
>

sure, I'm not defining a system of functions that is equivalent in power to all TMs
that all halt, I'm asserting that it has not been proven impossible.

If the standard TM architecture is used then my claim would be that a subset
(certain subsets) of those TMs are as powerful as the entire set.

Essentially you subtract the ones that don't halt. Meta functions like diagonal
constructs would be in the class that don't halt, if they map to a particular
TM number at all.

Halting proof was that you can't work out the halting value, its says nothing of
a rigged system that only contains halting functions. Where is the contradiction?
The halting function is impossible on any system, but if all functions halt the
halting function is not definable anyway. Perhaps Halt:output true will work
but the second function in the proof won't qualify as a function because it
is specifically constructed to go into an infinite loop, that is just dissallowed as
a valid function in this system obviously.

Herc

• Previous message: |-|erc: "Re: PROOF that emulators are impossible"
• In reply to: Will Twentyman: "Re: PROOF that numbers are countable"
• Next in thread: Will Twentyman: "Re: PROOF that numbers are countable"
• Reply: Will Twentyman: "Re: PROOF that numbers are countable"
• Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]