Re: First Proof That Infinitely Many Prime Numbers Come in Pairs

[Graham Cooper]

On May 18, 4:00 am, Gus Gassmann <no...@nospam.com> wrote:
> On 17/05/2013 1:45 PM, Aatu Koskensilta wrote:
>
> > dullr...@sprynet.com writes:
>
> >> Some years ago there was something in Scientific American about
> >> the difficulty of factoring large primes.
>
> >    I once read it somewhere that factoring large primes is what is known
> > as an NP ("Non-Polynomial") complete problem.
>
> ... which also settles that other notorious problem, proving that
> P = NP. Very impressive.
>

No Oracles allowed!


Herc

--

IS DOG+1 a NUMBER ?

http://blockprolog.com/nat-s-dog.png

[Graham Cooper]

On May 19, 7:36 am, 1treePetrifiedForestLane <Space...@hotmail.com>
wrote:
> ah, so, there could still be infinity of twin-primes.
>
>
>
>
>
>
>
> > It's easy to see that there are arbitrarily large gaps between some
> > consecutive primes.  Consider N factorial: the N-1 integers N!-N ... N!-2
> > are all composite.



N!-N
---- = (N-1)! - 1
N

^_^

[Richard Tobin]

In article <1a2603a5-a24b-4b98...@k8g2000pbf.googlegroups.com>,

Graham Cooper <graham...@gmail.com> wrote:

>N!-N
>---- = (N-1)! - 1
> N

Very good, but what is your point?

-- Richard

[Charlie-Boo]

Dogs dogs dog dog dogs.

C-B

> http://blockprolog.com/nat-s-dog.png

[Charlie-Boo]

On May 18, 4:48 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

Let P be a recursive set and f(I) be a recursive function.

Let Q be (all A)(exists B) LT(A,B) ^ P(B) ^ P(f(B))

Could Q be true and unprovable? Could Q be false and unrefutable?

For which P and f [specific or general] is Q one or the other of these
conditions?

C-B

[Charlie-Boo]

e.g. if Q == ZF is consistent then the 1st condition holds. But the
heck with silly ZF, and we aren't questioning sets anyway, say we are
using PA.

Then if Q = = PA is consistent the 1st condition holds (for us.)

Now what can be proven in PA? This question is really sad, because
apparently people don't know - does anyone? I say this because when
talking about when certain theorems known to be true in PA apply, they
always say "If it has Peano's axioms" or some dinky "You can express
arithmetic in it." PA means a+b=c = = |- a+b=c , a*b=c = = |- a*b=c
and |-(allx)TRUE(x) where TRUE(a) = = a is a natural number. ( TRUE
means everything as in {x|TRUE} not like a "true sentence".)

In general, if Q is a Godel sentence of PA. In other words . . . etc.

C-B

[Graham Cooper]

Define provable!


My Defn:

if THM is an axiom then THM is provable
if A is provable and B is provable
and A^B->THM
then THM is provable

[Charlie-Boo]

On May 19, 4:31 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On May 20, 2:46 am, Charlie-Boo <shymath...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On May 18, 4:48 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > On May 18, 4:00 am, Gus Gassmann <no...@nospam.com> wrote:
>
> > > > On 17/05/2013 1:45 PM, Aatu Koskensilta wrote:
>
> > > > > dullr...@sprynet.com writes:
>
> > > > >> Some years ago there was something in Scientific American about
> > > > >> the difficulty of factoring large primes.
>
> > > > >    I once read it somewhere that factoring large primes is what is known
> > > > > as an NP ("Non-Polynomial") complete problem.
>
> > > > ... which also settles that other notorious problem, proving that
> > > > P = NP. Very impressive.
>
> > > No Oracles allowed!
>
> > > Herc
>
> > > --
>
> > > IS DOG+1 a NUMBER ?
>
> > >http://blockprolog.com/nat-s-dog.png
>
> > Let P be a recursive set and f(I) be a recursive function.
>
> > Let Q be (all A)(exists B) LT(A,B) ^ P(B) ^ P(f(B))
>
> > Could Q be true and unprovable?  Could Q be false and unrefutable?
>
> > For which P and f [specific or general] is Q one or the other of these
> > conditions?
>
> > C-B
>
> Define provable!

Pay 'tension! I said "say we are using PA. . . . Now what can be
proven in PA? This question is really sad, . .."

This may make a constructive proof actually useful (Martin Davis
demanded that I make my axiomatized proofs of incompleteness in logic
constructive despite the loss from ignoring nonconstructive proofs,
much to my chagrin.)

If we can construct Godel sentences (as opposed to only proving they
exist), then as a bonus we can see if they are equivalent to our Q

Now what's the formula for constructing a Godel sentence? And don't
forget that while Godel thought of only 1, there are 3 in the seminal
"18 Word Proof of the Incompleteness Theorems of Godel, Rosser and
Smullyan".

C-B

[Charlie-Boo]

On May 18, 4:48 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

Dog is a number you cannot count to.

http://forums.cannabisculture.com/forums/ubbthreads.php?ubb=showflat&Number=576091

> http://blockprolog.com/nat-s-dog.png

[Graham Cooper]

I use a 10 symbol Godel numbering and a recursive proof predicate for
A0 based on the defn I gave you.


!a0(a1)

8203215

Note a 2nd line is required A1=8203215 (in binary)

A FULLY CODED GODEL STATEMENT

https://groups.google.com/group/alt.bible/browse_thread/thread/4f4dc0f290a90d12/aef683a9f8504edd


*******************************
a0(a11)=a11
a0(a11)=a0(a111)^a0(a110)^(!(a111^a110)va11)
a1=11111010010101111001111
!a0(a1)
*******************************

This is just my definition above encoded.

if THM is an axiom then THM is provable
if A is provable and B is provable
and A^B->THM
then THM is provable

proof( a11 ) <-> axiom( a11 )
proof( a11 ) <- proof( a111 ) & proof( a111 ) & not( a111 or a11 )

then

a11 = 8203215
not(proof(8203215)

Herc

[Graham Cooper]

...

> a1 = 8203215
> not(proof(8203215)
>


The above formula refers to itself !

Herc

[Charlie-Boo]

Yes, and a fine job indeed. But let us abstract UP and generalize.
We have neither digits, numerals nor constant considerations. We have
only sets and functions. We have Q, defined to be (all A)(exists B)
LT(A,B) ^ P(B) ^ P(f(B)) where P be a recursive set and f(I) be a
recursive function. You have the recursive relation x proves y, the
r.e. set x is provable, the recursive function f(x)=y iff x proves y.
Let us pay homage to Godel, for He created the undecidable sentence
and decided it is true. How did he construct that temple from the
wisdom in his head? When will the Godel Sentence G be the same as the
C-B Sentence Q?

You must express "wff number x with x substituted for its free
variable is not provable", or, expanding your possibilities, realize
that,

"When any one of these sets [the true, provable and unrefutable
sentences] P, is expressible or representable, the sentence
that expresses or represents, respectively, 'This is in P.' is
undecidable." This includes:

1. Since unprovability is expressible: The sentence that expresses
"This is not provable." is undecidable.
2. Since refutability is expressible: The sentence that expresses
"This is refutable." is undecidable.
3. Since refutability is representable: The sentence that represents
"This is refutable." is undecidable.

http://www.cs.nyu.edu/pipermail/fom/2010-July/014933.html

When is each of these Q?

C-B

[Graham Cooper]

On May 20, 8:48 pm, Charlie-Boo <shymath...@gmail.com> wrote:
> On May 20, 1:31 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > ...
>
> > > a1 = 8203215
> > > not(proof(8203215)
>
> > The above formula refers to itself !
>
> > Herc
>
> Yes, and a fine job indeed.  But let us abstract UP and generalize.
> We have neither digits, numerals nor constant considerations.  We have
> only sets and functions.

OK but you need SOME functionality in your parameters.

Be it
call by name sqrt(X)
call by value sqrt(2)
call by line number, godel number..

20 not proof(20)

or

G<->not(proof(G))

are fine too, it depends what TYPE the domain of your functions are.

> We have Q, defined to be (all A)(exists B)
> LT(A,B) ^ P(B) ^ P(f(B)) where P be a recursive set and f(I) be a
> recursive function. You have the recursive relation x proves y, the
> r.e. set x is provable, the recursive function f(x)=y iff x proves y.
> Let us pay homage to Godel, for He created the undecidable sentence
> and decided it is true.  How did he construct that temple from the
> wisdom in his head?  When will the Godel Sentence G be the same as the
> C-B Sentence Q?
>
> You must express "wff number x with x substituted for its free
> variable is not provable", or, expanding your possibilities, realize
> that,
>
> "When any one of these sets [the true, provable and unrefutable
> sentences] P, is expressible or representable, the sentence
> that expresses or represents, respectively, 'This is in P.' is
> undecidable."  This includes:
>
> 1. Since unprovability is expressible: The sentence that expresses
> "This is not provable." is undecidable.
> 2. Since refutability is expressible: The sentence that expresses
> "This is refutable." is undecidable.
> 3. Since refutability is representable: The sentence that represents
> "This is refutable." is undecidable.
>
> http://www.cs.nyu.edu/pipermail/fom/2010-July/014933.html
>
> When is each of these Q?
>
> C-B

You need to define provable in a more functional method.

GODEL = no PROVE(X) predicate
TARSKI = and no TRUE(X) either

isn't helping the poor old programmer!

As far as PROGRAMMING IN LOGIC is concerned.

2+1=3 ?

and

PROVE( 2+1=3 )

is just selecting the [TRACE=ON ] button!

----------------

That is why I simplified my provability predicate in Provable Set
Theory.

OLD

XeS<->p(X) <-> provable( XeS<->p(X) )

not(provable(X)) <-> provable(not(X))

provable(THM) <-> provable(A) ^ provable(B) ^ provable(A^B->THM)

-----------------


Godel was right in a way - YOU JUST CANT PROGRAM PROVABLE()!

it reduces down to the same as:

TRUE-WFF(X)
THEOREM(X)
DERIVE(X)
TRUE(X)
SOLVE(X)
THM(X)
T(X)


So THM(X) <-> NOT(PROVABLE(NOT(X)))

can just be done with double negation.

F<->F
~(F<->~F) %NO CONTRADICTIONS!

..

XeS<->p(X) <-> XeS<->p(X)

RUSSELLSET <-> (CONTRADICTION)

RUSSELLSET <-> FALSE

~EXIST(rs) rs<->x~ex


---------------------


I can't help any further because I simply don't have a predicate for
PROVE(THM)


t(R,z(Z)) :- if(L,R) , t(L,Z).


Where does it go?


Herc