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Re: On the diagonal argument again
Posted:
May 19, 2012 6:25 PM
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On May 20, 7:58 am, "K_h" <KHol...@SX729.com> wrote:
> "Graham Cooper" wrote in message
>
> news:6d8a9ebd-2051-46c7-ab90-59d307b90c8e@n5g2000pbg.googlegroups.com...
>
> On May 19, 10:20 am, "K_h" <KHol...@SX729.com> wrote:
>
> > > > Your point seems to be 'you made typos so you are wrong and confused'.
>
> > > Those are hardly just typos. You actually claimed that well-ordering is
> > > an
> > > axiom of ZFC! This indicates a big unfamiliarity with the subject.
>
> > I cut and pasted a text version from Transfer Principle of ZFC axioms
> > since they are images in Wikipedia.
>
> Odd. First you claimed you made typos then you claimed it is somebody else
> who made typos. No matter.
>
Not exactly.
>
> > > > It looks similar to ZFC AOI
>
> > > > FUNCTION P(x, INF)
> > > > BEGIN
> > > > IF (x = 0) RETURN TRUE
> > > > ELSE RETURN P(x-1, INF)
> > > > END
>
> > > It looks like you have replaced the bi-conditional with a conditional
> > > (as I
> > > recommended) so your P() function only counts down to zero and can never
> > > count up (assuming it starts with a natural). If your function starts
> > > with
> > > a natural number n then it will generate sets of the form
> > > {0,1,2,...,n-1,n}.
> > > It will also generate infinite sets: if it starts with X=pi then you
> > > will
> > > generate the infinite set {...,pi-5,pi-4,pi-3,pi-2,pi-1,pi}. It's not a
> > > well ordered set but it is a genuine set in ZF as well as satisfying
> > > your
> > > `set membership predicate'.
>
> > It's a DEFINITION OF Predicate P to be SUBSTITUTED into the
> > SPECIFICATION axiom.
>
> > There is no "->" suitable here.
>
> Previously you had a thing like P(X,INF)<-->(stuff) and I suggested you
> write P(X,INF)-->(stuff) so you only "count down" instead of "counting up".
> Now you write P(X,INF) as a function that only counts down instead of being
> part of a bi-conditional assertion. Now you write that "-->" is not
> suitable here yet the bi-conditional <--> is the conjunction of two
> conditionals -->. So is <--> suitable here and, if so, why and, if not, why
> not?
It's a SET MEMBERSHIP PREDICATE straight out of NAIVE SET THEORY.
The predicate P() is defined as "x ~e x"
when defining Russell's Set.
Because Predicates and their definitions both have truth values, a
biconditional is shorthand (and works) for predicate definitions.
e.g.
DEFINE PLUS(a,b) = "a + b" // a numeric value function
DEFINE IS-IN-RUSSELL-SET(a, RS) = "a ~e a" // uses truth values
or more concisely in LOGIC
IS-IN-RUSSELL-SET(a, RS) <-> a ~e a
Predicates can be defined using bi-conditionals in logic
Then we can safely SUBSTITUTE the formula for P for any P in other
formula.
Strictly when talking about many different Predicates we should name
them differntly.
P-RS(x, RS) <-> x ~e x
P-INF(n, INF) <-> ...
or P1() P2() ...
P just stands for any Predicate.
----
In Naive Set Theory any definable collection is a set.
N.S.T. Y = { x | P(x,Y) }
OR
x e Y <-> P(x,Y)
NOTICE NST is similar format to AXIOM OF SPECIFICATION
----
If you **define** P to be
P(x,Y) <-> x ~e x
then you get
Y e Y <-> Y ~e Y
Russell Set Contradiction
A weakness in NST making it INCONSISTENT.
that's why ZFC was invented to circumvent Predicate Set Instantiation
with subset instantiation only.
>
>
>
> > RIGHT!
>
> > The following must hold true for INF to stratify as a set.
>
> > INF = { x | (x = 0) v (E(z) x = S(z) & P(z, INF)) }
> > IFF
> > NOT(PROOF(NOT(EXIST(INF) INF = { x | (x = 0) v (E(z) x = S(z) & P(z,
> > INF)) } )))
>
> > So all you do is ASSUME the 3rd line just above!
>
> Yes. There is no proof that infinity does not exist because infinity does
Since Paradoxes are unavoidable falsehoods
the only Self Evident Truths I accept are anti-paradoxes!
'There is no Russell's Set!'
> exist. So the third line is more than a speculative assumption, it is a
> fact for reasonable domains like N. You've defined INF to be the set of all
> things satisfying :
>
> { x | (x = 0) v (E(z) x = S(z) & P(z, INF)) }
>
> and every natural number satisfies this. So, assuming your domain is N,
> then INF=N and obviously INF is not a member of N since it is not a natural
> number.
>
> > The Predicate version of INF is
>
> > P(x,INF) = (x = 0) v (E(z) x = S(z) & P(z, INF))
>
> > which only holds if x is a Natural Number, otherwise no z would exist
> > to satisfy E(z)
>
> ALEPH_0 is not a natural number so naturally it won't have a predecessor.
> But your relation above does hold true for hyper-natural numbers and surreal
> numbers. Consider the infinite hyper-natural number w-1:
>
> P(w-1, INF) = (x = 0) v (E(w-2) w-1 = S(w-2) & P(w-2, INF))
>
> So the domain is only the natural numbers if you specify it. So to specify
> it you need to add the statement AxAz((xeN)^(zeN)). So your predicate needs
> to be:
>
> P(x,INF) = (x = 0) v (E(z) x = S(z) & P(z, INF)) ^ (AxAz (xeN)^(zeN) )
>
>
>
> > P(Pi, INF) = (x = 0) v (E(z) x = S(z) & P(z, INF))
> > P(Pi, INF) = E(z) x = S(z) & P(z, INF)
> > P(Pi, INF) = FALSE
>
> > Pi is not the successor of a [successor of] 0
>
>
The central tenet of formalism is context-free deduction.
i.e. being able to do maths WITHOUT understanding why.
A computer won't know that N EXISTS because it's evident!
This is just SYNTAX not a SET.
Y = { x | P(x) }
This is what the computer sees:
x e y <-> P(x) <-> x e x
where e is a RELATION
TABLE e
------------
MEMBER SET
tom people
1 N
2 N
3 N
7 PRIMES
..
Herc
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