Cont: Deeper than primes - Continuation 2

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Which set are you defining and which set are you using? The numbers that I use for Godel Numbering do not have to be the set of natural numbers.

I always find it a bit funny that people would use Godel numbering, using natural numbers, to encode statements in logic when they don't actually need the machinery to prove Godel's theorem. It would seem much simpler to just encode them as strings of symbols (such as ASCII or Latex something).
 
We can use any base we want, there are trivial mappings between the sets of finite strings in various bases. Simplest of all we could just use base2, binary bits, where each set of 8 bits is a byte which maps to some symbol in the ASCII table.
 
jsfisher said:
Most of what you post should be ignored. Little of what you write follows any sort of logical argument, and you don't even bother to understand what most of the terms you use actually mean. "Completeness" is just the latest example.

At least take the time to find out what completeness of a formal system means.
Wow, this thread is still going - after 10 years - and Doron isn't making any more sense than he was then.

I admire your patience and tenacity - although I have to wonder why you still do it ;)
Both replies are no more than unsupported statements.

So, dlorde and jsfisher please support them.

Just saying them is not the same as also logically support them.

So the stage is yours and this time do not share only your opinions about http://www.internationalskeptics.com/forums/showpost.php?p=12784867&postcount=3357 but actually logically support your arguments according to its content.
 
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Both replies are no more than unsupported statements.

So, dlorde and jsfisher please support them.

Just saying them is not the same as also logically support them.

So the stage is yours and this time do not share only your opinions about http://www.internationalskeptics.com/forums/showpost.php?p=12784867&postcount=3357 but actually logically support your arguments according to its content.
I offer the entire thread as supporting evidence for both the fact that it's over 10 years old (I provided a link), and that you're making no more sense now than then (ask every contributor other than yourself).
 
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I offer the entire thread as supporting evidence for both the fact that it's over 10 years old (I provided a link), and that you're making no more sense now than then (ask every contributor other than yourself).

You offer your opinion, which does not actually logically deals with anything that is found in my posts.
 
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Doronshadmi, please define set A. You keep saying it is the set of natural numbers and then say it is not. Declare what set A is nice and clear. Don't define it with sub-definitions
Set A is a set in infinitely many axioms (where each axiom is written by finitely many symbols) which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself, such that Infinity is taken in terms of Platonic Infinity (By Platonic Infinity there exists a complete set of infinitely many things as a complete whole (without using any process)).

Some example: The set of all natural numbers is infinite in terms of Platonic infinity.
 
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Some correction of the beginning of my previous post.

It has to be corrected to: "Set A is a set of infinitely many axioms ..."

So let's write the previews post here in order to complete here my argument.

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Set A is a set of infinitely many axioms (where each axiom is written by finitely many symbols) which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself, such that Infinity is taken in terms of Platonic Infinity (By Platonic Infinity there exists a set of infinitely many things as a complete whole (without using any process)).

Some example: The infinite set of all natural numbers is taken in terms of Platonic infinity.

Now all we care (as written in http://www.internationalskeptics.com/forums/showpost.php?p=12784867&postcount=3357) is about the set of all infinitely many wffs (in terms of Platonic Infinity) that can be established in A.

Each wff has some Godel number, where at least one of these wffs, called G, states "There is no number m such that m is the Godel number of a proof in A, of G"

Since all wffs are already in A and all Godel numbers are already in A (because Infinity is taken in terms of Platonic Infinity) there is a Godel number of a proof of G in A, which contradicts G in A, exactly because A is complete and therefore inconsistent, since Infinity is taken in terms of Platonic Infinity.

So the problem is actually the notion of a complete set of infinity many things in terms of Platonic Infinity, and in order to save the consistency of A, ZF(C) Axiom Of Infinity is taken in terms of Potential Infinity (process is used, exactly as done in case of GIT).

But then ZF(C) Axiom Of Infinity can't be used in order to establish sets in terms of Platonic Infinity (for example: the notion of The infinite set of all natural number is logically inconsistent).

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Now, Gödel was a Platonist (he agreed with Actual infinity in terms of Cantor (which is actually Platonic Infinity)) and his main motivation behind his Incompleteness Theorems was to logically demonstrate that formal systems that are strong enough in order to deal with Arithmetic, can't be complete AND consistent and also can't prove their own consistency (which means that many "interesting" formal systems can't deal with Platonic realms).

But Gödel's Incompleteness Theorems also prove that the very notion of Actual infinity in terms of Platonism (which is also Actual infinity in terms of Cantor) does not hold logically (at least in the strong sense).
 
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Which set of infinitely many axioms? There are ma.ny such sets. Which one did you have in mind?
Any set that is strong enough in order to deal with Arithmetic.

For example: Set A is a set of infinitely many axioms (where each axiom is written by finitely many symbols) which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself.
 
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Any set that is strong enough in order to deal with Arithmetic.

Well, sets are not "strong enough" in order to deal with Arithmetic. That would require a formal system. I think you must have meant the set of axioms for some formal system. And since you don't care which one, how about the set of axioms for ZFC?

Conveniently enough, that set is infinite.


For example: Set A is a set of infinitely many axioms (where each axiom is written by finitely many symbols) which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself.

Using the Axiom of Infinity on ZF(C) itself? Multiple things are wrong, there. ZFC is a set theory, a formal system. Did you perhaps mean the set of axioms that establish ZFC?

The real problem, though, is with "using [the] Axiom of Infinity on ZF(C)". That does not mean anything. That particular axiom simply says a certain set with a certain property exists within the formal system. It is not something you can "use on" something; you don't, for example, "use it on" another set to establish yet another set with that same certain property mentioned in the Axiom of Infinity.

The Axiom of Infinity tells us a certain set exists. (It guarantees one, but as it turns out there are many such sets.) The von Neumann ordinal is the simplest example of such a set. It is not an operator, though. You can't use it on a set.
 
Well, sets are not "strong enough" in order to deal with Arithmetic.
A is a set of axioms, which are a formal system.

Using the Axiom of Infinity on ZF(C) itself? Multiple things are wrong, there. ZFC is a set theory, a formal system. Did you perhaps mean the set of axioms that establish ZFC?

The real problem, though, is with "using [the] Axiom of Infinity on ZF(C)". That does not mean anything.
It means a lot. A is a formal system with infinitely many axioms, which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself, such that A is complete in terms of Platonic Infinity.

The Axiom of Infinity tells us a certain set exists. (It guarantees one, but as it turns out there are many such sets.) The von Neumann ordinal is the simplest example of such a set. It is not an operator, though. You can't use it on a set.
No operator is involved by using ZF(C) Axiom Of Infinity on ZF(C) itself, if Infinity is taken in terms of Platonic Infinity.

If you reject the notion that ZF(C) Axiom Of Infinity is not taken in terms of Platonic Infinity, then by this axiom, even the infinite set of all natural numbers can't be established in terms of Cantorian Actual Infinity.
 
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Well, sets are not "strong enough" in order to deal with Arithmetic.
A is a set of axioms, which are a formal system, so when I wrote "Any set that is strong enough in order to deal with Arithmetic." it was about a set of axioms (in case that you are missing that the last discussion is about a set of axioms in terms of Platonic Infinity).

Now, there is a non-interesting solution about the discussed subject, as follows:

G states: "There is no number m such that m is the Godel number of a proof in A, of G"

If G is already an axiom in A (where A is an infinite set of axioms, such that Infinity is taken in terms of Platonic Infinity) it is actually a wff that is true in A, which does not have any Godel number that is used in order to encode G's proof (since axioms are true wff that do not need any proof in A).

But then no proof is needed and mathematicians are out of job (therefore it is an unwanted solution).
 
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A is a set of axioms, which are a formal system.

Then don't write "any set" when you don't mean "any set".

It means a lot. A is a formal system with infinitely many axioms, which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself, such that A is complete in terms of Platonic Infinity.

No operator is involved by using ZF(C) Axiom Of Infinity on ZF(C) itself, if Infinity is taken in terms of Platonic Infinity.

Then do tell us all what you mean by "using ZF(C) Axiom Of Infinity on ZF(C) itself". In that explanation you will need to stick to what the axiom actually says, not what you want it to mean.
 
Then don't write "any set" when you don't mean "any set".
Please stop read partially by also ignore the considered context of the discussed subject.

For example, please read all of http://www.internationalskeptics.com/forums/showpost.php?p=12785789&postcount=3372 before you replay to some part of it.

Thank you.

Then do tell us all what you mean by "using ZF(C) Axiom Of Infinity on ZF(C) itself".
I mean that A is a set of infinitely many axioms which is "strong enough" in order to deal with Arithmetic, where Infinity is taken in terms of Platonic Infinity.

In that explanation you will need to stick to what the axiom actually says, not what you want it to mean.
The considered subject is not about the axioms (that, by definition, are not proven) but about wffs in A that have to be proven in A, where the amount of such wffs is taken in terms of Platonic Infinity.
 
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